MATHS M.C.Q (1 Marks)

2. $ \frac{1}{2}$

4. does not exist

Solution:

Using,

$ \lim_\limits{\text{x} \rightarrow 0}|\text{x}|=-\text{x}$

$ \lim_\limits{\text{x} \rightarrow 0}|\text{x}|=+\text{x}$

we get $\lim_\limits{\text{x} \rightarrow \text{a}}-\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$

$\lim_\limits{\text{x} \rightarrow \text{a}}+\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$

Since, LHL is not equal to RHL, hence the limit does not exist.

1. $\cos9$ 

Solution:

$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$

Differentiate both the sides with respect to x, we get

$\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x})\frac{\text{d}}{\text{dx}}\sin(\text{x}+9)-\sin(\text{x}+9)\frac{\text{d}}{\text{dx}}(\cos\text{x})}{\cos^2\text{x}}$ (Quotient rule)

$=\frac{(\cos\text{x})(\cos(\text{x}+9))-(\sin(\text{x}+9))(-\sin\text{x})}{\cos^2\text{x}}$

$=\frac{(\cos\text{x})(\cos(\text{x}+9))+(\sin(\text{x}+9))(\sin\text{x})}{\cos^2\text{x}}$

$=\frac{\cos(\text{x}+9-\text{x})}{\cos^2\text{x}}$

$=\frac{\cos9}{\cos^2\text{x}}$

Thus, $\frac{\text{dy}}{\text{dx}}$ at x = 0 is $\cos9$

1. 5050

Solution:

Given f(x) = x¹⁰⁰ + x⁹⁹ + .... + x + 1

f'(x) = 100.x¹⁰⁰ + 99.x⁹⁸ + .... + 1

S0, f'(1) = 100 + 99 + 98 + ..... + 1

$=\frac{100}{2}[2\times100+(100-1)(-1)]$

$=50[200-99]=50\times101=5050$

$=5050$

3. $\frac{1}{2}$

Solution:

Given $​​\lim\limits_{\text{x} \rightarrow 0}\frac{\text{cosec}\text{x}-\cot\text{x}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}{\text{x}}$ 

$ =\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}=\frac{2\sin^{2}\frac{\text{x}}{2}}{\text{x}\cdot\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow1}\frac{\sin\frac{\text{x}}{2}}{\text{x}\cos\frac{\text{x}}{2}} =\frac{\tan\frac{\text{x}}{2}}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\frac{\text{x}}{2}}{2\times\frac{\text{x}}{2}}$

$=\frac{1}{2}\times1=\frac{1}{2}$

1. a = 1, b = 2

Solution:

$ \mathop {\lim }\limits_{\text{x} \to 0} {\left( {\cos \text{x} + \text{a}\sin \text{bx}} \right)^{\frac{1}{\text{x}}}}$ so its limit will be e^k, where

$\text{k}=\lim\limits_{\text{x}\to0}\frac{1}{\text{x}}(\text{cos x}+\text{asinbx}-1)=\lim\limits_{\text{x}\to0}\frac{-\sin\text{x}+\text{abcosbx}}{1}=\text{ab}=2$

Hence all possible combination of aa and bb are possible whose product is 2

1. 0

Solution:

We know that,

$= \lim_\limits{\text{x}→0} \text{x} = 0$

And

$= -1 ≤ \sin \frac{1}{\text{x}} ≤ 1$

By Sandwich theorem,

$$ $=\lim_\limits{\text{x}→0} \text{x} \sin \frac{1}{\text{x}} =0$

3. $\frac{-1}{2\sqrt{2}}$

Solution:

This is of the form $\frac{0}{0}$, therefore we use L ’Hospital’ s rule and differentiate the numerator and

denominator.

$ =\lim_\limits{a \rightarrow b} \frac{\text{2sin}\text{x cos}+\cos\text{x}\sqrt{\text{2}}}{\text{2x}−4\text{x}}$

$= \frac{0+\sqrt{2}}{-4}$

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

2. $\text{3x}2\ \text{cos}⁡ \ \text{2} \cos⁡ \text{x3} – \text{2x} \sin⁡\ \text{x3 sin x2}$

Solution:

We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g}.)=\text{g}.\frac{\text{d}}{\text{dx}}(\text{f})+\text{f}.\frac{\text{dy}}{\text{dx}}(\text{g})$

Here $ \text{f} = \sin⁡ \text{x}3 \text{ and g} = \cos⁡\text{x}2$

$\frac{\text{d}}{\text{dx}} (\text{f}) = \text{3x2} \text{ cos⁡ x3}$

$\frac{\text{d}}{\text{dx}} (\text{g}) = -2\text{x}\ \sin \text{x}^2$

We now substitute this in our main equation,

$= \text{cos⁡ x2 .3x2 cos⁡ x3 + sin⁡ x3.(-2x sin x2)}$

$=\text{ 3x2 cos x2 cos⁡ x3 – 2x sin⁡ x3 sin x2}$

1. Yes

Solution:

Obviously, f(x) is continuous at [1, 2]

And, f(x) differentiable at [1, 2]

Also, f(1) = f(2) = 2

Now, f(x) = 0

⇒ 2x - 3 = 0

$⇒ \text{x} = \frac{3}{2}$

Thus, x belongs to [1, 2]

Hence, it is verified.

2. $0<∣\text{x−a}∣<δ⇒∣\text{f(x)}−\text{L}∣<ϵ$

Solution:

It is fundamental concept that,

for limit of a function f(x) to exist at any point aa there exists a real number δ>0,

such that 0 < |x - a|< δ,

for which |f(x) - L| < ϵ, where ϵ > 0

1. $ \displaystyle \lim_{\text{x} \rightarrow \text{a}}\text{f(x)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(x)}|=∞$'

Solution:

f : (a, b) → R is differentiable.

If $ \lim _\limits{ \text{x}\rightarrow \text{a} }{ \text{f}(\text{x} )}$

1. greater than

Solution:

Given, $ (1.1)^{10000 }= (1 + 0.1)^{10000}$

$ 10000\text{C}_0 + 10000\text{C}_1 × (0.1) + 10000\text{C}_2 × (0.1)² + \text{other} + \text{ve terms}$

= 1 + 10000 × (0.1) + other + ve terms

= 1 + 1000 + other + ve terms

> 1000

So, (1.1)¹⁰⁰⁰⁰ is greater than 1000

2. $ \frac{1}{2}$

Solution:

(1 + x)^m = 1 + mx + $ \frac{\text{m}(\text{m - 1)}}2\text{x}^2$ + ........

Now, $ \frac{\text{m}(\text{m - 1)}}2\text{x}^2$ = $ \frac{1}{8}\text{x}^2$

⇒ $ \frac{\text{m}(\text{m - 1)}}2\text{x}^2$ = $ \frac{-1}{8}\text{x}$

⇒ 4m² - 4m = -1

⇒ 4m² - 4m + 1 = 0

⇒ (2m - 1)² = 0

⇒ 2m - 1 = 0

⇒ m = $ \frac{1}{2}$

2. 2x cos x - x² sin x

Solution:

$\frac{ \text{d}}{\text{dx}(\text{x}2 \text{cos x})}$

Using the formula $ \frac{\text{d}}{\text{dx} [\text{f(x) g(x)}]} = \text{f}(\text{x}) \Big[\frac{\text{d}}{\text{dx} \text{g}(\text{x})}\Big] + \text{g(x)} \Big[\frac{\text{d}}{\text{dx} \text{f(x})}\Big]$

$= \frac{\text{d}}{\text{dx}(\text{x}^2 \cos \text{x})} = \text{x}^2 \Big[\frac{\text{d}}{\text{dx} (\cos \text{x})}\Big] + \cos x \Big[\frac{\text{d}}{\text{dx } \text{x}^2}\Big]$

$ = \text{x}^2(-\sin \text{x}) + \cos\text{x}(2\text{x})$

$ = 2\text{x} \cos \text{x} – \text{x}2 \sin \text{x}$

2. 1

Solution:

Substitute x = 1 + t

$ \text{L.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$

= 0 + 1 + 0 = 1

$ \text{R.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$

= 0 + 2 - 1 = 1

L.H.S = R.H.S

4. an irrational number

Solution:

Since n is a positive integer, assume n = 1

$(\sqrt{3}+1)³ + (\sqrt{3}−1)³$

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

$ = 3\sqrt{3} + 1 + 9 + 3\sqrt{3} + 3\sqrt{3}- 1 - 9 + 3\sqrt{3}$

= $12\sqrt{3}$, which is an irrational number.

2. $ \frac{(2\text{n})!}{(\text{n}!)^2}$

Solution:

We have,

(1 - 2x + 3x² - 4x³ + ........)^-n = {(1 + x)^-2}^-n

= (1 + x)²ⁿ

So, the coefficient of xⁿC₃ = ²ⁿC_n

= $ \frac{(2\text{n})!}{(\text{n}!)^2}$

4. $\text{x}^{2}-10\text{x}+21=0$

Solution:

Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ 

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$

$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$

$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$

$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]=7$

Therefore, the quadratic equation whose roots are 3 and 7 is $\text{x}^{2}-10\text{x}+21=0$

4. 50

Solution:

$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$

Differentiate both the sides with respect to x, we get 

$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100})$

$=\frac{\text{d}}{\text{dx}}(1)-\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x}^3)+\dots-\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\frac{\text{d}}{\text{dx}}(\text{x}^{100})$

$=0-1+\text{2x}-\text{3x}^2+\dots-\text{99x}^{98}+100\text{x}^{99}$

Putting x = 1, we get 

$\text{f}'(1)=-1+2-3+\dots-99+100$

$=(-1+2)+(-3+4)+(-5+6)+\dots+(-99+100)$

$=1+1+1+\dots+1(50$terms$)$

$=50$

2. 1

Solution:

Given: $\text{f}(\text{x})=\text{x}-[\text{x}],\text{x}\in\text{R}$

Now,

For $0\le\text{x}<1,[\text{x}]=0$

$\therefore\text{f}(\text{x})=\text{x}-0=\text{x},\forall\text{x}\in[0,1)$

Differentiate with respect to x, we get

$\text{f}'(\text{x})=1,\forall\text{x}\in[0,1)$

$\therefore\text{f}'\Big(\frac{1}{2}\Big)=1$

2. $ \frac{10}{(5!)^2}$

Solution:

The coefficient of x^r in the expansion of (1 + x)¹⁰ is ¹⁰C_r and ¹⁰C_r is maximum for
r = 10 = 5

Hence, the greatest coefficient = ¹⁰C₅

= $ \frac{10}{(5!)^2}$

2. -7160 x⁹ × y³

Solution:

4th term in (x - 2y)¹² = T₄

= T_{3 + 1}

= ¹²C₃ (x)^{12 - 3} × (-2y)³

= ¹²C₃ x⁹ × (-8y³)

$ = {\frac{(12 × 11 × 10)}{(3 × 2 × 1)} × \text{x}^9 ×(-8\text{y}^³)}$

= - (2 × 11 ×10 × 8) × x⁹ × y³

= -1760 x⁹ × y³

3. 6

Solution:

Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.

Now T₁ = ⁿC₀ × a^{n - 0} × b⁰ = 729

⇒ aⁿ = 729 ................ 1

T₂ = ⁿC₁ × a^{n - 1} × b¹ = 7290

⇒ n

a^{n - 1} × b = 7290 ....... 2

T₃ = ⁿC₂ × a^{n - 2} × b² = 30375

$ ⇒ \text{n}\frac{(\text{n}-1)}{2}$

a^{n - 2} × b² = 30375 ....... 3

Now equation 2, equation 1

n

$=\text{a}^{\text{n}-1}\times\text{b}=\frac{7290}{729}$

⇒ $\frac{\text{n}×\text{b}}{\text{n}} = 10 ....... 4$

Now equation 3, equation 2

$ ⇒ \text{n}\frac{(\text{n}-1)}{2}$

$ \text{an}-2 × \frac{\text{b}^2}{\text{n}}$

$=\text{a}^{\text{n}-1}\times\text{b}=\frac{30375}{7290}$

$ ⇒ \text{n}\frac{(\text{n}-1)}{\text{2a}}=​​\frac{30375\times2}{7290}$

$ ⇒ \text{n}\frac{(\text{n}-1)}{\text{a}}=​​\frac{30375\times2}{7290}$

$ ⇒ \text{n}\frac{(\text{n}-1)}{\text{a}}-\frac{\text{b}}{\text{a}} = \frac{60750}{7290}$ 

$ ⇒ 10 - \frac{\text{b}}{\text{a}} = \frac{60750}{729}$ $$(60750 and 7290 is divided by 10)

$ ⇒ 10 - \frac{\text{b}}{\text{a}} = \frac{25}{3}$ (6075 and 729 is divided by 243)

$\Rightarrow 10 -\frac{25}{3} = \frac{\text{b}}{\text{a}}$

$ ⇒\frac{(30-25)}{3} = \frac{\text{b}}{\text{a}}$

$⇒ \frac{5}{3} =\frac{\text{b}}{\text{a}} $

$⇒\frac{\text{b}}{\text{a}} = \frac{5}{3}\ \dots\dots (5)$

Put this value in equation 4, we get

$\text{n} × \frac{5}{3} = 10$

$\Rightarrow 5\text{n} = 30$

$ ⇒ \text{n} = \frac{30}{5}$

$\Rightarrow \text{n} = 6$

So, the value of n is 6.

2. 100

Solution:

$\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}$

Differentiate both the sides with respect to x, we get 

$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big(1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}\Big)$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2}\Big)+\dots+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{100}}{100}\Big)$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\dots+\frac{1}{100}\frac{\text{d}}{\text{dx}}(\text{x}^{100})$

$=0+1+\frac{1}{2}\times\text{2x}+\dots+\frac{1}{100}\times100\text{x}^{99}$

$=1+\text{x}+\text{x}^2+\dots+\text{x}^{99}$

Putting x = 1, we get

$\text{f}'(\text{x})=1+1+1+\dots+1$ (100 terms)

$=100$

1. 5050

Solution:

$\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1$

Differentiate both the sides with respect to x, we get

$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1)$

$=\frac{\text{d}}{\text{dx}}(\text{x}^{100})+\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\dots+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(1)$

$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1+0$

$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1$

Putting x = 1, we get 

$\text{f}'(\text{x})=100+99+98+\dots+2+1$

$=\frac{100(100+1)}{2}\ \Big(\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big)$

$=50\times101$

$=5050$​​​​​​​

2. 6.4

Solution:

Use L ’Hospital’ s Rule,

and differentiate the numerator and denominator.

$ \lim_{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x} ^2-5\text{x}} ?$

$ =\frac{32}{5}$
= 6.4

2. 10c²

^Solution:

Given, binomial expression is $\Big(\text{y²} + \frac{\text{c}}{\text{y}})5$

Now, T_{r + 1} = ⁵C_r × (y²)^{5 - r} × $ \big(\frac{\text{c}}{\text{y}}\big)^\text{r}$

= ⁵C_r × y^{10 - 3r} × C^r

Now, 10 - 3r = 1

⇒ 3r = 9

⇒ r = 3

So, the coefficient of y = ⁵C₃ × c³ = 10c³

1. $\frac{5}{4}$

Solution:

$\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$

$=\frac{1}{2}\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}$

$=\frac{1}{2}\text{x}^{\frac{1}{2}}-\text{2x}^{-\frac{1}{2}}$

Differentiate both the sides with respect to x, we get 

$\text{f}'(\text{x})=\frac{1}{2}\times\frac{1}{2}\text{x}^{\frac{1}{2}-1}-2\times\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}\ [\text{f}(\text{x})=\text{x}^\text{n}\Rightarrow\text{f}'(\text{x})=\text{nx}^{\text{n}-1}]$

$\Rightarrow\text{f}'(\text{x})=\frac{1}{4}\text{x}^{-\frac{1}{2}}+\text{x}^{-\frac{3}{2}}$

$\therefore\text{f}'(\text{1})=\frac{1}{4}\times1+1=\frac{5}{4}$

1. $-\frac{4\text{x}}{(\text{x}^2-1)^2}$

Solution:

$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$

$=\frac{\text{x}^2+1}{\text{x}^2-1}$

Differentiate both the sides with respect to x, we get

$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2}$ (Quotient rule)

$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$

$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$

$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$

2. 0

Solution:

Given: $\text{f}(\text{x})= \left\{ \begin{matrix} \text{x} \\ 0 \\ \text{x}^ 2 \end{matrix}\begin{matrix}\quad \text{x}<0 \\\quad \text{x}=0 \\ \quad \text{x}>0 \end{matrix} \right\}$

 $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=?$

Sol: left hand $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})$ x = 0right hand limit $\rightarrow \lim_\limits{\text{x} \rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x} \rightarrow0},\text{x}=0 $

right hand limit → $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow 0}\text{x}^2=0$

LHL = RHL f(0) = 0

LHL = RHL = f(0) = 0

1. 0

Solution:

$\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\tan\text{x}}{\cot\text{x}}=\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\sin\text{x}}}$

$\lim_\limits{\text{x} \rightarrow 0}\text{x}$

$ = 0$

4. 0

Solution:

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$

$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$

Differentiate both the sides with respect to x, we get

$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$

$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$

Putting x = 1, we get

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$

Thus, $\frac{\text{dy}}{\text{dx}}$ at x = 1 is 0.

4. 2xe^x2

^Solution:

We apply chain rule.

First we differentiate x².

$ \frac{\text{d}}{\text{dx}} (\text{x}2 ) = 2\text{x}$ Now, we know that $\frac{\text{d}}{\text{dx}}$(ex ) = ex

We differentiate ex2 in the same manner and then,

multiply with the derivative of $ \frac{{\text{x}}^2 \text{d}}{\text{dx} (\text{e}^{\text{x}2})} = \text{2xe}^{\text{x}2}.$

1. -2

Solution:

$\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$

Differentiate both the sides with respect to x, we get

$=\frac{(\sin\text{x}-\cos\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^2}$

$=\frac{-(\cos^2\text{x}+\sin^2\text{x}-2\cos\text{x}\sin\text{x})-(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})}{(\sin\text{x}-\cos\text{x})^2}$

$=\frac{-1+2\cos\text{x}\sin\text{x}-1-2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^2}$

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

Putting x = 0 is -2

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=0}=\frac{-2}{(\sin0-\cos0)}=\frac{-2}{(0-1)^2}=-2$

Thus, $\frac{\text{dy}}{\text{dx}}$ at x = 0 is -2

3. y

Solution:

$\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots$

Differentiate both the sides with respect to x, we get 

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\Big)$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^4}{4!}\Big)+\dots$

$=\frac{\text{d}}{\text{dx}}(1)+\frac{1}{1!}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2!}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{1}{3 !}\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{1}{4!}\frac{\text{d}}{\text{dx}}(\text{x}^4)+\dots$

$=0+\frac{1}{1!}\times1+\frac{1}{2!}\times2\text{x}+\frac{1}{3!}\times3\text{x}^2+\frac{1}{4!}\times4\text{x}^3+\dots$

$=1+\frac{1}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\ \Big[\frac{\text{n}}{\text{n}!}=\frac{1}{(\text{n}-1)!}\Big]$

$=\text{y}$

1. 5050

Solution:

f(x) = x¹⁰⁰ + x⁹⁹ + … + x + 1

f(x) = 100x⁹⁹ + 99x⁹⁸ + …. + 1 + 0

f(1) = 100(1)⁹⁹ + 99(1)⁹⁸ + ….+ 1

= 100 + 99 + …. + 1

This is an AP with common difference -1, a = 100, n = 100 and l = 1.

So, the sum of this AP = $\Big(\frac{100}{2}\Big)$[100 + 1]

= 50(101)

= 5050

Therefore, f(1) = 5050

1. 2d - b

Solution:

$ \begin{matrix}\lim\\\text{h}\rightarrow 0^{-} \end{matrix}\ \text{f(x)}=2(\alpha -\text{h})+\text{b}=2\alpha +\text{b}=\text{L} ............(1)$

$ \begin{matrix}\lim\\\text{h}\rightarrow 0 ^{+} \end{matrix}\text{ f(x)}=(\alpha +\text{h})+\text{d = L}\quad \quad \ \alpha =\text{L - d}\ ................(2)$
 

Substituting value of euation (2)in (1), we get

$ 2\left (\text{L}-\text{d} \right )+\text{b}=\text{L}$

L = 2d - b

Hence, option A is correct.

3. 1

Solution:

$\lim_\limits{\text{x} \rightarrow 0}\frac{\sin^2}{\text{x}^2}\times\lim_\limits{\text{x} \rightarrow 0}\frac{\text{e}^x}{\text{x}}$

We apply L’Hospital’s rule and differentiate numerator and denominator.

$1\times\lim_\limits{\text{x} \rightarrow 0}\frac{\text{e}^x}{\text{1}}$

$= 1$

3. 2

Solution:

he limit can be written as,

$\lim_{\text{x} \rightarrow 0}\frac{32\text{x}^2}{\sin^2 4\text{x}}$

$2\times\lim_\limits{\text{x} \rightarrow 0}\frac{4\text{x}}{\sin 4\text{x}}\times\ \lim_\limits{\text{x} \rightarrow 0}\frac{4\text{x}}{\sin 4\text{x}}$

= 2 x 1 x 1

= 2

4. $\text{dose not exist}$

Solution:

Given: $\text{f(x)}=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}$

Now, f(x) is not difined at x = a. Therefore, f(x) is not differentiable at x = a.

So, f'(a) dose not exist.

Hence, the correct answer is option (d).

4. $ \dfrac {1}{\text{4a}\sqrt {\text{a}-\text{b}}}$

Solution:

$= \displaystyle \lim _{ \text{x}\rightarrow \text{a} }{ \frac { \sqrt { \text{x-b} } -\sqrt {\text{ a-b} } }{ { \text{x} }^{ 2 }-{ \text{a} }^{ 2 } } } ​​\text{(a > b)}:$
This is the $ \frac{0}{0}$  form.Apply L-hospital rule

 $= \lim_\limits{\text{x}\to \text{a}}\dfrac{\dfrac{1}{2\sqrt {\text{x}-b}}-0}{2\text{x}-0}$

$\lim_\limits{\text{x}\to \text{a}}\frac{1}{4\text{x}\sqrt {\text{x}-\text{b}}}=\dfrac{1}{4\text{a}\sqrt{\text{a}-\text{b}}}$

Hence, this is the answer.

3. 6

Solution:

$ \lim_\limits{\text{x} \rightarrow 0}\frac{3\sin(2\text{x}^2)}{\text{x}^2}=$

$ \lim_\limits{\text{x} \rightarrow 0}\frac{2\ *\ 3\sin(2\text{x}^2)}{\text{2x}^2}=6$

1. $\text{n}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$

$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{1+\text{x}-(2)}=\text{n}(1)^{\text{n}-1}$

 $=\text{n}$

3. $ \frac{-1}{2}$

Solution:

$=\lim_\limits{\text{x} \rightarrow -1\frac{\text{x}^2}{\text{x}-1}}$

Substituting x = -1

we get, $ \displaystyle \lim _{ \text{x}\rightarrow -1 } \frac{\text{x}^2}{\text{x}-1}= \frac { { (-1) }^2 1 }{ -1-1 } = -\frac {1}{2}$

1. $\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$

Solution:

Given, $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$ 

$\Rightarrow\text{y}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$

$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^{2}-1).2\text{x}-(\text{x}^{2}+1).2\text{x}}{(\text{x}^{2}-1)^{2}}$

$=\frac{2\text{x}(\text{x}^{2}-1-\text{x}^{2}-1)}{(\text{x}^{2}-1)^{2}}=\frac{2\text{x}(-2)}{(\text{x}^{2}-1)^{2}}$

$=\frac{-4\text{x}}{(\text{x}^{2}-1)^2{}}$

4. 6

Solution:

When x tends to 3, both the numerator and,

the denominator become 0 and it becomes of the form, 0.

Therefore, we use L’Hospital’s rule,

which states the we differentiate the numerator and the denominator,

until a definite answer is reached.

On differentiating once we get,

$\lim_{\text{x} \rightarrow 3}\frac{2\text{x}}{1}$

Since, this not an indeterminate form now, we can substitute the value of x.

= 2 x 3

= 6

3. -1

Solution:

$ \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{x}+\cos \text{x}}{\sin \text{x}-\cos\text{x}}$

Substituting x = 0, we get

$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{x}+\cos \text{x}}{\sin \text{x}-\cos\text{x}}$

$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{0}+\cos \text{0}}{\sin \text{0}-\cos\text{0}}$

$ = \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\ \text{0}+\text{1}}{0-1}$

4. 6

Solution:

The denominator becomes 0, as x approaches 4.

$ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$ Here, if we factorize the numerator we get

$ \lim_{\text{x} \rightarrow 4}\frac{(\text{x}-4) (\text{x}+2)}{\text{x}-4}$

We can now cancel out (x - 4) from both the numerator and denominator.

We get, $ \lim_{\text{x} \rightarrow 4}(\text{x}+2)=6$

4. 1

Solution:

$\lim _\limits{ \text{x}\rightarrow 1 }{ \frac { \sin { \left( { \text{e} }^{ \text{x}-1 }-1 \right) } }{ \log { \text{x} } } }$

$ =\lim _\limits{ \text{h}\rightarrow 0 }{ \frac { \sin { \left( { \text{e} }^{\text{ h }}-1 \right) } }{ \log { \left( 1+\text{h} \right) } } }$

$ =\displaystyle\lim _{\text{ h}\rightarrow 0 }{ \frac { \sin { \left( {\text{ e} }^{\text{ h} }-1 \right) } }{ \left( { \text{e} }^{ \text{h} }-1 \right) } } \times \frac { \left( {\text{ e} }^{ \text{h} }-1 \right) }{ \log { \left( 1+\text{h} \right) } }​$

$ =1\times\lim _\limits{ \text{h}\rightarrow 0 }{ \frac { \left( \text{h}+\frac { {\text{ h} }^{ 2 } }{ 2! } +\dots \right) }{ \left(\text{ h}-\frac { {]\text{ h} }^{ 2 } }{ 2 } +\dots \infty \right) } }$

= 1 × 1 = 1

1. $\cos9$

Solution:

Given $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$ 

$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}.\cos(\text{x}+9)-\sin(\text{x}+9)(-\sin\text{x})}{\cos^{2}\text{x}}$

$=\frac{\cos\text{x}\cos(\text{x}+9)+\sin\text{x}\sin(\text{x}+9)}{\cos^{2}\text{x}}$

$=\frac{\cos(\text{x}+9-\text{x})}{\cos^{2}\text{x}}=\frac{\cos9}{\cos^{2}\text{x}}$

$=\frac{\cos9}{(1)^{2}}=\cos9$ 

4. 2(xy - y)

Solution:

Since, x = et sint and y = et cost Therefore,

$ \frac{\text{dx}}{\text{dt}} $ = e^t sint + e^t cost = y + x And,

$ \frac{\text{dx}}{\text{dt}} $ = et cost - et sint = y - x So,

$=\text{y}= \frac{\text{dx}}{\text{dt}} $

$ =\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}$

$ = \frac{(\text{y} - \text{x})}{(\text{y} + \text{x})}$

Thus, $\text{y}= \frac{[(\text{x} + \text{y})(\text{y} - 1) - (\text{y} - \text{x})(\text{y} + 1)]}{(\text{x} + \text{y})^2}$

Or, (x + y)² y = (x + y - y + x)y - x - y + x = 2xy - 2y = 2(xy - y) 10.

If, y = (sin-1x)² , then what is the value o.

4. $ \sqrt {{\text{e}^{\text{i}\alpha }}}$

3. $\frac{1}{2}$

3. Does not exist.

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$

$\text{L}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{-\sin\text{x}}{\text{x}}=-1$

$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}=1$

$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$ 

So, the limit does not exist.

1. 40

Solution:

$\lim_{\text{y} \rightarrow 4}\text{f}(\text{y}) =\text{y}^ 2+6\text{y}$

f(4) = 4² + 6(4)

f(4) = 16 + 24

f(4) = 40

1. $\frac{2}{π}$

Solution:

$ \sin \frac{⁡π}{2} = 1$

$\lim_{\text{y} \rightarrow \frac{\pi}{2}}\frac{\text{sin}}{\text{x}}=\frac{\sin\frac{\pi}{2}}{\frac{pi}{2}}$

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

$ = \frac{2}{π}$

4. 0

Solution:

Given that $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ 

$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}^{\frac{3}{2}}}$

$\big(\frac{\text{dy}}{\text{dx}}\big)=\frac{1}{2}-\frac{1}{2}=0$

1. 4

Solution:

$ \lim\limits_{\text{x}→3}\ ​2\text{x}^2−3\text{x}−5 $

$ = 2(3)^2 - 3(3) - 5$

= 18 - 9 - 5

= 4

2. 100

Solution:

Given $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^{2}}{2}+.....+\frac{\text{x}^{100}}{100}$ 

$\text{f}(\text{x})=1+\frac{2\text{x}}{2}+......+\frac{100\text{x}}{100}$

$\therefore \text{f'}(1)=1+11+....+1=100$ 

3. 0

Solution:

Let $\lim\limits_{\text{x} \rightarrow 2} \text{x}^2 - 5\text{x} + 6$ This is not an indeterminate form

Therefore, $\text{L}=(2)2−5(2)+6\Rightarrow \text{L}=0$.

2. 1

Solution:

Given,

f(x) = x - [x]

f(x) = 1 - 0 {[x] = integer less than or equal to x}

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

1. 9

Solution:

$\text{f}(\text{x})=\frac{3\text{x}^2+\text{ax}+\text{a}+1}{\text{x}^2+\text{x}-2}$

As 0 x → -2, D^r → 0.

Hence, as x → -2, N^r →0. Therefore,

12 - 2a + a + 1 = 0 or a = 13

Hence, option A is correct.

1. $\frac{5}{4}$

Solution:

Given that $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$ 

$\therefore\text{f}(\text{x})=\frac{1}{2}\bigg[\frac{\sqrt{\text{x}.}1-(\text{x}-4).\frac{1}{2\sqrt{\text{x}}}}{\text{x}}\bigg]$

$=\frac{1}{2}\Big[\frac{2\text{x}-\text{x}+4}{2\sqrt{\text{x}.\text{x}}}\Big]$

$=\frac{1}{2}\Bigg[\frac{\text{x}+4}{2(\text{x})^{\frac{3}{2}}}\Bigg]$

$\therefore\text{f}(\text{x})\ \text{x}=1=\frac{1}{2}\Big[\frac{1+4}{2\times1}\Big]=\frac{5}{4}$

2. 1

Solution:

We need to use product rule in both the terms to get the answer.

$\frac{\text{d}}{\text{dx}} (\text{f.g}) = \text{g}.\frac{\text{d}}{\text{dx}} (\text{f})+ (\text{f}).\frac{\text{dy}}{\text{dx}} (\text{g})$

Here f = e^x and g = tan ⁡x

$ \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \tan \text{x}) = \tan⁡ \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}) + \text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}}(\tan ⁡\text{x})$

$\frac{d}{dx} (\text{e}^{\text{x}} \tan \text{x}) = \tan⁡ \text{e}^{\text{x}} + \text{e}^{\text{x}}.\sec2\text {⁡x}$

At x = 0 we get,

$ = \tan ⁡0.\text{e}0 + \text{e}0.\sec2⁡0$

= 0.(1) + 1.(1)

= 1

3. Does not exist

Solution:

Given $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$

$\text{f'}(\text{x})=\frac{(\text{x}-\text{a})(\text{n.}\text{x}^{\text{n-1}}-(\text{x}^{\text{n}-\text{a}^{\text{n}}})1}{(\text{x}-\text{a})^{2}}$

So, $\text{f}(\text{a})=\frac{0}{0}$ = Does not exist.

2. 6.4

Solution:

Use L’Hospital’s Rule, and differentiate the numerator and denominator.

$ \lim_\limits{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x}^2-5\text{x}}$

$ =\frac{32}{5}$

= 6.4

4. does not exist

Solution:

Left hand limit is$\lim_\limits{\text{x} \rightarrow \text{a}}​\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}​​​​​=-\infty$

And Right hand limit is $\lim_\limits{\text{x} \rightarrow \text{a}}​\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}​​​​​=+\infty$

$\text{L.H.L.}\neq \text{R.H.L.}$

Therefore, the given limit does not exist.

Hence, the option D is correct.

3. 4

Solution:

$=\lim_\limits{\text{n} \rightarrow \infty}​\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$

$ = \displaystyle \lim_{\text{n}\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{\text{n}} \right)^2}{\left(1+\Large \frac{2}{\text{n}} \right)\left(1+\Large \frac{3}{\text{n}} - \Large \frac{1}{\text{n}^2} \right)} }=\text{n}$

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

4. $\text{y}=\sin\text{x}+\text{c}$

4. $\text{2ex.cos⁡ x}$

Solution:

We need to use product rule in both the terms to get the answer.

$=\frac{d}{dx}(\text{f.g})=\text{g}\frac{d}{dx}(\text{f})+\text{f}\frac{d}{dx}(\text{fg})$

$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{⁡x}) = ({\text{e}^{\text{x}},.\frac{\text{d}}{\text{dx}}} (\sin⁡\text{x}) \\+ \text{sin ⁡x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}})) + (\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}} (\cos \text{⁡x}) + \cos⁡ \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}))$

$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{⁡x}) =(\text{e}^{\text{x}}.\cos⁡ \text{x} + \sin \text{⁡x} . \text{e}^{\text{x}}) + (\text{e}^{\text{x}}.(-\sin ⁡\text{x}) + \cos ⁡\text{x}.\text{e}^{\text{x}})$

$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text {x}} \sin \text{x} + \text{e}^{\text {x}} \cos \text{⁡x}) = \text{e}^{\text {x}}.\cos⁡ \text{x} + \sin⁡ \text{x} . \text{e}^{\text {x}} – \text{ex}.\sin⁡ \text{x} + \cos \text{⁡x}.\text{e}^{\text {x}}$

$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = 2\text{e}^{\text{x}}.\text{cos}⁡ \text{x}$

3. 6

Solution:

We have, $ \text{y} = (\sin^{-1}\text{x})^2 $ ………..(1)

Differentiating with respect to x,

we get, $ \text{y} = \frac{(\sin^{-1}\text{x})^2}{1-\text{x}^2} $ 1/2 or,

Squaring both sides,

(1 - x2 )(y)2 = 4(sin - 1x)2 From (1),

(1 - x2 )( y)2 = 4y Differentiating with respect to x,

4 = 2 + 4 = 6

4. does not exisz

Solution:

$\text{L.H.L}.=\lim​ \tan\text{x}=+∞ \ \text{x}→\Big(\frac{\pi}{2}\Big)^-$

$\text{R.H.L}.=\lim​ \tan\text{x}=-∞ \ \text{x}→\Big(\frac{\pi}{2}\Big)^+$

Clearly left hand $ \text{limit} \neq$ right hand limit.

Hence given limit does not exist.

1. does not exist

Solution:

$ \lim_\limits{\text{t} \rightarrow 0}\frac{\sqrt{1-\cos2\text{t}}}{\text{t}}$

Clearly R.H.L. = $ \sqrt{2}$​

L.H.L. = $ -\sqrt{2}$

Since R.H.L.$ \neq$ L.H.L. So, limit does not exist.

Hence, option A is correct.

1. $ \pi$

Solution:

$=\lim_\limits{\text{x} \rightarrow 1}(1+\sin\pi)π\text{x}$

$= (1+\sin \pi(1))\\ $

$=\pi(1+0)$

$= \pi$

2. $1$

Solution:

$\text{f(x)}=\text{x}\sin\text{x}$

Differentiating both sides with respect to x, we get

$\text{f}'\text{(x)}=\text{x}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\times\frac{\text{d}}{\text{dx}}\text{(x)}$ (Product rule)

$=\text{x}\times\cos\text{x}+\sin\text{x}\times1$

$=\text{x}\cos\text{x}+\sin\text{x}$

Putting $\text{x}=\frac\pi{2},$ we get

$=\frac{\pi}{2}\times0+1$

$=1$

Hence, the correct answer is option (b)

3. 9

Solution:

$ \displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$

$ =\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{x}} \times\frac{\sin 3\text{x}}{\text{x}}$

$= \displaystyle\lim_{\text{x} \rightarrow 0} 3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg) \times3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg)$

$=3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}\times3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}$

$ = 3\times3=9$

4. 0

Solution:

$ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}^2\sec\text{x}}{\sin\text{x}}​​​\times​\ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\cos\text{x}} $$$

= 1 × 0

= 0

1. 0

Solution:

Using direct substitution, we obtain,$ =\displaystyle\lim_{\text{x}\rightarrow 2} \dfrac{\text{x}^2-4}{\text{x}+3}$

$ =\dfrac{4-4}{2+3}=0$

1. 2

Solution:

As their is not any x term in the denominator,

we can directly substitute the value of x as 0.

Thus, we have $\lim_\limits{\text{x} \rightarrow 0}\frac{2\text{x}^2+3\text{x}+4}{2}$

$ =\frac { 2.0+3.0+4 }{ 2 }$

$ =\frac{4}{2}=2$

1. $\frac{1}{2}$

3. 0

Solution:

$ = \displaystyle \underset{\text{x}\rightarrow 2}{\lim} \text{x}^2-5\text{x}+6$

$ ={ 2 }^{ 2 }-5\times 2+6$

= 4 - 10 + 6

= 0

1. sin⁡ x + tan⁡ x sec⁡ x

Solution:

We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})=\text{g.}\frac{\text{d}}{\text{dx}}(\text{f})+(\text{f})\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})$

Here, f = sin⁡ x and g = tan⁡ x

$\frac{\text{d}}{\text{dx}}$ (sin⁡ x tan⁡ x) = cos⁡ x tan⁡ x + sec²⁡ x sinx

$\frac{\text{d}}{\text{dx}}$ (sin⁡ x tan⁡ x) = sin⁡ x + tan⁡ x sec⁡ x

2. $\frac{4}{9}$

Solution:

Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$

 $=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$

$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]=\Big[\frac{2\theta}{2\theta}\Big]^{2}=\Big(\frac{2}{3}\Big)^{2}=\frac{4}{9}$

$=\frac{4}{9}$

1. 1

Solution:

for $ \text{x}=30^+,$

$ \text{ x}-3 > 0$

Let $\text{L}=\displaystyle \lim_{\text{x}-3^+}\dfrac {|\text{x}-3|}{\text{x}-3}$

$ =\displaystyle \lim_{\text{x}\rightarrow 3^+}$

$ =\dfrac{(\text{x}-3)}{(\text{x}-3)}$

$ = \lim\limits_{\text{x}\rightarrow 3^+}(1) =1$

1. $\frac{7}{3}$

1. 2

Solution:

Let $ \lim\limits_{\text{x} \rightarrow 0} \frac{2\text{x}^2 + 3\text{x} + 4}{2}$ ​This is not an indeterminate form,

Therefore, $\text{L}=\lim\limits_{\text{x}\rightarrow 0}\dfrac {2(0)+3(0)+4}{2}=\dfrac {4}{2}\text{L}=2$

1. $-2$

Solution:

Given $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$ 

$\frac{\text{dy}}{\text{dx}}=\frac{-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$

$=\frac{-(\sin\text{x}+\cos\text{x})^{2}(\sin\text{x}+\cos\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$

$=\frac{\sin^{2}\text{x}+\cos^{2}\text{x}+2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^{2}}$

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

$\therefore \Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-2}{(\sin\text{0}-\cos0)^{2}}=\frac{-2}{(-1)^{2}}=2$

3. n

Solution:

$= {\lim _\limits{\text{x}\to 1}}\frac{{{\text{x}}^{\text{n}}}-1}{\text{x}-1}$

$ =\lim_\limits{\text{x}\to 1}\frac{\left( \text{x}-1 \right)\left( {{\text{x}}^{\text{n}-1}}+{{\text{x}}^{\text{n} -2}}+.....+\text{x}+1 \right)}{\text{x}-1}$

$ =\lim_\limits{\text{x}\to 1}\sum\limits_{\text{i}=0}^{\text{n}-1}{{{\text{x}}^{\text{i}}}}$

$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{i}}}}$

$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{}}}}$

$ = \text{n}$

Hence, this is the answer.

4. 2

Solution:

Given $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$ 

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

$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)=\tan\frac{\pi}{4}+1$

$=1+1=2$

$=2$

3. $1$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$ 

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$

$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$

Taking limit, we get

$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]=\frac{1}{2}\times1\times2$

$=1$

2. $-\frac{1}{10}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ 

$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$

$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$

Taking limit, we get

$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}=\frac{-1}{2\times5}=\frac{-1}{10}$

1. 1

Solution:

Since it is of the form $ \frac{∞}{∞}$

we use L 'Hospital' s rule and differentiate the numerator and denominator

$\text{L}=\lim_{\text{x} \rightarrow \infty}\frac{\text{x}^2-9}{\text{x}^2-3\text{x} +2}$

On differentiating once, we get $\text{L}=\lim_{\text{x} \rightarrow \infty}\frac{2\text{x}}{2\text{x}}$

Which is equal to, $\lim_{\text{x} \rightarrow \infty}1 = 1.$

3. -1

Solution:

Given, $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}=\lim\limits_{\text{x} \rightarrow\pi}\frac{\sin(\pi)-\text{x}}{-(\pi-\text{x})}$ 

$=-1$

1. 1

Solution:

$ \lim_\limits{\text{x} \rightarrow\infty}\frac{\sqrt{\text{x}^2+1}-^3\sqrt{\text{x}^2+1}}{^4\sqrt{\text{x}^4+1}-^5\sqrt{\text{x}^4+1}}$

$ =\displaystyle \lim_{\text{x}\rightarrow \infty}\frac {\sqrt {1+1/\text{x}^2}-\sqrt [3]{1/\text{x}+1/\text{x}^3}}{\sqrt [4]{1+1/\text{x}^4}-\sqrt [5]{1/\text{x}-1/\text{x}^5}}$

$ =\frac{1-0}{1-0}=1$

3. $\frac{1}{2}$

2. $\frac{1}{10}$

2. $\frac{1}{2}$

Solution:

Given $ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}} =\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}\big[\frac{\tan2\text{x}}{\text{x}}-1\big]}{\text{x}\big[3-\frac{\sin\text{x}}{\text{x}}\big]}$

$ =\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\tan2\text{x}}{2\text{x}}\times2-1}{3-\frac{\sin\text{x}}{\text{x}}}=\frac{1.2-1}{3-1}$

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

4. 50

Solution:

Given that f(x) = 1 - x + x² - x³ + .......-x⁹⁹ + x¹⁰⁰

f'(x) = -1 + 2x - 3x² + ... -99.x⁹⁸ + 100.x⁹⁹

f'(x) = -1 + 2 - 3 + ....-99 + 100

=(- 1 - 3 - 5 ....-99) + (2 + 4 + 6 + ....100)

$=\frac{50}{2}\big[2\times1+(50-1)(-2)\big]+\frac{50}{2}\big[2\times2(50-1)2\big] $

$=25\big[-11+102\big]=25\times2=50$

4. Does not exist

Solution:

$=\lim_\limits{\text{x} \rightarrow 0}\frac{\sin|\text{x}|}{\text{x}}$

$\text{ LHL} =-1,\text{RHL}=1$

Limit does not exist.

4. $ (\text{abc})^{\frac{2}{3}}$

Solution:

$ \displaystyle \lim _{ \text{x} \rightarrow 0 }{ { \left( \frac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } \right) }^{ \frac{ 2 }{ \text{x} } } } ={ \left( \frac { 3 }{ 3 } \right) }^{\frac { 2 }{ 0 } }$

$ =1∞\text{ form}={ \text{e} }^{ \displaystyle \lim _{ \text{x}\rightarrow 0 }{ { \left( \cfrac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } -1 \right) }^{\frac{ 2 }{ \text{x} } } } }$

$ =\text{e}\frac{2}{3}(\log \text{a}+\log \text{b}+\log \text{c})$

$= (\text{abc})^{\frac{2}{3}}$

2. $\frac{\text{m}}{\text{n}}$

Solution:

Given $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}=\lim\limits_{\text{x} \rightarrow 1}\frac{\frac{\text{x}^{\text{m}-(1)^\text{m}}}{\text{x}-1}}{\frac{\text{x}^{\text{n}-(1)^{\text{n}}}}{\text{x}-1}}$ 

$=\frac{\text{m}(1)^{\text{m}-1}}{\text{n}(1)^{\text{n}-1}}=\frac{\text{m}}{\text{n}}$

$=\frac{\text{m}}{\text{n}}$

4. 2xe^x2

Solution:

We apply chain rule. First we differentiate x².

$\frac{\text{d}}{\text{dx}} (\text{x}^2) = 2\text{x}$

Now, we know that $\frac{\text{d}}{\text{dx}} (\text{e}^x) = \text{e}^\text{x}$

We differentiate e^x2 in the same manner and then multiply with the derivative of x²

^{$\frac{\text{d}}{\text{dx}} (\text{e}^\text{x}) = 2\text{x}\text{e}^\text{x}$}

2. $ \text{e}^4$

Solution:

$ \lim\limits_{\text{x} \to 0}\left(\frac{(1+\text{x})^2}{\text{e}^{\text{x}}}\right)^{\frac{4}{\sin \text{x}}}=\lim\limits_{\text{x} \to 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\frac{\text{x}}{\sin \text{x}}}\right)^8}{\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$

We have

$ \lim\limits_{\text{x} \to 0}\frac{\sin \text{x}}{\text{x}}=1$

and $ \lim\limits_{\text{x} \to 0}(1+\text{x})^{\frac{1}{\text{x}}}=$

both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing,

$=\lim_\limits{\text{x} \rightarrow 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\sin \text{x}}}\right)^8}{\lim_\limits{\text{x} \rightarrow 0}\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$

[Using division property of limits]

$=\frac{\left(\lim\limits_{\text{x}\to 0}\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\left(\lim\limits_{\text{x}\to 0}\frac{\text{x}}{\sin \text{x}}\right)}\right)^8}{\text{e}^{\left(\lim\limits_{\text{x}\to 0}\dfrac{4\text{x}}{\sin \text{x}}\right)}}$ [Using limit property]

$$$= \text{e}^4$

3. 3

Solution:

We know that in the range of $ (0, 2π)$ the graph of sinx and cosx intersects each other in three points.

And we know that these points of intersection are only the critical points

Thus, there are 3 critical points.

1. 0

Solution:

We have,$\lim\limits_{\text{n} \rightarrow \infty} \dfrac{\text{n}!}{(\text{n}+1)!-\text{n}!}$

$=\lim\limits_{\text{n} \rightarrow \infty} \dfrac{\text{n}!}{(\text{n}+1)\text{n}!-\text{n}!}$

$ =\lim\limits_{\text{n}\rightarrow \infty} \dfrac{1}{\text{n}+1-1}$

$ =\lim\limits_{\text{n}\rightarrow \infty}\frac{1}{\text{n}}=0$

1. 0

Solution:

This is of the form $ \frac{∞}{∞}$,

therefore we use L ’Hospital’ s rule and,

differentiate the numerator and denominator.

3. 4

Solution:

$=\lim_\limits{\text{n} \rightarrow \infty}​\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$

$ = \displaystyle \lim_{\text{n}\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{\text{n}} \right)^2}{\left(1+\Large \frac{2}{\text{n}} \right)\left(1+\Large \frac{3}{\text{n}} - \Large \frac{1}{\text{n}^2} \right)} }=\text{n}$

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

2. 10

Solution:

Given, $ \displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}:$
 

Substituting x = 1 we get

$ \displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}=$

$ =\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{(1)}^{2}+4\text{(1)}+4}{2\text{(1)}-1}$

$ =\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{}+4\text{}+4}{2\text{}-1}=10$

2. 39

Solution:

$ =\displaystyle \lim _{\text{x}\rightarrow 3 }{ \left( 4{ \text{x} }^{ 2 }+3 \right) } =4{ \left( 3 \right) }^{ 2 }+3=36+3=39$

2. 1

Solution:

Given f(x) = x - [x]

we have ti first check for differentiability of f(x) at $\text{x}=\frac{1}{2}$

$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}=\frac{-\text{h}}{-\text{h}}=1$

$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$

$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}=\frac{\text{h}}{\text{h}}=1$

Since, LHD = RHD

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

2. 2

Solution:

Given,$ \lim\limits_{\text{x}\to \text{a}}\frac{\text{x}^5-\text{a}^5}{\text{x - a}}=80$

or, $ 5\text{a}^4=80$ [ Using direct formula]or, $\text{a}^4=16$ or, $\text{a}=2.$

4. None of these.

Solution:

Given $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ 

$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $

$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}=-1$

$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}=1$

$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$

So, the limit does not exist.

3. $ \sqrt{2}$

Solution:

$ \lim_\limits{\text{x} \rightarrow \infty}\frac{2\sqrt{\text{x}}+3\sqrt[3]{\text{x}}+4\sqrt[4]{\text{x}}+...+\text{n}\sqrt[\text{n}]{\text{x}}}{\sqrt{(2\text{x}-3}+{\sqrt[3]{(2\text{x}-3)}+{\sqrt[4]{(2\text{x}-3)}+...+}{\sqrt[\text{n}]{(2\text{x}-3)}}}}$

Dividing numerator and denominator by$ \sqrt{\text{x}}$

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

2. $ ^{2\text{n}}{\text{C}}_\text{n}$

Solution:

$ \lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2}$

$ \lim_\limits{\text{x} \rightarrow 1}\frac{\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{2\text{n}})}{(1-\text{x})}}{\Big[\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{\text{n}})}{(1-\text{x})}\Big]^2}$

$ =\frac{1\times2\times3\times\ldots(2\text{n})}{(1\times2\times3\ldots \text{n})^2} = \frac{(2n)!}{\text{n}!\text{n}!}={}^{2\text{n}}\text{C}_\text{n}$

Hence, option B is correct.

1. 2

Solution:

Given $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}^{2}\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\text{x}^{2}}{4}\times4\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\text{x}}{2}\Big)\cdot2\cos\text{x}}{\sin^{2}\frac{\text{x}}{2}}$

$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\bigg)\cdot2\cos\text{x}$

$=2\cos\text{x}0=2\times1=2$

2. $ \frac{5}{3}$

Solution:

$ \mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$

$=\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{5\text{x}}}\times\frac{\text{3x}}{\tan\text{x}}\times\frac{5}{3}$

we know that $ =\displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=1$
$=\mathop {\lim }\limits_{\text{x} \to 0}\times\frac{\text{3x}}{\tan\text{x}}=1$

$= \text{L}=1\times 1\times \dfrac {5}{3}$

$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=\frac{5}{3}$