Maths M.C.Q (1 Marks)

and $\mathop {\lim }\limits_{x \to 1 + } \,\,\frac{1}{{|\,\,1 - x\,\,|}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h - 1}} = \infty $

Hence $\mathop {\lim }\limits_{x \to 1} \,\frac{1}{{|\,\,1 - x\,\,|}} = \infty .$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^3}\,\left( {4 + \frac{4}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^3}\left( {1 + \frac{5}{n} + \frac{5}{{{n^2}}} - \frac{2}{{{n^3}}}} \right)}} = 4$

$\mathop {\lim }\limits_{x \to \,0 + } \,f(x) = \mathop {\lim }\limits_{h \to \,0} \left( {\frac{{{e^{1/h}} - 1}}{{{e^{1/h}} + 1}}} \right) $

$= \mathop {\lim }\limits_{h \to \,0} \frac{{{e^{1/h}}\left( {1 - \frac{1}{{{e^{1/h}}}}} \right)}}{{{e^{1/h}}\left( {1 + \frac{1}{{{e^{1/h}}}}} \right)}} = 1$

Similarly $\mathop {\lim }\limits_{x \to \,0 - } f(x) = - 1$. 

Hence limit does not exist.

Aliter : Apply $L-$ Hospital’s rule

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1 + \frac{{3{x^2}}}{6}}}{{5{x^4}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin x + \frac{{6x}}{6}}}{{20{x^3}}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \cos x + 1}}{{60{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{120\,x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{120}} = \frac{1}{{120}}.$

Aliter : Apply $L-$ Hospital’s rule, 

$\mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{x - \log \,(1 + x)}}{{{x^2}}}} \right]$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \frac{1}{{1 + x}}}}{{2x}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,{\left( {\frac{1}{{1 + x}}} \right)^2} = \frac{1}{2}.$

Note : Students should remember that

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum n}}{{{n^2}}} = \frac{1}{2},\,\,\,\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^2}}}{{{n^3}}} = \frac{1}{3}$ 

and $\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^3}}}{{{n^4}}} = \frac{1}{4}.$

$ = f(a)g'(a) - g(a)f'(a) = 2 \times 2 - ( - 1)\,(1) = 5.$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{x}{y}\,\left\{ {\frac{{\cos x - \cos \,(x + y)}}{{\cos \,(x + y)\,\cos x}}} \right\}} \right] + \mathop {\lim }\limits_{y \to 0} \sec \,(x + y)$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{{x\sin \,\left( {x + \frac{y}{2}} \right)}}{{\cos \,(x + y)\,.\,\,\cos x}}\,.\,\frac{{\sin \,\left( {\frac{y}{2}} \right)}}{{\,\,\,\frac{y}{2}}}} \right] + \sec x$

$= x\ tan\ x\ sec\ x + sec\ x = sec\ x\ (x\ tan\ x+1).$

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y) - x\,\sec x}}{y}$

$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y)\tan \,(x + y) + \sec \,(x + y) - 0}}{1}$

{Differentiating w.r.t.$y$ assuming $x$ as constant}

$ = x\sec x\tan x + \sec x. = \sec x(x\tan x + 1)$

$\mathop {\lim }\limits_{x \to 1} \,\frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \sqrt {\cos y} }}{{{y^2}}}$

Now rationalizing it, we get

$\mathop {\lim }\limits_{y \to 0} \,\frac{{(1 - \cos y)}}{{{y^2}(1 + \sqrt {\cos y} )}}$

$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \cos y}}{{{y^2}}}\,.\,\mathop {\lim }\limits_{y \to 0} \,\frac{1}{{1 + \sqrt {\cos y} }} $

$= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

Therefore, given function $ = f'(a) + kf'(e) = 1$

$ \Rightarrow \,\,\frac{1}{a} + \frac{k}{e} = 1\,\, $

$\Rightarrow \,\,k = e\,\left( {\frac{{a - 1}}{a}} \right)$

Aliter : Apply $ L-$ Hospital’s rule to find both the limits.

$ = \mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{1}{{x + 1}}} \right)^{x + 3}}$

$ = \mathop {\lim }\limits_{x \to \infty } \,{\left[ {{{\left( {1 + \frac{1}{{x + 1}}} \right)}^{x + 1}}} \right]^{\frac{{\,(x + 3)}}{{(x + 1)}}}} = e$

$\left\{ \because \,\,\underset{x\to \infty }{\mathop{\lim }}\,\,{{\left( 1+\frac{1}{x+1} \right)}^{x+1}}=e \right.$

and $\mathop {\lim }\limits_{x \to \infty } \frac{{\,(x + 3)}}{{(x + 1)}} = \left. {\mathop {\lim }\limits_{x \to \infty } \frac{{\,\left\{ {1 + (3/x)} \right\}}}{{\left\{ {1 + (1/x)} \right\}}} = 1} \right\}$.

Apply $L-$ Hospital’s rule, we get

$\mathop {\lim }\limits_{x \to \pi /2} \,\frac{{\cos x - \sin 2x}}{{ - \sin x}} = 0$.

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x - \frac{1}{{1 - x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \sin x - \frac{1}{{{{(1 - x)}^2}}}}}{2} = - \frac{1}{2}$.

Aliter : $\mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x + \log \,(1 - x)}}{{{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\left( {x - \frac{{{x^3}}}{{3\,\,!}} + \frac{{{x^5}}}{{5\,\,!}} - ...} \right)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\left( { - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...} \right)}}{{{x^2}}}$

$\left( {\because \sin x = x - \frac{{{x^3}}}{{3\,!}} + \frac{{{x^5}}}{{5\,!}} - ..} \right.$ 

and  $\left. {\log \,(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - ..} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{{ - {x^2}}}{2} - {x^3}\left( {\frac{1}{{3\,\,!}} + \frac{1}{3}} \right) - \frac{{{x^4}}}{4}...}}{{{x^2}}} = - \frac{1}{2}.$

$= \mathop {\lim }\limits_{x \to \infty } \,{\left\{ {{{\left( {1 + \frac{1}{{a + bx}}} \right)}^{a + bx}}} \right\}^{\frac{{c + dx}}{{a + bx}}}} = {e^{d/b}}$

$\left\{ {\because \,\,\mathop {\lim }\limits_{x \to \infty } \,{{\left( {1 + \frac{1}{{a + bx}}} \right)}^{a + bx}} = e} \right.$  and 

$\left. {\mathop {\lim }\limits_{x \to \infty } \frac{{c + dx}}{{a + bx}} = \frac{d}{b}} \right\}$.

$ = \mathop {\lim }\limits_{n \to \infty } \,5\,{\left[ {{{\left\{ {1 + {{\left( {\frac{4}{5}} \right)}^n}} \right\}}^{{{(5/4)}^n}}}} \right]^{(1/n)\,.\,{{(4/5)}^n}}} = 5\,.\,{e^0} = 5$.

$\left( {\because \,\,{{\left( {\frac{4}{5}} \right)}^n} \to 0\,{\text{as }}\,n \to \infty } \right)$

$= \mathop {\lim }\limits_{x \to \infty } \,\sqrt {\left( {\frac{{1 + \frac{{\sin x}}{x}}}{{1 - \frac{{\cos x}}{x}}}} \right)} = \mathop {\lim }\limits_{x \to \infty } \sqrt 1 = 1$

$[\,\because \,\,\mathop {\lim }\limits_{x \to \infty } \,\frac{{\sin x}}{x}$ and $\mathop {\lim }\limits_{x \to \infty } \,\frac{{\cos x}}{x}$ both are equal to  $0$ ]

$ = y\mathop {\lim }\limits_{n \to \infty } \,\,{\left[ {{{\left( {1 + {{\left( {\frac{x}{y}} \right)}^n}} \right)}^{{{\left( {\frac{y}{x}} \right)}^n}.}}} \right]^{\frac{1}{n}.{{\left( {\frac{x}{y}} \right)}^n}}}$

$ = y{e^0} = y$, 

$\left[ {\because \,\,\frac{x}{y} < 1\, \Rightarrow \,\,{{\left( {\frac{x}{y}} \right)}^n} \to 0\,\,{\text{as}}\,\,n \to \infty } \right]$.

$\therefore \,\,\,\mathop {\lim }\limits_{x \to - 1} \,\frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }} = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$

$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\cos \,(y/2)}}\, $

$= \mathop {\lim }\limits_{y \to \pi } \,\,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}\frac{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}$

$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{1}{{\frac{{\sqrt 2 }}{2}(\sqrt \pi + \sqrt y )}}.\frac{1}{{\frac{{\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}}} = \frac{1}{{\sqrt {2\pi } }}.$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {x + \sqrt x } }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \,\frac{{\sqrt {1 + {x^{ - 1/2}}} }}{{\sqrt {1 + \sqrt {{x^{ - 1}} + {x^{ - 3/2}}} } + 1}} = \frac{1}{2}$.

$ = \frac{{{x^2} - 8x + 15}}{{{x^2} + x - 12}} = \frac{{(x - 3)\,(x - 5)}}{{(x - 3)\,(x + 4)}}$

$\therefore \,\,\mathop {\lim }\limits_{x \to 3} \,[f(x) + g(x) + h(x)] = \mathop {\lim }\limits_{x \to 3} \,\,\frac{{(x - 3)\,\,(x - 5)}}{{(x - 3)\,\,(x + 4)}} = - \frac{2}{7}$.

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{1 + \sqrt {2 + x} - 3}}{{(\sqrt {1 + \sqrt {2 + x} + \sqrt 3 )\,\,(x - 2)} }}$

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {2 + x} - 2}}{{(\sqrt {1 + \sqrt {2 + x} + \sqrt 3 )\,\,(x - 2)} }}$

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{(x - 2)}}{{(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3 )\,\,(\sqrt {2 + x} + 2)\,\,(x - 2)}}$

$ = \frac{1}{{(2\sqrt 3 )\,4}} = \frac{1}{{8\sqrt 3 }}.$

$ = \frac{1}{{\sqrt 2 }}\,\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{\frac{{\sin \,({x^2}/2)}}{{{x^2}/2}}}}{{{{\left( {\frac{{\sin \,(x/2)}}{{x/2}}} \right)}^2}}}} \right).\frac{{{x^2}/2}}{{{x^2}/4}} = \sqrt 2 $. 

$= \mathop {\lim }\limits_{x \to \infty } \,\,\frac{{{x^2}\left( {2 - \frac{3}{3}\,} \right)\,\left( {3 - \frac{4}{x}} \right)}}{{{x^2}\left( {4 - \frac{5}{x}} \right)\,\left( {5 - \frac{6}{x}} \right)}} = \frac{6}{{20}} = \frac{3}{{10}}$

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{\log \,(1 + t)}}$,          $\{$Putting $x = 2 + t\} $

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\frac{{{e^t} - 1}}{t}.\frac{t}{{\log \,(1 + t)}}$

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\left( {\frac{1}{{1\,\,!}} + \frac{t}{{2\,\,!}} + ...} \right) \times \left[ {\frac{1}{{\left( {1 - \frac{1}{2}t + \frac{1}{3}{t^2} - ...} \right)}}} \right]$

$ = 1\,\,.\,\,1\,\,.\,\,1 = 1,\,\,\,\,(\because \,\,{\text{As}}\,\,t \to 0,\,\,{e^t} - 1 \to 0).$

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{4}{{\left( {\sqrt {1 + \frac{8}{x} + \frac{3}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} } \right)}} = 2$.

$\therefore$ $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = k{(5)^{k - 1}}$; 

But given, $\mathop {\lim }\limits_{x \to 5} \,\frac{{{x^k} - {5^k}}}{{x - 5}} = 500$,

$\therefore$ $k{(5)^{k - 1}} = 500$; $k\,{(5)^{k - 1}} = 4\,{(5)^{4 - 1}},$

$\therefore \,\,\,k = 4$.

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{(1 - {x^2} - 1 - {x^2})}}{{{x^2}\,(\sqrt {1 - {x^2}} + \sqrt {1 + {x^2})} }}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - 2}}{{\,(\sqrt {1 - {x^2}}  + \sqrt {1 + {x^2})} }} = \frac{{ - 2}}{{1 + 1}} =  - 1$

and $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h) = \mathop {\lim }\limits_{h \to 0} \,\,\, - (0 + h) = 0$

$\therefore \,\,\,\mathop {\lim }\limits_{x \to 0} \,\,f(x) = 0$,    $\left( {\because \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)} \right)$ .

Applying $ L-$ Hospital's rule,

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^x} + x\,{e^x} - \frac{1}{{1 + x}}}}{{2x}}$,   $\left( {\frac{0}{0}\,{\rm{form}}} \right)$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,\left[ {{e^x} + {e^x} + x\,{e^x} + \frac{1}{{{{(1 + x)}^2}}}} \right]$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,[1 + 1 + 0 + 1] = \frac{3}{2}$.

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {4\,{{( - 1/h)}^2} + 5\,( - 1/h) + 8} }}{{4\,( - 1/h) + 5}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{(1/h)\sqrt {4\, - 5h + 8{h^2}} }}{{(1/h)\,( - \,4 + 5h)}} = \frac{{\sqrt 4 }}{{ - 4}} = - \frac{1}{2}$.

$ \Rightarrow \,\,y = {e^{1/m}},\,\,\,\left( {\because \mathop {\lim }\limits_{x \to \,\infty } \,{{\left( {1 + \frac{1}{x}} \right)}^x} = e} \right)$

$ = \mathop {\lim }\limits_{h \to 0} \,\,(3 - 3h) = 3 - 3\,.\,0 = 3$

$R.H.L.$$ = \mathop {\lim }\limits_{x \to 1 + 0} f(x) = \mathop {\lim \,\,}\limits_{h \to 0} \,f\,(1 + h) = \mathop {\lim }\limits_{h \to 0} \,\,[5 - 3(1 + h)]$

$ = \mathop {\lim }\limits_{h \to 0} \,\,(2 - 3h) = 2 - 3\,.\,0 = 2$

Hence $\mathop {\lim }\limits_{x \to 1} \,\,f(x)$ does not exist.

Applying $L-$ Hospital's rule, we get

$\mathop {\lim }\limits_{x \to 2} \,\,\frac{{3{x^2}}}{{2x}} = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{3 \times 2 \times 2}}{{2 \times 2}} = 3$.

Applying  $ L-$ Hospital's rule, we get

$y = \mathop {\lim }\limits_{x \to 3} \,\,3{x^2} - 2x = (27 - 6) = 21$.

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{1 + {x^2}}}}}{1} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{{1 + {x^2}}} = \frac{1}{{1 + 0}} = 1$.

==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{({x^{1/2}} - 2)(x + 4 + 2\sqrt x )}}{{(\sqrt x - 2)(\sqrt x + 2)}}$

==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{(x + 4 + 2\sqrt x )}}{{(\sqrt x + 2)}}$$ = \frac{{4 + 4 + 2\sqrt 4 }}{{\sqrt 4 + 2}}$$ = \frac{{12}}{4} = 3$.

Trick : Applying  $ L-$ Hospital’s rule, we get

$\mathop {\lim }\limits_{x \to 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1}$$ = \frac{3}{2}{(4)^{1/2}} = 3.$

Applying $ L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{3{x^2}}}$,      $\left( {\frac{0}{0}} \,\,form \,\, \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} \times \frac{{ - 2x}}{{{{(1 - {x^2})}^{3/2}}}} + \frac{{2x}}{{{{(1 + {x^2})}^2}}}}}{{6x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{6}\left[ {\frac{1}{{{{(1 - {x^2})}^{3/2}}}} + \frac{2}{{{{(1 + {x^2})}^2}}}} \right]\, = \,\frac{1}{2}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to \infty } \,{\left\{ {{{\left( {1 + \frac{{a - b}}{{x + b}}} \right)}^{\frac{{x + b}}{{a - b}}}}} \right\}^{a - b}} = {e^{a - b}}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to \pi /2} {a^{\cos x}}\left( {\frac{{{a^{\cot x - \cos x}} - 1}}{{\cot x - \cos x}}} \right)$

$ = {a^{\cos (\pi /2)}}\mathop {{\rm{lim}}}\limits_{x \to \pi /2} \left( {\frac{{{a^{\cot x - \cos x}} - 1}}{{\cot x - \cos x}}} \right)$$ = 1.\log a = \log a$.

$ = {x^2}\sin x - {x^3}\cos x - {x^3}\tan x + 2{x^2}\cos x - x\sin x$

Hence $\mathop {\lim }\limits_{x \to 0} \frac{{f(x)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \,\left( {\sin x - x\cos x - x\tan x + 2\cos x - \left. {\frac{{\sin x}}{x}} \right)} \right.$

$ = 0 - 0 - 0 + 2 - 1 = 1$.

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {1 + \frac{1}{{n(n - 1)}}} \right)}^{n(n - 1)}}}}{{{{\left( {1 - \frac{1}{{n(n - 1)}}} \right)}^{n(n - 1)}}}}$$ = \frac{e}{{{e^{ - 1}}}} = {e^2}$.

$g(x) = {\cos ^{ - 1}}\left\{ {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right\}$

Put $x = \tan \theta $ in both equations

$f(\theta ) = {\cot ^{ - 1}}\left\{ {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right\}$$ = {\cot ^{ - 1}}\left\{ {\tan 3\theta } \right\}$

$f(\theta ) = {\cot ^{ - 1}}\cot \left( {\frac{\pi }{2} - 3\theta } \right) = \frac{\pi }{2} - 3\theta \Rightarrow f'(\theta ) = - 3$ .….(i)

and $g(\theta ) = {\cos ^{ - 1}}\left\{ {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right\}$$ = {\cos ^{ - 1}}(\cos 2\theta ) = 2\theta $

$ \Rightarrow g'(\theta ) = 2$ …..(ii)

Now $\mathop {\lim }\limits_{x \to a} \left( {\frac{{f(x) - f(a)}}{{g(x) - g(a)}}} \right) = \mathop {\lim }\limits_{x \to a} \left( {\frac{{f(x) - f(a)}}{{x - a}}} \right)\frac{1}{{\mathop {\lim }\limits_{x \to a} \left( {\frac{{g(x) - g(a)}}{{x - a}}} \right)}}$

$ = f'(x).\frac{1}{{g'(x)}} = - 3 \times \frac{1}{2} = - \frac{3}{2}$.

Using $ L-$ Hospital’s rule

==> $y = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {\frac{1}{{\sqrt {1 - {{(x + 2)}^2}} }}} \right)}}{{2x + 2}}$

==> $y = \frac{1}{{ - 4 + 2}} = - \frac{1}{2}$.

$ = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{{x + 1}}} \right)^{\frac{{x + 1}}{2}.2}}$

$ = {\left\{ {\mathop {\lim }\limits_{x \to \infty } {{\left( {1 + \frac{2}{{x + 1}}} \right)}^{\frac{{x + 1}}{2}}}} \right\}^2}$$ = {e^2}$.

$ = \mathop {{\rm{lim}}}\limits_{n \to \infty } \,{({4^n})^{\frac{1}{n}}}{\left[ {\frac{{{3^n}}}{{{4^n}}} + 1} \right]^{\frac{1}{n}}}$

$ = \mathop {{\rm{lim}}}\limits_{n \to \infty } 4\,{\left[ {1 + \frac{1}{{{{\left( {\frac{4}{3}} \right)}^n}}}} \right]^{1/n}}$

$ = 4\mathop {{\rm{lim}}}\limits_{n \to \infty } \,{\left[ {1 + \frac{1}{{{{\left( {\frac{4}{3}} \right)}^n}}}} \right]^{1/n}}$

$ = 4{\left[ {1 + \frac{1}{\infty }} \right]^0} = 4 \times {(1)^0}$ $ = 4 \times 1 = 4$.

$ = {e^{\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{ - 4\left( {3 - \frac{1}{x}} \right)}}{{\left( {1 - \frac{1}{x}} \right)}}} \right]}} = {e^{ - 12}}$.

Using $ L-$ Hospital’s rule three times, then

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}.\cos x}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}{{\cos }^2}x + \sin x.{e^{\sin x}}}}{{\sin x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}.{{\cos }^3}x + {e^{\sin x}}2\cos x\sin x + {e^{\sin x}}.\cos x\sin x + {e^{\sin x}}.\cos x}}{{\cos x}}$

$ = 1$.

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{{(10)}^n}\left[ {{{\left( {\frac{1}{{10}}} \right)}^n} - 1} \right]}}{{{{(10)}^{n + 1}}\left( {1 + \frac{1}{{{{10}^{n + 1}}}}} \right)}} = - \frac{1}{{10}}$

$\therefore \alpha = 1$.

$ = \mathop {\lim }\limits_{x \to 0} \frac{{3{x^2}/(1 + {x^3})}}{{3{{\sin }^2}x\cos x}}$

[By using $ L-$  Hospital rule]

$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{{1 + {x^3}}}{{\left( {\frac{x}{{\sin x}}} \right)}^2}.\frac{1}{{\cos x}}} \right]$

$ = \frac{1}{{1 + 0}}.{(1)^2}.\frac{1}{1} = 1$.

$\left. { + \left( {\frac{1}{{(2n - 1)}} - \frac{1}{{(2n + 1)}}} \right)} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right] = \frac{1}{2}$.

$ = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{1}{6}\frac{{n\,(n + 1)\,(2n + 1)}}{{1 + {n^3}}}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{6}\frac{{\left( {1 + \frac{1}{n}} \right)\,\left( {2 + \frac{1}{n}} \right)}}{{\left( {\frac{1}{{{n^3}}} + 1} \right)}}$

$ = \frac{1}{6}\,.\,1\,.\,\frac{2}{{(1)}} = \left( {\frac{1}{3}} \right).$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{{2\,(1 - n)}}{n}$

$ = \mathop {\lim }\limits_{n \to \infty } 2\,\left( {\frac{1}{n} - 1} \right)$$ = 2(0 - 1) = - 2$.

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{10}}\left[ {{{\left( {1 + \frac{1}{x}} \right)}^{10}} + {{\left( {1 + \frac{2}{x}} \right)}^{10}} + ... + {{\left( {1 + \frac{{100}}{x}} \right)}^{10}}} \right]}}{{{x^{10}}\left[ {1 + \frac{{{{10}^{10}}}}{{{x^{10}}}}} \right]}} = 100$.

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{n(n + 1)}}{{2({n^2} + 100)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{2{n^2}\left( {1 + \frac{{100}}{{{n^2}}}} \right)}} = \frac{1}{2}$.

Applying  $ L- $ Hospital rule, we get

$\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos {x^2}}}{1} = 1$.

$ = \frac{1}{3}\,\left[ {\frac{{2\sqrt 2 }}{2}} \right] = \frac{{\sqrt 2 }}{3},\,\,0 < \frac{{\sqrt 2 }}{3} < \frac{1}{2}.$

Aliter : Apply $ L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }} $

$= \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }}}}{{\frac{3}{{2\sqrt {2 + 3x} }} + \frac{3}{{2\sqrt {2 - 3x} }}}}$

$ = \frac{{\frac{1}{2} + \frac{1}{2}}}{{\frac{3}{{2\sqrt 2 }} + \frac{3}{{2\sqrt 2 }}}} = \frac{{2\sqrt 2 }}{6} = \frac{{\sqrt 2 }}{3}$.

$ = \mathop {\lim }\limits_{h \to 0} \,\,(1 - (1 - h) + [1 - h - 1] + [1 - (1 - h)])\,$

$ = \mathop {\lim }\limits_{h \to 0} \,(h + [ - h] + [h]) = \mathop {\lim }\limits_{h \to 0} \,(h - 1 + 0) = - 1$

and $\mathop {\lim }\limits_{x \to 1 + } \,(1 - x + [x - 1] + [1 - x])\,$

$ = \mathop {\lim }\limits_{h \to 0} \,(1 - (1 + h) + [1 + h - 1] + [1 - (1 + h)])$

$ = \mathop {\lim }\limits_{h \to 0} \,( - h + [h] + [ - h]) = \mathop {\lim }\limits_{h \to 0} \,( - h + 0 - 1) = - 1$

$\therefore$  $\mathop {\lim }\limits_{x \to 1} \,f(x) = - 1$.

$ - 1 = \mathop {\lim }\limits_{x \to a} \,\,\frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = \mathop {\lim }\limits_{x \to a} \,\frac{{{a^x}{{\log }_e}a - a{x^{a - 1}}}}{{{x^x} + {a^a}{{\log }_e}a}}$

$ \Rightarrow \,\, - 1 = \frac{{{a^a}{{\log }_e}a - a\,.\,{a^{a - 1}}}}{{{a^a} + {a^a}{{\log }_e}a}} = \frac{{{{\log }_e}a - 1}}{{{{\log }_e}a + 1}}$.....$(i)$

Now $(i)$ is satisfied only when $a = 1.$

$I$. $\lim _{n \rightarrow \infty} \frac{2^n+(-2)^n}{2^n}=\lim _{n \rightarrow \infty} 1+(-1)^n=$ does not exist

$11$. $\lim _{n \rightarrow \infty} \frac{3^n+(-3)^n}{4^n}=\lim _{n \rightarrow \infty}\left(\frac{3}{4}\right)^n+\left(\frac{-3}{4}\right)^n$

$=0+0=0$

Given, $S_n=\sum_{k=\bullet}^n \frac{1}{\sqrt{n^2+k}}$

$\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\ldots+\frac{1}{\sqrt{n^2+n}}$

$\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\ldots+\frac{1}{\sqrt{n^2+n}}$ $\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+\ldots+\frac{1}{\sqrt{n^2}}$

$\lim _{n \rightarrow \infty} \frac{n+1}{\sqrt{n^2+n}} \leq \lim _{n \rightarrow \infty} S_n \leq \lim _{n \rightarrow \infty} \frac{n+1}{\sqrt{n^2}}$

$1 \leq \lim _{n \rightarrow \infty} S_n \leq 1 \Rightarrow \lim _{n \rightarrow \infty} S_n=1$

We have, $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$

$=\lim _{x \rightarrow-\infty} \frac{4 x^2-x-4 x^2}{\sqrt{4 x^2-x-2 x}}$

$=\lim _{x \rightarrow-\infty} \frac{-x}{-\sqrt{\left.\sqrt{4-\frac{x}{x^2}}+2\right)}}=\frac{1}{2+2}=\frac{1}{4}$

$=\lim _{x \rightarrow-\infty} \frac{1}{\sqrt{4-\frac{1}{x}+2}}$

We have,

$x_n =\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$\Rightarrow \quad \lim _{n \rightarrow \infty} x_n =\lim _{n \rightarrow \infty}\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$=\lim _{n \rightarrow \infty} 3^{n / 2 n}\left[\left(\frac{2}{3}\right)^n+1\right]^{\frac{1}{2 n}}$

$=3^{3 / 2} \times 1=\sqrt{3}$

Let $L=\lim _{x \rightarrow 2} \frac{a^x+a^{3-x}-\left(a^2+a\right)}{a^{3-x}-a^{x / 2}}$

Apply L-Hospital rule

$L=\lim _{x \rightarrow 2} \frac{a^x \log a-a^{3-x} \log a}{-a^{3-x} \log a-\frac{1}{2} e^{x / 2} \log a}$

$L=\lim _{x \rightarrow 2} \frac{a^x-a^{3-x}}{-a^{3-x}-\frac{a^{x / 2}}{2}}$

$L=\frac{a^2-a}{-a-\frac{a}{2}}=\frac{2}{3}(1-a)$

Let

$p =\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^2}$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6}{x^2} \log \left(\frac{x}{\sin x}\right)$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6 \log \left(\frac{x}{\sin x}\right)}{x^2}$

Apply $L-$ Hospital rule

$\log p=\lim _{x \rightarrow 0}$ 

$6 \frac{\frac{\sin x}{x} \frac{(\sin x-x \cos x)}{\sin ^2 x}}{2 x}$

$\log p=\lim _{x \rightarrow 0} 6 \frac{\sin x(\sin x-x \cos x)}{x \cdot 2 x \sin ^2 x}$

$\log p=\lim _{x \rightarrow 0} 3 \frac{\sin x}{x} \times \lim _{x \rightarrow 0}$

$\frac{\sin x-x \cos x}{\frac{\sin ^2 x}{x^2}} \times x^3$

$\log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

$\Rightarrow \quad \log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

Again apply L.Hospital's rule

$\log p=3 \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^2}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\therefore \quad p=e^1=e$

Let

$p =\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^2}$

$\log p =\lim _{x \rightarrow 0} \frac{6}{x^2} \log \left(\frac{x}{\sin x}\right)$

$\log p =\lim _{x \rightarrow 0} \frac{6 \log \left(\frac{x}{\sin x}\right)}{x^2}$ 

Apply L-Hospital rule

$\log p=\lim _{x \rightarrow 0} 6 \frac{\frac{\sin x}{x} \frac{(\sin x-x \cos x)}{\sin ^2 x}}{2 x}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} 6 \frac{\sin x(\sin x-x \cos x)}{x \cdot 2 x \sin ^2 x}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} 3 \frac{\sin x}{x} \times \lim _{x \rightarrow 0}$

$\frac{\sin x-x \cos x}{\frac{\sin ^2 x}{x^2}} \times x^3$

$\Rightarrow \quad \log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

Again apply L.Hospital's rule

$\quad \log p=3 \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^2}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\therefore \quad p=e^1=e$

$\lim _{x \rightarrow \infty} f(x)=M > 0$

(a) $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) f(x)=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}} f(x)$ $=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}} \times \lim _{x \rightarrow \infty} f(x)=1 \times M=M$

Hence, option (a) is true.

(b) $\lim _{x \rightarrow \infty} \sin (f(x))$

$\sin \left(\lim _{x \rightarrow \infty} f(x)\right)=\sin M$

Option (b) is also true.

(c) $\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x)$

$\lim _{x \rightarrow \infty} x e^{-x} \frac{\sin \left(e^{-x}\right)}{e^{-x}} f(x)$

$=0 \times 1 \times M=0$

Hence, option (c) is false.

(d) $\lim _{x \rightarrow \infty} \frac{\sin x}{x} f(x)=0 \times M=0$

Option (d) is also true.

We have,

$a_n = a _1 \cdot a_2 \cdot a_3 \ldots a_{n-1}+4, a_1=5$

$a_2 =a_1+4=5+4=9$

$a_3 =a_1 a_2+4=5 \times 9+4=49$

$a_4 =a_1 a_2 a_3=5 \times 9 \times 49+4=2209$

$a_4=\left(a_3-2\right)^2=(49-2)^2=47^2=2209$

$a_3 =\left(a_2-2\right)^2=(9-2)^2=49$

$a_n =\left(a_{n-1}-2\right)^2$

$\therefore \quad \lim _{n \rightarrow \infty} \frac{\sqrt{ a _n}}{a_{n-1}} =\lim _{n \rightarrow \infty} \frac{\sqrt{\left(a_{n-1}-2\right)^2}}{a_{n-1}}$

$=\lim _{n \rightarrow \infty}\left(\frac{ a _{n-1}-2}{a_{n-1}}\right)=1$

We have,

$a_n=\sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1, a_0=\cos \theta$

$a_1=\sqrt{\frac{1+a_0}{2}}=\sqrt{\frac{1+\cos \theta}{2}}$

$a_1=\sqrt{\frac{2 \cos ^2 \frac{\theta}{2}}{2}}=\cos \frac{\theta}{2}$

$a_2=\sqrt{\frac{1+\cos \frac{\theta}{2}}{2}}=\cos \frac{\theta}{2^2}$

$a_n=\cos \frac{\theta}{2^n}$

$\lim _{n \rightarrow \infty} 4^n\left(1-a_n\right)$

$=\lim _{n \rightarrow \infty} 4^n\left(1-\cos \frac{\theta}{2^n}\right)$

$=\lim _{n \rightarrow \infty} 4^n\left(2 \sin ^2 \frac{\theta}{2^{n+1}}\right)$

$=\lim _{n \rightarrow \infty} 4^n \cdot 2\left(\frac{\sin \frac{\theta}{2^{n+1}}}{\frac{\theta}{2^{n+1}}}\right)^2 \times \frac{\theta^2}{2^{2 n+2}}$

$=\lim _{n \rightarrow \infty} \frac{4^n}{4^n \cdot 2}\left(\frac{\sin \frac{\theta}{2^{n+1}}}{\frac{\theta}{2^{n+1}}}\right)^2 \times \theta^2=\frac{\theta^2}{2}$

We have all natural numbers whose decimal expansion has only even digits $0,2,4,6,8$.

$a_n=n$th number of sequence if these are arranged in increasing orders.

Let $n=5^K$, then $5$ based representation of $n$ is a $1$ followed by $K$ zeroes,

$\therefore$ Decimal representation and doubling its value, we obtain a $2$ followed by $K$ zeroes that is $210^K$.

$\therefore \quad \log a_n=K+\log _{10} 2$

$\Rightarrow \log n =K \log _{10} 5$

$\Rightarrow \quad \lim _{n \rightarrow \infty} \frac{\log a_n}{\log n} =\frac{K+\log _{10} 2}{K \log _{10} 5}$

$K$ goes to infinity

$\therefore \quad \lim _{n \rightarrow \infty} \frac{\log a_n}{\log n}=\frac{1}{\log _{10} 5}=\log _5 10$

$ =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} $

$ =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)}$

Applying limit $\mathrm{a}=\frac{1}{4 \sqrt{2}}$

$ b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} $ $=\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} $

$ b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$

$ \Rightarrow \lim _{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3} $

$ \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3} $

$ \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} $

$ \Rightarrow \beta=-3, \alpha=1 $

$ \Rightarrow 2 \alpha-\beta=2+3=5$

$\therefore a=1$

$\mathrm{b}=\lim _{\mathrm{x} \rightarrow 0} \frac{\int_0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$

Using $L' HOPITAL$ Rule

$b=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\left(1+x^{2024}\right)} \times \frac{1}{2 x}=\frac{1}{2}$

$\begin{array}{ll}\text { Now, } \mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, & \mathrm{x}^2+\mathrm{x}+4=0 \\ & (\mathrm{D}<0)\end{array}$

$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$

$lim _{x \rightarrow 0} \frac{e^{2 s i n x}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}$

Let $|\sin x|=t$

$\lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2}$

$=\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2$

$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$

$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$

$ \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} $

$ \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} $

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$

$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$

$ \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} $

$ \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} $

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)$

$=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)}$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right)}{h}\left(\frac{\sin ^{-1} 1}{1}\right)$

Let $\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}}$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}}$

$=\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}}$

$\mathrm{R}=\frac{\pi}{\sqrt{2}}$

$=\frac{8}{3} \frac{\sqrt{5}}{6^{2 / 3}} \quad \begin{gathered}\mathrm{m}=8 \\ \mathrm{n}=3\end{gathered}$

$ 8 m+12 n=100$

Using L Hopital Rule.

$\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}$ (Again L Hopital)

Using $L.H.$ Rule

$ =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} $

$ =-\frac{1}{6}$

$ =16 \times \frac{2}{\mathrm{a}^2}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)^2 $

$ =\frac{32}{\mathrm{a}^2}\left(\frac{17}{4}\right)=\frac{17.8}{\mathrm{a}^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} $

$ =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} $

$ =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 $

$ =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} $

$ \alpha+\beta=153+17=170$

$ \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(-r^3+r^2(n+1)-n r\right)}{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}} $

$ \lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)} $

$ \lim _{n \rightarrow \infty} \frac{n(n-1)\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}} $

$ \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} $

$ \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3}$

By expansion

$\lim _{x \rightarrow 0} \frac{2\left(1-\left(1-\frac{x^2}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 x^2}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 x^2}{2}\right) \ldots \ldots\left(1-\frac{1}{10} \cdot \frac{100 x^2}{2}\right)}{x^2}$

$\lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{x^2}{2}\right)\left(1-\frac{2 x^2}{2}\right)\left(1-\frac{3 x^2}{2}\right) \ldots\left(1-\frac{10 x^2}{2}\right)}{x^2}\right)$

$\lim _{x \rightarrow 0} \frac{2\left(1-1+x^2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right)\right)}{x^2}$

$ 2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots . .+\frac{10}{2}\right) $

$ 1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55$

$ =1 $

$ \beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x} $

$ =e^{1 / 2} $

$ x^2-(1+\sqrt{e})+\sqrt{e}=0 $

$ a x^2+b x-\sqrt{e}=0$

On comparing

$ a=-1, b=\sqrt{e}+1 $

$ 12 \ln (a+b)=12 \times \frac{1}{2}=6$

$ \mathrm{x}=\frac{1-\mathrm{c}^2}{2+\mathrm{c}-3 \mathrm{c}^2}, \mathrm{y}=\frac{1-3 \mathrm{x}}{5}=\frac{\mathrm{c}-1}{5\left(2+\mathrm{c}-3 \mathrm{c}^2\right)} $

$ \mathrm{h}=\lim _{\mathrm{c} \rightarrow 1} \frac{(1-\mathrm{c})(1+\mathrm{c})}{(1-\mathrm{c})(2+3 \mathrm{c})}=\frac{2}{5} $

$ \mathrm{~K}=\lim _{\mathrm{c} \rightarrow 1} \frac{\mathrm{c}-1}{-5(\mathrm{c}-1)(3 \mathrm{c}+2)}=-\frac{1}{25} $

$ \text { Centre }\left(\frac{2}{25},-\frac{1}{25}\right), $

$ \mathrm{r}=\sqrt{\left(2-\frac{2}{5}\right)^2+\left(0-\frac{1}{25}\right)^2}=\sqrt{\frac{64}{25}+\frac{1}{625}} $

$ \mathrm{r}=\frac{\sqrt{161}}{25} $

$ \left(\mathrm{x}-\frac{2}{5}\right)^2+\left(\mathrm{y}+\frac{1}{25}\right)^2=\frac{161}{125} $

$ \Rightarrow 25 \mathrm{x}^2+25 \mathrm{y}^2-20 \mathrm{x}+2 \mathrm{y}-60=0$

$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} $

$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} $

$ =(-e) \times(-1) \frac{4}{2 \times 2}=e$

$ =\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^2}{2\left(x-\frac{\pi}{2}\right)} $

$ =\lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{2 \sin x \sin \left(\frac{\pi}{2}-x\right)}{2\left(x-\frac{\pi}{2}\right)}+\frac{\sin \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right\} 3 x^2 $

$ =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 $

$ =\frac{9 \pi^2}{8}$

$\lim _{x \rightarrow a}([x]-5-[2 x]-2)=0$

$\lim _{x \rightarrow a}([x]-[2 x])=7$

${[a]-[2 a]=7}$

$a \in I, \quad a=-7$

$a \notin I, \quad a=I+f$

$\text { Now, }[a]-[2 a]=7$

$\quad-I-[2 f]=7$

$\text { Case-I: f } \in\left(0, \frac{1}{2}\right)$

$2 f \in(0,1)$

$-I=7$

$I=-7 \Rightarrow a \in(-7,-6.5)$

Case$-II:$ $f \in\left(\frac{1}{2}, 1\right)$

$2 f \in(1,2)$

$- I -1=7$

$I =-8 \Rightarrow a \in(-7.5,-7)$

Hence, $a \in(-7.5,-6.5)$

$\operatorname{Lim}_{n \rightarrow \infty} \frac{3 n(n-1)}{2\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)}$

$=\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)$

$\lim _{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^2-4 p(2 p+h)+q^2+82+16}{h^2}\right)^2=\frac{1}{2}$

Using L'Hospital's

$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$

Applying L'Hospitals Rule

$48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}$

$=12$

$\lim _{x \rightarrow \infty} x^3 \times\left\{\frac{x^3\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6\right\}}{x^6\left\{\left(1+\sqrt{1-\frac{1}{x^2}}\right)^6+\left(1-\sqrt{1-\frac{1}{x^2}}\right)^6\right\}}\right\}$

$=\frac{(2 \sqrt{3})^6+0}{2^6+0}=3^3=(27)$

$\lim _{x \rightarrow 0} 2\left(\frac{1}{2} \times \frac{9}{1} \times \frac{1 \times 64}{1 \times 32}\right)=18$

$\therefore x^2+b x+a=a(1-\alpha x)(1-\beta x)$

$\therefore \lim _{x \rightarrow \frac{1}{\alpha}}\left\{\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right\}^{\frac{1}{2}}=\lim _{x \rightarrow \frac{1}{2}}\left\{\frac{1-\cos a(1-\alpha x)(1-\beta x)}{2\{a(1-\alpha x)(1-\beta x)\}^2} \cdot a^2(1-\beta x)^2\right\}^{\frac{1}{2}}$

$=\left[\frac{1}{2} \cdot \frac{1}{2} a ^2\left(1-\frac{\beta}{\alpha}\right)^2\right]^{\frac{1}{2}}$

$=\frac{1}{2} \frac{1}{\alpha \beta}\left(1-\frac{\beta}{\alpha}\right)=\frac{1}{2}\left(\frac{1}{\alpha \beta}-\frac{1}{\alpha^2}\right)$

$=\frac{1}{2 \alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)=\frac{1}{ k }\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$

$\therefore k =2 \alpha$

On expansion,

$\lim _{x \rightarrow 0} \frac{\left(1+a x+\frac{a^2 x^2}{2}\right)-\left(1-\frac{b^2 x^2}{2}\right)-\frac{c x}{2}(1-c x)}{2 x^2}=17$

$\lim _{x \rightarrow 0} \frac{\left(a-\frac{c}{2}\right) x+x^2\left(\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}\right)}{2 x^2}=17$

For limit to be exist a $-\frac{c}{2}=0$

$a=\frac{c}{2}$

and $\frac{a^2+b^2+c^2}{4}=17$

$a^2+b^2+4 a^2=17 \times 4$

$5 a^2+b^2=68$

On rationalising each term

$\lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_n}-\sqrt{a_1}}{d}\right)$

$\lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{(n-1) d}{\left(\sqrt{a_n}+\sqrt{a_1}\right) d}\right)=1$

$--\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right) < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$

$\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$

$\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 \text { and } \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n=0$

$\Rightarrow \lim _{n \rightarrow \infty} L=0$

$\lim \limits_{x \rightarrow 0^{-}} f(x)=\alpha-\frac{4}{3} \Rightarrow 0+\frac{\alpha^{-1}-2}{-1}=\alpha-\frac{4}{3}$

$\Rightarrow 2-\frac{1}{\alpha}=\alpha-\frac{4}{3}$

$\Rightarrow \alpha+\frac{1}{\alpha}=\frac{10}{3}$

$\Rightarrow \alpha=3 ; \alpha \in I$

$\lim _{x \rightarrow 0} \frac{\left(\frac{\ln \left(1+x^{2}+x^{4}\right)}{x^{2}+x^{4}}\right) x^{2}\left(1+x^{2}\right) \cos x}{\left(\frac{\sin ^{2} x}{x^{2}}\right) x^{2}}=1$

$\therefore k =1$

$=\lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\tan ^{2} x\left[\sin ^{2} x-3 \sin x+2\right]}{\sqrt{9}+\sqrt{9}}$

$=\lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\tan ^{2} x(\sin x-1)(\sin x-2)}{6}$

$=\frac{1}{6} \lim \limits_{x \rightarrow \frac{\pi}{2}} \tan ^{2} x(1-\sin x)$

$=\frac{1}{6} \lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\sin ^{2} x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\frac{1}{12}$

$\lim \limits_{x \rightarrow 0}\left(\frac{2 \cdot \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}}\right)$

$\lim\limits _{x \rightarrow 0} 2\left(\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\left(\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right)\left(\frac{\left.\frac{x+\sin x}{2}\right)}{x^{4}}\left(\frac{x-\sin x}{2}\right)\right.\right.$

$\lim\limits _{x \rightarrow 0}\left(\frac{x^{2}-\sin ^{2} x}{2 x^{4}}\right):\left(\frac{0}{0}\right)$

Apply $L-Hopital\; Rule$ :

$\lim \limits_{x \rightarrow 0} \frac{2 x-2 \sin x \cos x}{2.4 \cdot x^{3}}$

$\lim\limits _{x \rightarrow 0} \frac{2 x-\sin 2 x}{8 x^{3}} ; \frac{0}{0}:$ Again apply $L-Hopital \;rule$

$\lim \limits_{x \rightarrow 0} \frac{2-2 \cos (2 x)}{8(3) x^{2}}$

$\lim \limits_{x \rightarrow 0} \frac{2(1-\cos (2 x))}{24\left(4 x^{2}\right)} \times 4 \Rightarrow \frac{2}{24} \times \frac{1}{2} \times 4 \Rightarrow \frac{1}{6}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)-x}{1-\tan \left(\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right)}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^{2}}-x}{1-\left(\frac{\sqrt{1-x^{2}}}{x}\right)}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}}(-x)=-\frac{1}{\sqrt{2}}$

$L.H.L.$ $\lim \limits_{x \rightarrow 7-} \frac{18-[1-x]}{[x]-3 a}$

$=\frac{18-(-6)}{6-3 a}$

$=\frac{24}{6-3 a}$

$R.H.L.$ $\lim \limits_{x \rightarrow 7^{+}} \frac{18-[1-x]}{[x]-3 a}$

$=\frac{18-(-7)}{7-3 a}$

$=\frac{25}{7-3 a}$

Now $L.H.L. \;= \;R.H.L.$

$\frac{24}{6-3 a}=\frac{25}{7-3 a}$

$\Rightarrow 168-72 a=150-75 a$

$\Rightarrow 18=-3 a$

$\Rightarrow a =-6$

For finite limit

$a+b-5=0$...............$(1)$

Apply $L'H $rule

$\lim\limits _{x \rightarrow 1} \frac{\cos \left(3 x^{2}-4 x+1\right)(6 x-4)-2 x}{\left(6 x^{2}-14 x+a\right)}=-2$

For finite limit

$6-14+a=0$

$a=8$

From $(1)\,\,\,\,\,\, b=-3$

Now $(a-b)=11$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$ using $L-H$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{56(\cos x-\sin x)}{2 \sqrt{2} \cos 2 x}\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-56(\sin x+\cos x)}{-4 \sqrt{2} \sin 2 x} \quad$ using L-H Rule

$=7 \sqrt{2} \cdot \sqrt{2}=14$

$\lim _{n \rightarrow \infty}\left\{1-\frac{1}{2}\left(\frac{n+1}{n^{2}}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2 !}\left(\frac{n+1}{n^{2}}\right)^{2}+\ldots .\right\}+\alpha n+\beta=0$

$\lim _{n \rightarrow \infty} n-\frac{1}{2}+\frac{1}{n}+\ldots+n \alpha+\beta=0$

$\alpha=-1, \beta=\frac{1}{2}$

$8(\alpha+\beta)=-4$

$\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[n k \cdot n+1+2+\ldots+n]$

$=\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot\left[n^{2} k+\frac{n(n+1)}{2}\right]$

$=\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1} \cdot n^{2}\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}{n^{k+1}}$

$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)$

$\left(k+\frac{1}{2}\right)$

$RHS$

$\lim _{ n \rightarrow \infty} \frac{1}{ n ^{ k +1}}\left(1^{ k }+2^{ k }+\ldots+ n ^{ k }\right)=\frac{1}{ k +1}$

$\text { LHS }=\text { RHS }$

$k +\frac{1}{2}=33 \cdot \frac{1}{ k +1}$

$(2 k +1)( k +1)=66$

$( k -5)(2 k +13)=0$

$k =5 \text { or }-\frac{13}{2}$

$\beta=\lim _{x \rightarrow 0} \frac{1+\alpha x-\left[1+3 x+\frac{9 x^{2}}{2 !}+\ldots .\right]}{(\alpha x) \frac{\left(e^{3 x}-1\right)}{3 x} 3 x}$

$\beta=\lim _{x \rightarrow 0} \frac{(\alpha x-3 x)-\frac{9 x^{2}}{2 !}-\ldots \ldots .}{3 \alpha x^{2}}$

For existence of limit $\alpha-3=0$ $\alpha=3$

Limit $\beta=\frac{-3}{2 \alpha}$

$\beta=-\frac{1}{2}$

Now,$\alpha+\beta=\frac{5}{2}$

From $1^{\infty}$

$=e^{\lim _{x \rightarrow 0}\left[\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)-1\right] \times \frac{100}{x}}$

$=e^{\lim _{x \rightarrow 0}\left[\frac{100}{x}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)-\left((x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)\right)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)\right]}$

$=e^{\lim _{x \rightarrow 0} \frac{100}{x}\left[\left(\frac{(x+2 \cos x)^{3}+(x+2)^{3}+2(x+2 \cos x)^{2}-2(x+2)^{2}+3 \sin (x+2 \cos x)-3 \sin (x+2)}{8+8+3 \sin ^{2}}\right)\right]}$

$=e^{\frac{100}{16+3 \sin ^{2}}} \lim _{x \rightarrow 0} \frac{3(x+2 \cos x)^{2} \times(1+2 \sin x)-3(x+2)^{2}-4(x+2 \cos x)}{x(1-2 \sin x)-4(x+2)+3 \cos (x+2 \cos x) \times(1-2 \sin x)-3 \cos (x+2)}$

$=e^{\frac{100}{16+3 \sin 2}}\left(\frac{12-3(4)+8 \times 1-8+3 \cos 2-3 \cos 2}{1}\right)$

Using $L'H$ rule.

$=e^{o}=1$

constant terms should be zero

$a+\beta=0$

coeff of $x$ should be zero

$\alpha-\beta+\gamma=0$

$\operatorname{coeff}$ of $x^{2}$ should be zero

$\lim _{x \rightarrow 0} \frac{x^{3}\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)+x^{4}\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)}{x}=\frac{2}{3}$

$\frac{\alpha}{2}+\frac{\beta}{2}=0$

$\frac{\alpha}{6}-\frac{\beta}{6}-\frac{\gamma}{6}=2 / 3$

$\alpha=1, \beta=-1, \gamma=-2$

$=\lim _{\theta \rightarrow 0} \frac{-\tan \left(\pi \sin ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$

$=\lim _{\theta \rightarrow 0}-\left(\frac{\tan \left(\pi \sin ^{2} \theta\right)}{\pi \sin ^{2} \theta}\right)\left(\frac{2 \pi \sin ^{2} \theta}{\sin \left(2 \pi \sin ^{2} \theta\right)}\right) \times \frac{1}{2}$

$=\frac{-1}{2}$ 

$=\lim _{n \rightarrow a} \tan \left(\sum_{r=1}^{n} \tan ^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)\right)$

$=\tan \left(\lim _{n \rightarrow a} \sum_{r=1}^{n}\left[\tan ^{-1}(r+1)-\tan ^{-1}(r)\right]\right)$

$=\tan \left(\lim _{n \rightarrow \infty}\left(\tan ^{-1}(n+1)-\frac{\pi}{4}\right)\right)$

$=\tan \left(\frac{\pi}{4}\right)=1$

$\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$

$\Rightarrow \lim _{x \rightarrow a} \frac{f(a)-a f^{\prime}(x)}{1} \quad$ (Lopitals rule)

$=f(a)-a f^{\prime}(a)$

$=4-2 a$

$S=\sum_{n=1}^{9} \frac{2}{4\left(n^{2}+3 n+2\right)}=\frac{1}{2} \sum_{n=1}^{9}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$

$S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{44}$

$\lim _{x \rightarrow 2}\left(\frac{2 x f(2)-4 f^{\prime}(x)}{1}\right)$

$=\frac{4(4)-4}{1}=12$

$L=\lim _{h \rightarrow 0} \frac{4 \sinh }{3 h}$

$\Rightarrow L=\frac{4}{3}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{a\left(1+x+\frac{x^{2}}{2 !} \ldots\right)-b\left(1-\frac{x^{2}}{2 !}+\ldots\right)+c\left(1-x+\frac{x^{2}}{2 !}\right)}{\left(\frac{x \sin x}{x}\right) x}=2$

$a-b+c=0$  $.......(1)$

$a-c=0$  $............(2)$

$\frac{a+b+c}{2}=2$

$\Rightarrow a+b+c=4$

So $6 L+1=2$

and

$\begin{array}{c} r \leq[ r ]< r +1 \\ 2 r \leq[2 r ]<2 r +1 \end{array} $$ $$ \begin{array}{ccc} 3 r & \leq[3 r ] & <3 r +1 \\ \vdots & \vdots & \vdots \\ nr & \leq[ nr ] & < nr +1 \end{array}$

$r +2 r +\ldots+ nr \leq[ r ]+[2 r ]+\ldots+[ nr ]<( r +2 r +\ldots+ nr )+ n$

$\frac{\frac{n(n+1)}{2} \cdot r}{n^{2}} \leq \frac{[r]+[2 r]+\ldots . .+[n r]}{n^{2}}<\frac{\frac{n(n+1)}{2} r+n}{n^{2}}$

Now,

$\lim _{n \rightarrow \infty} \frac{n(n+1) \cdot r}{2 \cdot n^{2}}=\frac{r}{2}$

and

$\lim _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}+n}{n^{2}}=\frac{r}{2}$

So, by Sandwich Theorem, we can conclude that

$\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^{2}}=\frac{r}{2}$

$\text { So, } l=\exp \left(\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots .+\frac{1}{n}}{n}\right)$

Now,

$0 \leq 1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n} \leq 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}$

$\leq 2 \sqrt{n}-1$

So, $l=\exp (0)$ (from sandwich theorem)

$=1$

$=\lim _{x \rightarrow 0} \frac{\operatorname{ax}-\left(e^{4 x}-1\right)}{\operatorname{ax} \cdot 4 x} \quad$ Use $\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1$

Apply L'Hospital Rule

$=\lim _{x \rightarrow 0} \frac{a-4 e ^{4 x }}{8 ax } \quad\left(\frac{ a -4}{0}\right.$ form $)$

limit exists only when $a-4=0 \Rightarrow a=4$

$=\lim _{x \rightarrow 0} \frac{4-4 e^{4 x}}{32 x}$

$=\lim _{x \rightarrow 0} \frac{1-e^{4 x}}{8 x}$

$\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow 0} \frac{- e ^{4 x } \cdot 4}{8}=-\frac{1}{2} \Rightarrow b =-\frac{1}{2}$

$a-2 b=4-2\left(-\frac{1}{2}\right)$

$=5$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{1\left(1+\frac{2\left(x^{2}+b x+c\right)}{1 !}+\frac{2^{2}\left(x^{2}+b x+c\right)^{2}}{2 !}+\ldots\right)-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2\left(x^{2}+b x+1\right)^{2}}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2(x-\alpha)^{2}(x-\beta)^{2}}{(x-\beta)^{2}}$

$\Rightarrow 2(\beta-\alpha)^{2}=2\left(b^{2}-4 c\right)$

$\Rightarrow a>0$

Now, $\lim _{x \rightarrow \infty} \frac{\left(x^{2}-x+1-a^{2} x^{2}\right)}{\sqrt{x^{2}-x+1}+a x}=b$

$\Rightarrow \quad \lim _{x \rightarrow \infty} \frac{\left(1-a^{2}\right) x^{2}-x+1}{\sqrt{x^{2}-x+1}+a x}=b$

$\Rightarrow \lim _{x \rightarrow \infty} \frac{\left(1-a^{2}\right) x^{2}-x+1}{x\left(\sqrt{\left.1-\frac{1}{x}+\frac{1}{x^{2}}+a\right)}=b\right.}$

Now, $\lim _{x \rightarrow \infty} \frac{-x+1}{x\left(\sqrt{\left.1-\frac{1}{x}+\frac{1}{x^{2}}+a\right)}\right.}=b$

$\Rightarrow \frac{-1}{1+a}=b \Rightarrow b=-\frac{1}{2}$

$(a, b)=\left(1,-\frac{1}{2}\right)$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi \cos ^{4} x\right)}{2 x^{4}}$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi-2 \pi \cos ^{4} x\right)}{\left[2 \pi\left(1-\cos ^{4} x\right)\right]^{2}} 4 \pi^{2} \cdot \frac{\sin ^{4} x}{2 x^{4}}\left(1+\cos ^{2} x\right)^{2}$

$=\frac{1}{2} \cdot 4 \pi^{2} \cdot \frac{1}{2}(2)^{2}=4 \pi^{2}$

Using L Hopital rule

$\alpha=\lim _{x \rightarrow \frac{\pi}{4}} \frac{3 \tan ^{2} x \sec ^{2} x-\sec ^{2} x}{-\sin \left(x+\frac{\pi}{4}\right)}$

$\Rightarrow \alpha=-4$

$\beta=\lim _{x \rightarrow 0}(\cos x)^{\cot x}=e^{\lim _{x \rightarrow 0} \frac{(\cos x-1)}{\tan x}}$

$\beta=e^{\lim _{x \rightarrow 0} \frac{-(1-\cos x)}{x^{2}} \frac{x^{2}}{\left(\frac{\tan x}{x}\right)^{x}}}$

$\beta=e^{\lim _{x \rightarrow 0}\left(\frac{-1}{2}\right)^{\frac{x}{1}}}=e^{0} \Rightarrow \beta=1$

$\alpha=-4 ; \beta=1$

If $a x^{2}+b x-4=0$ are the roots then $16 a-4 b-4=0\, \&\, a+b-4=0$

$\Rightarrow \mathrm{a}=1 \,\&\, \mathrm{~b}=3$

$\lim _{x \rightarrow 1} \frac{x^{n} f(1)-f(x)}{x-1}=44$

$\lim _{x \rightarrow 1} \frac{9 x^{n}-\left(x^{6}+2 x^{4}+x^{3}+2 x+3\right)}{x-1}=44$

$\lim _{x \rightarrow 1} \frac{9 n x^{n-1}-\left(6 x^{5}+8 x^{3}+3 x^{2}+2\right)}{1}=44$

$\Rightarrow 9 n-(19)=44$

$\Rightarrow 9 n=63$

$\Rightarrow n=7$

form $1^{\infty}$

$e^{\lim _{x \rightarrow 0}}\left(\frac{1-\cos x \sqrt{\cos 2} x}{x^{2}}\right) \times(x+2)$

Now limt ${x \rightarrow 0} \frac{\lim _{x \rightarrow 0} \frac{1-\cos x \sqrt{\cos 2} x}{}}{x^{2}}$

$\lim _{x \rightarrow 0} \frac{\sin x \sqrt{\cos 2} x-\cos x \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)}{x^{2}}$

(by L' Hospital Rule)

$\lim _{x \rightarrow 0} \frac{\sin x \cos 2 x+\sin 2 x \cdot \cos x}{2 x}$

$=\frac{1}{2}+1=\frac{3}{2}$

So, $e^{\operatorname{limit_x \rightarrow 0}\left(\frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}\right)(x+2)}$

$=e^{\frac{3}{2} \times 2}=e^{3}$

$\Rightarrow a=3$

$\lim _{x \rightarrow 0} \frac{x(\alpha-\beta)+x^{2}\left(\alpha+\frac{\beta}{2}+\gamma\right)+x^{3}\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)}{x^{3}}$

For limit to exist

$\alpha-\beta=0, \alpha+\frac{\beta}{2}+\gamma=0$

$\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 \ldots \text { (i) }$

$\beta=\alpha, \gamma=-3 \frac{\alpha}{2}$

Put in $(i)$

$\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10$

$\frac{\alpha}{6}+\frac{3 \alpha}{2}=10 \Rightarrow \frac{\alpha+9 \alpha}{6}=10$

$\Rightarrow \alpha=6$

$\alpha=6, \beta=6, \gamma=-9$

$\alpha+\beta+\gamma=3$

$\lim _{x \rightarrow 0} \frac{x\left\{(1-\sin x)^{1 / 8}+(1+\sin x)^{1 / 8}\right\}\left\{(1-\sin x)^{1 / 4}+(1+\sin x)^{1 / 4}\right\}\left\{(1-\sin x)^{1 / 2}+(1+\sin x)^{1 / 2}\right\}}{(1-\sin x-1-\sin x)}$

$\lim _{x \rightarrow 0} \frac{8 x}{-2 \sin x}=-4$

Divide by $3^{2 r }$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{\left(\frac{2}{3}\right)^{2 r} \cdot 2+3}\right)$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{3\left(\left(\frac{2}{3}\right)^{2 r+1}+1\right)}\right)$

Let $\left(\frac{2}{3}\right)^{r}=t$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3} t^{2}}\right)$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{t-\frac{2 t}{3}}{1+t \cdot \frac{2 t}{3}}\right)$

$\sum_{ r =1}^{ k }\left(\tan ^{-1}( t )-\tan ^{-1}\left(\frac{2 t }{3}\right)\right)$

$\sum_{ r =1}^{ k }\left(\tan ^{-1}\left(\frac{2}{3}\right)^{ r }-\tan ^{-1}\left(\frac{2}{3}\right)^{ r +1}\right)$

$S _{ k }=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{ k +1}$

$S _{\infty}=\lim _{ k \rightarrow \infty}\left(\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{ k +1}\right)$

$=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}(0)$

$\therefore \quad S _{\infty}=\tan ^{-1}\left(\frac{2}{3}\right)=\cot ^{-1}\left(\frac{3}{2}\right)$

$\Rightarrow \quad \lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}+\frac{x^{2}-1}{x-1}+\ldots . . \frac{x^{n}-1}{x-1}\right)=820$

$\Rightarrow 1+2+\ldots .+n=820$

$\Rightarrow \quad n(n+1)=2 \times 820$

$\Rightarrow \quad n(n+1)=40 \times 41$

since $\mathrm{n} \in \mathrm{N},$ so $n=40$

$=\lim \limits_{x \rightarrow 0} \frac{1}{x}\left\{\tan \left(\frac{\pi}{4}+x\right)-1\right\}$

$=\lim \limits_{x \rightarrow 0}\left(\frac{1+\tan x-1+\tan x}{x(1-\tan x)}\right)$

$=e^{\lim \limits_{x \rightarrow 0} \frac{2 \tan x}{x(1-\tan x)}}$

$=e^{2}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(1-\cos \frac{x^{2}}{2}\right)\left(1-\cos \frac{x^{2}}{4}\right)}{4\left(\frac{x^{2}}{2}\right)^{2}}=\frac{1}{16\left(\frac{x^{2}}{4}\right)^{2}}=\frac{1}{8} \times \frac{1}{32}=2^{-k}$

$\Rightarrow 2^{-8}=2^{-k} \Rightarrow k=8$

$\mathrm{k}=3+2=5$

$=\lim _{x \rightarrow 2} \frac{\left(3^{x}-9\right)\left(3^{x}-3\right)}{\left(3^{x / 2}-3\right)}$

$=\lim _{x \rightarrow 2} \frac{\left(3^{x / 2}+3\right)\left(3^{x / 2}-3\right)\left(3^{x}-3\right)}{\left(3^{x / 2}-3\right)}$

$=36$

$=e^{\lim _{x \rightarrow 0}\left(\frac{-4}{7 x^{2}+2}\right)}=\frac{1}{e^{2}}$

$f(\mathrm{x})=\left[\mathrm{x}^{2}\right] \sin (\pi \mathrm{x})$ will be discontinuous at nonintegers

$\therefore \mathrm{x}=\sqrt{\mathrm{A}+1}$ i.e. $\sqrt{5}$

$ L =\lim _{h \rightarrow 0} \frac{(a+2(a+h))^{1 / 3}-(3(a+h))^{1 / 3}}{(3 a+a+h)^{1 / 3}-(4(a+h))^{1 / 3}} $

$=\lim _{h \rightarrow 0} \frac{(3 a)^{1 / 3}\left(1+\frac{2 h}{3 a}\right)^{1 / 3}-(3 a)^{1 / 3}\left(1+\frac{h}{a}\right)^{1 / 3}}{(4 a)^{1 / 3}\left(1+\frac{h}{4 a}\right)^{1 / 3}-(4 a)^{1 / 3}\left(1+\frac{h}{a}\right)^{1 / 3}} $

$=\lim _{h \rightarrow 0}\left(\frac{3^{1 / 3}}{4^{1 / 3}}\right)\left[\frac{\left(1+\frac{2 h}{9 a}\right)-\left(1+\frac{h}{3 a}\right)}{\left(1+\frac{h}{12 a}\right)-\left(1+\frac{h}{3 a}\right)}\right]$

$=\left(\frac{3}{4}\right)^{1 / 3} \frac{\left(\frac{2}{9}-\frac{1}{3}\right)}{\left(\frac{1}{12}-\frac{1}{3}\right)}=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{8-12}{3-12}\right)$

$=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{-4}{-9}\right)=\frac{4^{1-\frac{1}{3}}}{3^{2-\frac{1}{3}}}=\frac{4^{2 / 3}}{3^{5 / 3}}$

$=\frac{(8 \times 2)^{1 / 3}}{(27 \times 9)^{1 / 3}}=\frac{2}{3}\left(\frac{2}{9}\right)^{1 / 3}$

$\operatorname{RHL}: \lim _{x \rightarrow 0^{+}}\left|\frac{1-x+x}{\lambda-x+1}\right|=\left|\frac{1}{\lambda}\right|$

For existence of limitt

$LHL = RHD$

$\Rightarrow \frac{1}{|\lambda-1|}=\frac{1}{|\lambda|} \Rightarrow \lambda=\frac{1}{2}$

$\therefore \quad L=\frac{1}{|\lambda|}=2$

$\because \lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}+x^{4}}-1}{x}\left(\frac{0}{0}\right.$ from $)$

$\lim _{x \rightarrow 0} \frac{\left(1+x^{2}+x^{4}\right)-1}{x\left(\sqrt{1+x^{2}+x^{4}}+1\right.}$

$\lim _{x \rightarrow 0} \frac{x\left(1+x^{2}\right)}{\left(\sqrt{1+x^{2}+x^{4}}+1\right)}=0$

$\lim _{x \rightarrow 0} \frac{e^{\frac{\sqrt{1+x^{2}+x^{4}-1}}{x}-1}}{\left(\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}\right)}=1$

roots are $2 -1$

$\Rightarrow \lim _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos \left(x^{2}-x-2\right)}}{(x-2)}$

$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2 \sin ^{2} \frac{\left(x^{2}-x-2\right)}{2}}}{(x-2)}$

$=\lim _{x \rightarrow 2^{+}} \frac{\sqrt{2} \sin \left(\frac{(x-2)(x+1)}{2}\right)}{(x-2)}$

$=\frac{3}{\sqrt{2}}$

So, $\sqrt {1 + {y^4}}  = 1 + \frac{{{y^4}}}{2}$

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {2 + \frac{{{y^4}}}{2}}  - \sqrt 2 }}{{{y^4}}}$

$ = \frac{{\sqrt 2 \left( {1 + \frac{{{y^4}}}{8} - 1} \right)}}{{{y^4}}} = \frac{{\sqrt 2 }}{8} = \frac{1}{{4\sqrt 2 }}$

$x \to {0^ - }$

$\left. \begin{array}{l}
\left[ x \right] =  - 1\\
\left| x \right| =  - x
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \frac{{x\left( { - x - 1} \right)\sin \left( { - 1} \right)}}{{ - x}} =  - \sin 1$

$\mathop {\lim }\limits_{x \to 1 + } \frac{{\left( {1 - x + \sin \left( {1 - x} \right)} \right)\sin \left( { - \frac{\pi }{2}} \right)}}{{\left( {x - 1} \right)\left( { - 1} \right)}}$

$\mathop {\lim }\limits_{x \to 1 + } \frac{{ - \left( {x - 1} \right)\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} =  - 1 + 1 = 0$

$1\left( \pi  \right).{\left( 1 \right)^2} + \mathop {Lt}\limits_{x \to 0} {\left( {1 - \frac{{\sin x\left[ x \right]}}{{x\left[ x \right]}}.\frac{{x\left[ x \right]}}{{\left| x \right|}}} \right)^2}\,\,\,\,\,\,\,\,......\left( i \right)$

$\mathop {Lt}\limits_{x \to {0^ - }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{x}{{ - x}}\left( { - 1} \right) = 1$

$\mathop {Lt}\limits_{x \to {0^ + }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{{x\left( 0 \right)}}{x} = 0$

Put in equation $(i)$

$\therefore $ Limit does not exist.

$ = \frac{{4x}}{{\sin 4x}}.\frac{{\cos 4x}}{{{{\cos }^2}2x}}{\cos ^2}x$

$ \Rightarrow 1\,\,\,\,\,as\,\,\,x \to 0$

$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{3{{\cot }^2}x\left( { - \cos \,e{c^2}x} \right) - {{\sec }^2}x}}{{ - \sin \left( {x + \frac{\pi }{4}} \right)}} = 8$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - x} \left( {\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt \pi  }}$

Assuming $x = \cos \theta $

$\mathop {\lim }\limits_{\theta  \to {0^ + }} \frac{{2\theta }}{{\sqrt 2 \sin \left( {\frac{\theta }{2}} \right)}}.\frac{1}{{2\sqrt \pi  }} = \sqrt {\frac{2}{\pi }} $

$ = \frac{{{{\left( 1 \right)}^2}.\left( {2\sqrt 2 } \right)}}{{\frac{1}{2}}} = 4\sqrt 2 $

$k = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{f\left( {3 + x} \right) - f\left( {2x} \right) - f\left( 3 \right)\left( {f\left( 2 \right)} \right)}}{{x\left( {1 + f\left( {2 - x} \right) - f\left( 2 \right)} \right)}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)}}{{\left( {1 + f\left( {2 - x} \right) - f\left( 2 \right)} \right) - xf'\left( {2 - x} \right)}}$

$ = 0$

$ \Rightarrow {e^k} = 1$

$\frac{0}{0}$ from, so we use $L-$ Hopital Rule

$ = \mathop {\lim }\limits_{x \to 2} \frac{{f'\left( x \right).2f\left( x \right)}}{1}$

$ = f'\left( 2 \right).2f\left( 2 \right)$

$ = 12f'\left( 2 \right)$

$\mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \frac{{{k^2} + {k^2} + {k^2}}}{{2k}}.........\left( 2 \right)$

$(1)=(2)$

$ \Rightarrow k = \frac{8}{3}$

$ = \int\limits_0^1 {{{\left( {1 + x} \right)}^{1/3}}} dx = \frac{3}{4}\left( {{2^{4/3}} - 1} \right)$

$1 - a + b = 0\,\,\,\,\,\,\,.......\left( i \right)$

$2 - a = 5\,\,\,\,\,\,\,.....\left( {ii} \right)$

$ \Rightarrow a + b =  - 7$

$f\left( x \right)$ attains maximum value when $\left| {x - 2} \right| = 0 \Rightarrow x = 2 = \alpha $

$g\left( x \right) = \left| {x + 1} \right|$

$g\left( x \right)$ attins minimum value of $x =  - 1 = \beta $

$\mathop {\lim }\limits_{x \to  - \alpha \beta } \frac{{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)}}{{{x^2} - 6x + 8}}$

$ = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}}$

$ = \frac{{\left( {2 - 1} \right)\left( {2 - 3} \right)}}{{\left( {2 - 4} \right)}} = \frac{1}{2}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{{x^2} + 2\sin x + 1 - {{\sin }^2}x - x + 1}}$ $\left( {\sqrt {{x^2} + 2\sin x + 1}  + \sqrt {{{\sin }^2}x - x + 1} } \right)$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{x + 2\sin x}}{{{x^2} + 2\sin x - {{\sin }^2}x + x}}.\left( 2 \right)$

Applying $L'H$ Rule

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{2.\left( {1 + 2\cos x} \right)}}{{2x + 2\cos x - 2\sin x\cos x + 1}}$

$ = \frac{{2\left( 3 \right)}}{{2 + 1}} = 2$

Hence the correct answer is option $(A)$.

$ = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + .... + 15}}{x}} \right) - \left( {\left\{ {\frac{1}{x}} \right\} + \left\{ {\frac{2}{x}} \right\} + .. + \left\{ {\frac{{15}}{x}} \right\}} \right)$

$\because $ $0 \le \left\{ {\frac{r}{x}} \right\} < 1\,\,\,\,\,\,\,\,\, \Rightarrow 0 \le x\left\{ {\frac{r}{x}} \right\} < x$

$\therefore \mathop {\lim }\limits_{x \to {0^ + }} x\left( {\frac{{1 + 2 + 3 + .... + 15}}{x}} \right) = \frac{{15 \times 16}}{2} = 120$

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\left( {x\tan 2x - 2x\tan \,x} \right)}}{{{{\left( {1 - \cos \,2x} \right)}^2}}} = \mathop {\lim }\limits_{x \to 0} K$ (say)

$ \Rightarrow K = \frac{{x\left[ {\frac{{2\tan \,x}}{{1 - {{\left( {\tan \,x} \right)}^2}}}} \right] - 2x\,\tan x}}{{{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$

$ = \frac{{2x\,\tan x - \left[ {2x\,\tan x - 2x\,{{\tan }^3}x} \right]}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {1 - {{\tan }^2}x} \right)}}$

$ = \frac{{2x\,{{\tan }^3}x}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ = \frac{{2x\,\frac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{4{{\sin }^4}\,x \times \left( {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} \right)}}$

$ \Rightarrow K = \frac{x}{{2\sin \,x \times \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\cos x}}$

$L = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2\sin \,x}} \times \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos 0\left( {{{\cos }^2}0 - {{\sin }^2}0} \right)}}$

$ = \frac{1}{2}$

Here $'L'$ is in the indeterminate from i.e.,$\frac{0}{0}$

$\therefore $ usinh the $L'$ Hosoital rule we get:

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3}{{\left( {27 + x} \right)}^{\frac{{ - 2}}{3}}}}}{{ - \frac{2}{3}{{\left( {27 + x} \right)}^{\frac{{ - 1}}{3}}}}} = \frac{{\frac{1}{3} \times {{\left( {27} \right)}^{\frac{{ - 2}}{3}}}}}{{\frac{{ - 2}}{3} \times {{\left( {27} \right)}^{\frac{{ - 1}}{3}}}}} =  - \frac{1}{6}$

Put $\frac{\pi }{2} - x = t \Rightarrow $ as $x \to \frac{\pi }{2} \Rightarrow t \to 0$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\cot \left( {\frac{\pi }{2} - t} \right)\left( {1 - \sin \left( {\frac{\pi }{2} - t} \right)} \right)}}{{8{t^3}}}$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t\left( {1 - \cos \,t} \right)}}{{8{t^3}}}$

$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t}}{{8t}}.\frac{{1 - \cos t}}{{{t^2}}}$

$ = \frac{1}{8}.1.\frac{1}{2} = \frac{1}{{16}}$

Rationalise 

$ \Rightarrow A = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{\left( {3x - 9} \right) \times \left( {2x - 4 + \sqrt 2 } \right)}}{{\left\{ {\left( {2x - 4 - 2} \right)} \right\} \times \left( {\sqrt {3x}  + 3} \right)}}$

        $ = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{3\left( {x - 3} \right)}}{{2\left( {x - 3} \right)}} \times \frac{{\sqrt {2x - 4}  + \sqrt 2 }}{{\left( {\sqrt {3x}  + 3} \right)}}$

         $ = \frac{3}{2} \times \frac{{2\sqrt 2 }}{6} = \frac{1}{{\sqrt 2 }}$

$\,\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\frac{1}{n}} \right)}^a} + {{\left( {\frac{2}{n}} \right)}^a} + ..... + {{\left( {\frac{n}{n}} \right)}^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {{n^a} + \frac{{n\left( {n + 1} \right)}}{2}} \right]}}.$

$ = \,\frac{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{{{\left( {1 + \frac{1}{n}} \right)}^{a - 1}}\left[ {a + \frac{1}{2}\left( {1 + \frac{1}{n}} \right)} \right]}}\, = \frac{1}{{60}}$

$ = \frac{{\int\limits_0^1 {{x^n}dx} }}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} = \frac{{\frac{1}{{a + 1}}}}{{a + \frac{1}{2}}} = \frac{1}{{60}}$

$ \Rightarrow \frac{{\frac{1}{{a + 1}}}}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$

$ \Rightarrow 2{a^2} + 3a - 119 = 0$

$ \Rightarrow 2{a^2} + 17a - 14a - 119 = 0$

$ \Rightarrow \left( {a - 7} \right)\left( {2a + 17} \right) = 0$

                                               $a = 7, - \frac{{17}}{2}$

$\log p = \frac{1}{2}$

$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$

        $\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$

$\therefore 2a = 3 \Rightarrow a = 3/2$

$\, = \,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{{\sin }^2}x} \right)}^2}}}{{2x\left( {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ....} \right) - x\left( {2x + \frac{{{2^3}{x^3}}}{3} + \frac{{{2^5}{x^5}}}{{15}} + ....} \right)}}$

             $\, = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{{x^4}\left( {\frac{2}{3} - \frac{8}{5}} \right) + {x^6}\left( {\frac{4}{{15}} - \frac{{64}}{{15}}} \right)}}$

             $ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\left( {1 + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - ....} \right)}^4}}}{{ - 2 + {x^2}\left( { - \frac{{60}}{{15}}} \right) + ......}}$

        (dividing numerator & denominator by ${{x^4}}$)

          $=2$

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{\sin x}}{e^{{x^2}}} + \frac{1}{2}} \right)\frac{1}{{\cos x}} = 1 + \frac{1}{2} = \frac{3}{2}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi \left( {1 - {{\sin }^2}x} \right)} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{{x^2}}}$

$ = \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}} \cdot \frac{{\pi {{\sin }^2}x}}{{{x^2}}}$

We know,  $\mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {f\left( x \right)} \right)}}{{f\left( x \right)}} = 1$

So, our limits becomes,

$ = 1 \cdot \pi \left( 1 \right) = \pi $

$\therefore \mathop {lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = \pi $

$\,\mathop {\lim }\limits_{x \to 0} f\left( {\frac{{1 - \cos 3x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{1 - \cos 3x}}} \right)$

$\, = \,\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}}}{{2{{\sin }^2}\frac{{3x}}{2}}}} \right)$

$\, = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\frac{9}{4}.{x^2}.\frac{4}{9}}}{{{{\sin }^2}\frac{{3x}}{2}}}} \right)$

$ = \frac{4}{{9 \times 2}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\frac{{{{\sin }^2}\frac{{3x}}{2}}}{{{{\left( {\frac{{3x}}{2}} \right)}^2}}}}}} \right)$

$ = \frac{2}{9}\frac{{\mathop {\lim }\limits_{x \to 0} }}{{\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{{3x}}{2}}}{{{{\left( {\frac{{3x}}{2}} \right)}^2}}}}}$

$\left\{ {\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right\}$

$ = \frac{2}{9}\left[ {\frac{1}{1}} \right] = \frac{2}{9}$

$\, \Rightarrow \,\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {{x^2} + kx - 2x - 2k} \right\}}}{{{{\left( {x - 2} \right)}^2}}} = 5$

$\, \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {x\left( {x - 2} \right) + k\left( {x - 2} \right)} \right\}}}{{\left( {x - 2} \right) \times \left( {x - 2} \right)}} = 5$

$\, \Rightarrow \mathop {\lim }\limits_{x \to 2} \left( {\frac{{\tan \left( {x - 2} \right)}}{{\left( {x - 2} \right)}}} \right) \times \mathop {\lim }\limits_{x \to 2} \left( {\frac{{\left( {k + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)}}} \right) = 5$

$ \Rightarrow 1 \times \mathop {\lim }\limits_{x \to 2} \left( {k + x} \right) = 5$

{$\because $ $\mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h}}{h} = 1$}

or $k + 2 = 5$

$ \Rightarrow \boxed{k = 3}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x\left( {3 + \cos x} \right)}}{{x \times \frac{{\tan 4x}}{{4x}} \times 4x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{{x^2}}} \times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {3 + \cos x} \right)}}{4} \times \frac{1}{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan 4x}}{{4x}}}}$

$ = 2 \times \frac{4}{4} \times 1$              ($\because $ $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\sin \theta }}{\theta } = 1$ and $\mathop {\lim }\limits_{\theta  \to 0} \frac{{\tan \theta }}{\theta } = 1$)

$=2$

$\, = \,\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}} \right)\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{2x + 1}}} \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right]$

$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}} \right).{\tan ^{ - 1}}\left( {\frac{{\frac{{x + 1}}{{2x + 1}} - 1}}{{1 + \frac{{x + 1}}{{2x + 1}}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{x}.{\tan ^{ - 1}}\left( {\frac{{ - x}}{{3x + 2}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{{\tan }^{ - 1}}\left( {\frac{x}{{3x + 2}}} \right)}}{{\frac{x}{{3x + 2}}}} \times \frac{1}{{3x + 2}}} \right] =  - \frac{1}{2}$

$\lim _{x \rightarrow \infty}\left(\frac{\log _0 x}{\log _0(x+1)}\right)^\beta \cdot \frac{\left(\log _0 x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0$

$\lim _{x \rightarrow \infty} \frac{\left(\log _8 x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t$

$\lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0$

As we know $\lim _{ x \rightarrow \infty} \frac{ x }{ e ^{ x }}=0$

$\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

$\because \int_{x^2}^x \sqrt{\frac{1-t}{t}} d t=\sin ^{-1} \sqrt{x}+\sqrt{x} \sqrt{1-x}-\sin ^{-1} x-x \sqrt{1-x^2}$

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} \sqrt{x}+\sqrt{x} \sqrt{1-x}-\sin ^{-1} x-x \sqrt{1-x^2}}{\sqrt{x}} \leq f(x) g(x) \leq \frac{2 \sqrt{x}}{\sqrt{x}}\right)$

$\Rightarrow 2 \leq \lim _{x \rightarrow 0} f(x) g(x) \leq 2$

$\Rightarrow \lim _{x \rightarrow 0} f(x) g(x)=2$

use expansion

$\beta=\lim _{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-\frac{x^3}{3}\right)}{x^3}+\lim _{x \rightarrow 0} \frac{\left(\left(1-\frac{x^2}{2}\right)-1\right)}{x^2} \frac{\sin x}{x}$

$\beta=\lim _{x \rightarrow 0} \frac{4 x^3}{3 x^3}+\lim _{x \rightarrow 0} \frac{-x^2}{2 x^2}$

$\beta=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$

$6 \beta=5$

$\therefore$ Using Lopital rule,

$=2 \lim _{x \rightarrow a^{+}} \frac{\left(\frac{1}{\sqrt{x}-\sqrt{\alpha}}\right) \cdot \frac{1}{2 \sqrt{x}}}{\left(\frac{1}{e^{\sqrt{x}}-e^{\sqrt{a}}}\right) \cdot e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}}$

$=\frac{2}{e^{\sqrt{a}}} \lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}}-e^{\sqrt{a}}\right)}{(\sqrt{x}-\sqrt{\alpha})}\left(\frac{0}{0}\right)$

$=\frac{2}{e^{\sqrt{a}}} \lim _{x \rightarrow a^{+}} \frac{\left(e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}-0\right)}{\left(\frac{1}{2 \sqrt{x}}-0\right)}=2$

$\lim _{x \rightarrow a^{+}} f(g(x))=\lim _{x \rightarrow a^{+}} f(2)$

$=f(2)=\sin \frac{\pi}{6}=\frac{1}{2}$

$=0.50$

Now $\int_0^{\pi / 3} g(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \cos 3 x d x=0$

Hence $g ( x )=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(B)$ Let $h ( x )=f( x )-3 \sin 3 x +\frac{6}{\pi}$

Now $\int_0^{\pi / 3} h(x) d x=\int_0^{\pi / 3} f(x) d x-3 \int_0^{\pi / 3} \sin 3 x d x+\int_0^{\pi / 3} \frac{6}{\pi} d x$ $=0-2+2=0$

Hence $h(x)=0$ has a root in $\left(0, \frac{\pi}{3}\right)$

$(C)$ $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{1- e ^{x^2}}=\lim _{x \rightarrow 0} \underbrace{\left.\frac{x^2}{1-e^{x^2}}\right)}_{-1} \underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=-1 \lim _{x \rightarrow 0} \frac{f( x )}{1}=-1$

$(D)$ $\lim _{ x \rightarrow 0} \frac{(\sin x ) \int_0 f( t ) dt }{ x ^2}$

$=\lim _{x \rightarrow 0} \underbrace{\left(\frac{\sin x}{x}\right)}_1\underbrace{\frac{\int_0^x f(t) d t}{x}}_{\text {Apply }}$

$=1 \lim _{x \rightarrow 0} \frac{f(x)}{1}=1$

Ans. $A,B,C$

$\frac{e^{\left(\frac{\ln (1-x)}{x}\right)}-\frac{1}{e}}{x^2}$

$=\lim _{x-0^{+}} \frac{1}{e} \frac{e^{\left(1+\frac{(n(1-x)}{x}\right)}-1}{x^{ a }}$

$=\frac{1}{e} \lim _{ x -0^{+}} \frac{1+\frac{\ell \operatorname{n}(1-x)}{ x }}{ x ^{ a }}$

$=\frac{1}{e} \lim _{x-0^{+}} \frac{\ln (1-x)+x}{x^{(a+1)}}$

$=\frac{1}{e} \lim _{x \rightarrow 0^{+}} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots \ldots . .\right)+x}{x^{a+1}}$

Thus, $a=1$

$=\frac{1}{ n } \sum_{ r =1}^{ n } \log \left(1+\frac{ r }{ n }\right)$

$\lim _{ n \rightarrow \infty} \ell n y _{ n }=\lim _{ n \rightarrow \infty} \frac{1}{n}$ $\sum_{ r =1}^{ n } \log \left(1+\frac{ r }{ n }\right)$

$\ell n\left(\lim _{n \rightarrow \infty} y_n\right)=\int_0^1 \log (1+x) d x$

$\ell n ( L )=[\log (1+ x ) \cdot x ]_0^1-\int_0^1 \frac{ x }{1+ x } dx$

$=\log 2-\left[\left. x \right|_0 ^1-\ell n \mid 1+ x \|_0^1\right]$

$=\log 2-1+[\ell n 2-0]$

$=\ell n \left(\frac{4}{ e }\right)$

$L =\frac{4}{ e }$

${[ L ]=1}$

$\Rightarrow \lim _{t \rightarrow x} \frac{f(x) \cos (t)-f^{\prime}(t) \sin (x)}{1}=\sin ^2(x)$

$\Rightarrow f(x) \cos x-f^{\prime}(x) \sin (x)=\sin ^2(x)$

$\Rightarrow \frac{d}{d x}\left(\frac{f(x)}{\sin x}\right)=-1$

$\Rightarrow f(x)=-x \sin (x) \quad(\because C=0)$

For $f(x)=-x \sin (x)$, option $(B), (C), (D)$ are correct.

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} \frac{-h(2+h)}{h} \cos \frac{1}{h} \text { does not exist }$

$\Rightarrow g(x)=\left.f(t) \operatorname{cosec}(t)\right|_x ^{\pi / 2}=3-\frac{f(x)}{\sin x}$

$\Rightarrow \lim _{x \rightarrow 0} g(x)=\lim _{x \rightarrow 0}\left(3-\frac{f(x)}{\sin x}\right)$

$=3-f^{\prime}(0)=2$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta x-\frac{\beta^3 x^3}{3!}+\ldots\right)}{a x-\left(x-\frac{x^3}{3!}+\ldots\right)}=1$

$\lim _{x \rightarrow 0} \frac{x^3\left(\beta-\frac{\beta^3 x^2}{3!}+\ldots\right)}{(\alpha-1) x+\frac{x^3}{3!}-\ldots}=1$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta-\frac{\beta^3 x^2}{3!}+\ldots\right)}{(\alpha-1)+\frac{x^2}{3!}-\ldots}=1$

$\Rightarrow a-1=0 \Rightarrow a=1$

$\text { At } a=1$

$\lim _{x \rightarrow 0} \frac{x^2\left(\beta-\frac{\beta^3 x^2}{3!}+\cdots\right)}{\frac{x^2}{3!} \ldots}=1$

$\Rightarrow \beta \times 3!=1 \Rightarrow \beta=\frac{1}{6}$

$\therefore 6(\alpha+\beta)=6\left(1+\frac{1}{6}\right)=7$

$f(x)=\lim _{n \rightarrow \infty}\left\{\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \ldots\left(x+\frac{n}{2}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) \ldots\left(x^2+\frac{n^2}{n^2}\right)}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \log \left\{\prod_{r=1}^n \frac{\left(x+\frac{n}{r}\right)}{\left(x^2+\frac{n^2}{r^2}\right)} \cdot \frac{n}{r}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{2} \sum_{r=1}^n \log \left\{\left(\frac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}}\right) \cdot \frac{1}{\frac{r}{n}}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\}$

$\Rightarrow \log f(x)=x \lim _{n \rightarrow \infty} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\} \frac{1}{n}$

$\Rightarrow \log f(x)=x \int_0^1 \log \left\{\frac{1+t x}{1+(t x)^2}\right\} d t$

$\Rightarrow \log f(x)=\int_0^x \log \left(\frac{1+u}{1+u^2}\right) d u \text {, where } u=t tx$

$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=\log \left(\frac{1+x}{1+x^2}\right)$

We observe that $f(x)>0$ for all $a>0$.

Also,

$\log \left(\frac{1+x}{1+x^2}\right) > 0 \Leftrightarrow \frac{1+x}{1+x^2} > 1 \Leftrightarrow x>x^2 \Leftrightarrow 0$

$ < x < 1$

$\therefore \frac{f^{\prime}(x)}{f(x)}>0 \text { for } 0

$ < 0 \text { for } x > 1$

$\Rightarrow f(x)$ is increasing on $(0,1)$ and decreasing on $(1, \infty)$.

$\Rightarrow f\left(\frac{1}{2}\right) \leq f(1) \text { and } f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$

So option$(b)$ is correct and option $(a)$ is incorrect.

From$(ii)$, we obtain

$f^{\prime}(2) < 0$. So,option $(c)$ is correct.

From$(i)$, we obtain

$\frac{f^{\prime}(3)}{f(3)}-\frac{f^{\prime}(2)}{f(2)}=\frac{\log (4)}{10}-\frac{\log (3)}{5}=\log \left(\frac{2}{3}\right) < 0$

$\Rightarrow \frac{f^{\prime}(3)}{f(3)} < \frac{f^{\prime}(2)}{f(2)} \text {. So, option (d) is not correct. }$

$f(x)=\sin \left(\frac{1}{3} g(g(x))\right)$

$\Rightarrow g(g(x)) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \forall x \in R$

Also, $g(g(g(x))) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \forall x \in R$

Hence, $f(x)$ and $f(g(x)) \in\left[-\frac{1}{2}, \frac{1}{2}\right]$

$\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{\sin \left(\frac{1}{3} g(g(x))\right)}{\frac{1}{3} g(g(x))} \cdot \frac{\frac{1}{3} g(g(x))}{g(x)}$

$\lim _{\alpha \rightarrow 0} \frac{e\left(e^{\left(\cos (\alpha)^n-1\right)}-1\right)\left(\cos \alpha^n-1\right)}{\left(\cos \left(\alpha^n\right)-1\right) \alpha^m \alpha^{2 n}} \alpha^{2 n}=-\frac{e}{2} \text { if and only if } 2 n-m=0$

$f(-x)=-f(x)$

$\text { Given } f(1)=\frac{1}{2}$

$\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\lim _{x \rightarrow 1} \frac{\frac{F(x)-F(1)}{x-1}}{\frac{G(x)-G(1)}{x-1}}=\frac{f(1)}{|f(f(1))|}=\frac{1}{14}$

$\Rightarrow \frac{1 / 2}{|f(1 / 2)|}=\frac{1}{14}$

$\Rightarrow f\left(\frac{1}{2}\right)=7 .$