2 Marks Questions

Question 512 Marks

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ where $f(x) = \left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$

Answer

Here f(x) = $\left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$
Now $\mathop {\lim }\limits_{x \to 0} f(x)$ = $\mathop {\lim }\limits_{x \to 0} 2x + 3$ = 2 $\times$ 0 + 3 = 3
$\mathop {\lim }\limits_{x \to 1} f(x)$ = $\mathop {\lim }\limits_{x \to 1} 3(x + 1) $ = 3(1 + 1) = 3 $\times$ 2 = 6

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Question 522 Marks

Evaluate $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$

Answer

Here $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \frac{\pi }{2} + y$ as $x \to \frac{\pi }{2},\;y \to 0$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{\tan 2\left( {\frac{\pi }{2} + y} \right)}}{{\frac{\pi }{2} + y - \frac{\pi }{2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan (\pi + 2y)}}{y}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{{2y}} \times 2 = 1 \times 2 = 2$

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Question 532 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$

Answer

Here $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$
$= \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right)$
$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x/2}}{{2\sin x/2\cos x/2}} = \mathop {\lim }\limits_{x \to 0} \tan x/2 = 0$

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Question 542 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}};$ a, b, a + b $\neq$ 0.

Answer

We have,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{x} + \frac{{bx}}{x}}}{{\frac{{ax}}{x} + \frac{{\sin bx}}{x}}}$
[dividing both numerator and denominator by x]
$ = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin (ax)}}{{ax}} \times a + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + \mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}} \times b}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}{\text{ and }}} {\mathop {\lim }\limits_{x \to a} f(x) + g(x) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)} \right]$
$ = \frac{{a\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + b\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}}$
$= \frac { a ( 1 ) + b } { a + b ( 1 ) }$ $\left[ {\because \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]$
$= \frac { a + b } { a + b } = 1$

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Question 552 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} x\sec x$

Answer

Here $\mathop {\lim }\limits_{x \to 0} x\sec x$
$ = \mathop {\lim }\limits_{x \to 0} x \times \frac{1}{{\cos x}}\rightarrow\mathop {\lim }\limits_{x \to 0} \frac{x}{{\cos x}} = \frac{0}{1} = 0$

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Question 562 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}.$

Answer

We have, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ax}}{{b\sin x}} + \frac{{x\cos x}}{{b\sin x}}} \right)$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} + \frac{1}{b}\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x}}{{\sin x}}$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\mathop {\lim }\limits_{bx \to 0} \frac{{\cos x}}{{\left( {\frac{{\sin x}}{x}} \right)}}$
$ = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}}$$\left[\because {\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$
$ = \frac{a}{b} \times \frac{1}{1} + \frac{1}{b} \times \frac{1}{1}$$\left[\because {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
$= \frac { a + 1 } { b }$

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Question 572 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$

Answer

Here $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{1 - \cos x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{2{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{(2\sin x/2\cos x/2)}^2}}}{{{{\sin }^2}x/2}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}x/2{{\cos }^2}x/2}}{{{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} 4{\cos ^2}x/2$ = 4/2 = 2

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Question 582 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$

Answer

Here $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}} = \frac{1}{\pi }$

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Question 592 Marks

Evaluate $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}$

Answer

Let y= $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \pi + y$, as $x \to \pi ,\;y \to 0$
$\therefore y=\;\mathop {\lim }\limits_{y \to 0} \frac{{\sin [\pi - \pi - y]}}{{\pi [\pi - \pi - y]}}=\mathop {\lim }\limits_{y \to 0} \frac{{\sin ( - y)}}{{ - \pi y}}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{ - \sin y}}{{ - \pi y}} = \frac{1}{\pi }\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = \frac{1}{\pi } \times1 = \frac{1}{\pi }$

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Question 602 Marks

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},\;a,\;b \ne 0$

Answer

Here $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}}$$= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times ax \times \frac{1}{{\frac{{\sin bx}}{{bx}} \times bx}}} \right]$
$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times \frac{1}{{\frac{{sinbx}}{{bx}}}} \times \frac{{ax}}{{bx}}} \right] = \frac{a}{b}$$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}}\frac{1}{{\frac{{\sin bx}}{{bx}}}}} \right]$
$ = \frac{a}{b} \times 1 \times 1 = \frac{a}{b}$

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Question 612 Marks

Evaluate $\lim \limits_{x \rightarrow 0} \frac{\sin a x}{b x}$

Answer

Given, $\lim _{x \rightarrow 0} \frac{\sin a x}{b x}$
$$Applying the limits in the given expression we get,$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{0}{0}$
Multiplying and dividing the given expression by a we get,
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{b x} \times \frac{a}{a}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{a x} \times \frac{a}{b}$
We know that: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$= \frac{a}{b} \lim _{a x \rightarrow 0} \frac{\sin a x}{a x}=\frac{a}{b} \times 1=\frac{a}{b}$

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Question 622 Marks

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

Answer

Here $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{\frac{{x + 2}}{{2x}}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{2x}} \times \frac{1}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2x}} = \frac{1}{{2 \times - 2}} = \frac{{ - 1}}{4}$

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Question 632 Marks

Evaluate $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}},\;a + b + c \ne 0$

Answer

Here $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}}$
$ = \frac{{a \times {{(1)}^2} + b \times 1 + c}}{{c \times {{(1)}^2} + b \times 1 + a}} = \frac{{a + b + c}}{{c + b + a}} = 1$

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Question 642 Marks

Evaluate $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}$

Answer

Here $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}\left[ {\frac{0}{0}{\text{form}}} \right]$
$= \mathop {\lim }\limits_{z \to 1} \frac{{{{({z^{1/6}} - 1)}^2}}}{{{z^{1/6}} - 1}}$
$ = \mathop {\lim }\limits_{z \to 1} \frac{{({z^{1/6}} + 1)({z^{1/6}} - 1)}}{{({z^{1/6}} - 1)}} = \mathop {\lim }\limits_{z \to 1}$
$= z^{1/6} + 1$
$= (1)^{1/6} + 1 = 1 + 1 = 2$

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Maths 2 Marks Questions