Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3} & =\lim _{x \rightarrow 3} \frac{x^{2}-(3)^{2}}{x-3} \\ & =\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)}=\lim _{x \rightarrow 3}(x+3) \\ & =3+3=6 \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 1 / 2} \frac{4 x^{2}-1}{2 x-1} & =\lim _{x \rightarrow 1 / 2} \frac{(2 x)^{2}-(1)^{2}}{2 x-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{(2 x+1)(2 x-1)}{(2 x-1)}=\lim _{x \rightarrow 1 / 2}(2 x+1) \\ & =2 \times \frac{1}{2}+1=1+1=2 \end{aligned} $

Solution

Given, $\quad \lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim _{h \rightarrow 0} \frac{(x+h)^{1 / 2}-(x)^{1 / 2}}{x+h-x}$

$ =\lim _{h \rightarrow 0} \frac{(x+h)^{1 / 2}-(x)^{1 / 2}}{(x+h)-x}\quad $ $ [\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

$ =\frac{1}{2} x^{\frac{1}{2}-1}=\frac{1}{2} x^{-1 / 2} \quad$ $ {[\because h \rightarrow 0 \Rightarrow x+h \rightarrow x]}$

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

Solution

Given, $\quad \lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x}=\lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{(x+2)-2}$

$ =\frac{1}{3} \times 2^{\frac{1}{3}-1} $

$ =\frac{1}{3} \times(2)^{-2 / 3} [ \because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}]\quad$

$ =\frac{1}{3(2)^{2 / 3}} {[\because x \rightarrow 0 \Rightarrow x+2 \rightarrow 2}] $

Solution

Given, $\lim _{x \rightarrow 0} \frac{(1+x)^{6}-1}{(1+x)^{2}-1}=\lim _{x \rightarrow 0} \frac{\frac{(1+x)^{6}-1}{x}}{\frac{(1+x)^{2}-1}{x}} \quad$ [dividing numerator and denominator by $x$ ]

$ =\lim _{x \rightarrow 0} \frac{\frac{(1+x)^{6}-1}{(1+x)-1}}{\frac{(1+x)^{2}-1}{(1+x)-1}}$ ${[\because x \rightarrow 0 \Rightarrow 1+x \rightarrow 1]}$

$ =\frac{\lim _{x \rightarrow 0} \frac{(1+x)^{6}-(1)^{6}}{(1+x)-1}}{\lim _{x \rightarrow 0} \frac{(1+x)^{2}-(1)^{2}}{(1+x)-1}}\quad [ \because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}] $

$ =\frac{6(1)^{6-1}}{2(1)^{2-1}} \quad[\therefore \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

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

Solution

Given, $\lim _{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{x-a}=\lim _{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{(2+x)-(a+2)}$

$ =\frac{5}{2}(a+2)^{\frac{5}{2}-1} [ \because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

$ =\frac{5}{2}(a+2)^{3 / 2} {[\because x \rightarrow a \Rightarrow x+2 \rightarrow a+2]} $

Solution

Given, $\quad \lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}=\lim _{x \rightarrow 1} \frac{\sqrt{x}[(x)^{7 / 2}-1]}{\sqrt{x}-1}$

$ \begin{matrix} =\lim _{x \rightarrow 1} \frac{(x)^{7 / 2}-1}{\sqrt{x}-1} \cdot \lim _{x \rightarrow 1} \sqrt{x} & [\because \lim _{x \rightarrow a} f(x) \cdot g(x)=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)] \\ =\lim _{x \rightarrow 1} \frac{\frac{x^{7 / 2}-1}{x-1}}{\frac{(x)^{1 / 2}-1}{x-1}} \cdot 1 \\ =\frac{\lim _{x \rightarrow 1} \frac{x^{7 / 2}-1}{x-1}}{\lim _{x \rightarrow 1} \frac{(x)^{1 / 2}-1}{x-1}} & \\ =\frac{\frac{7}{2}(1)^{\frac{7}{2}-1}-\frac{7}{2}}{\frac{1}{2}}=7 & \\ \frac{1}{2}(1)^{\frac{1}{2}-1} \frac{1}{2} & \end{matrix} $

Solution

Given, $\lim _{x \rightarrow 2} \frac{x^{2}-4}{\sqrt{3 x-2}-\sqrt{x+2}}=\lim _{x \rightarrow 2} \frac{(x^{2}-4) \sqrt{3 x-2}+\sqrt{x+2})}{(\sqrt{3 x-2}-\sqrt{x+2})(\sqrt{3 x-2}-\sqrt{x+2})}$

$ =\lim _{x \rightarrow 2} \frac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{(\sqrt{3 x-2})^{2}-(\sqrt{x+2})^{2}} $

$ \begin{aligned} & \quad[\because(a+b)(a-b)=a^{2}-b^{2}] \\ & =\lim _{x \rightarrow 2} \frac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{(3 x-2)-(x+2)} \\ & =\lim _{x \rightarrow 2} \frac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{3 x-2-x-2} \\ & =\lim _{x \rightarrow 2} \frac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{2 x-4} \\ & =\lim _{x \rightarrow 2} \frac{(x+2)(x-2)(\sqrt{3 x-2}+\sqrt{x+2})}{2(x-2)} \\ & =\lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{3 x-2}+\sqrt{x+2})}{2} \\ & =\frac{(2+2)(\sqrt{6-2}+\sqrt{2+2})}{2} \\ & =\frac{4(2+2)}{2}=8 \end{aligned} $

Solution

Given, $\quad \lim _{x \rightarrow \sqrt{2}} \frac{x^{4}-4}{x^{2}+3 \sqrt{2} x-8}=\lim _{x \rightarrow \sqrt{2}} \frac{(x^{2})^{2}-(2)^{2}}{x^{2}+3 \sqrt{2} x-8}$

$ \begin{aligned} & =\lim _{x \rightarrow \sqrt{2}} \frac{(x^{2}-2)(x^{2}+2)}{x^{2}+4 \sqrt{2} x-\sqrt{2} x-8} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})(x^{2}+2)}{x(x+4 \sqrt{2)}-\sqrt{2}(x+4 \sqrt{2})} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})(x^{2}+2)}{(x-\sqrt{2})(x+4 \sqrt{2})} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})(x^{2}+2)}{(x+4 \sqrt{2})} \\ & =\frac{(\sqrt{2}+\sqrt{2})[(\sqrt{2})^{2}+2]}{(\sqrt{2}+4 \sqrt{2})} \\ & =\frac{2 \sqrt{2}(2+2)}{5 \sqrt{2}}=\frac{8}{5} \end{aligned} $

Solution

Given,

$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2} \\ = & \lim _{x \rightarrow 1} \frac{x^{7}-x^{5}-x^{5}+1}{x^{3}-x^{2}-2 x^{2}+2} \\ = & \lim _{x \rightarrow 1} \frac{x^{5}(x^{2}-1)-1(x^{5}-1)}{x^{2}(x-1)-2(x^{2}-1)} \end{aligned} $

On dividing numerator and denominator by $(x-1)$, then

$ \begin{aligned} & =\lim _{x \rightarrow 1} \frac{\frac{x^{5}(x^{2}-1)}{(x-1)}-\frac{1(x^{2}-1)}{(x-1)}}{\frac{x^{2}(x-1)}{(x-1)}-\frac{2(x^{2}-1)}{(x-1)}} \\ & =\frac{\lim _{x \rightarrow 1} x^{5}(x+1)-\lim _{x \rightarrow 1} \frac{x^{5}-1}{x-1}}{\lim _{x \rightarrow 1} x^{2}-\lim _{x \rightarrow 1}(x+1)} \\ & =\frac{1 \times 2-5 \times(1)^{4}}{1-2 \times 2}=\frac{2-5}{1-4} \\ & =\frac{-3}{-3}=1 \end{aligned} $

Solution

Given, $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{3}}-\sqrt{1-x^{3}}}{x^{2}}=\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{3}}-\sqrt{1-x^{3}}}{x^{2}} \cdot \frac{\sqrt{1+x^{3}}+\sqrt{1-x^{3}}}{\sqrt{1+x^{3}}+\sqrt{1-x^{3}}}$

$=\lim _{x \rightarrow 0} \frac{(1+x^{3})-(1-x^{3})}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$ $=\lim _{x \rightarrow 0} \frac{1+x^{3}-1+x^{3}}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$=\lim _{x \rightarrow 0} \frac{2 x^{3}}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$=\lim _{x \rightarrow 0} \frac{2 x}{(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$=0$

Solution

Given, $\quad \lim _{x \rightarrow-3} \frac{x^{3}+27}{x^{5}+243}=\lim _{x \rightarrow-3} \frac{\frac{x^{3}+27}{x+3}}{\frac{x^{5}+243}{x+3}}$

$ \begin{matrix} =\lim _{x \rightarrow-3} \frac{\frac{x^{3}-(-3)^{3}}{x-(-3)}}{\frac{x^{5}-(-3)^{5}}{x-(-3)}}=\frac{\lim _{x \rightarrow-3} \frac{x^{3}-(-3)^{3}}{x-(-3)}}{\lim _{x \rightarrow-3} \frac{x^{5}-(-3)^{5}}{x-(-3)}} & [\because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}] \\ =\frac{3(-3)^{3-1}}{5(-3)^{5-1}}=\frac{3}{5} \frac{(-3)^{2}}{(-3)^{4}} & [\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}]\\ =\frac{3}{5(-3)^{2}}=\frac{3}{45}=\frac{1}{15} & \end{matrix} $

Solution

Given, $\lim _{x \rightarrow 1 / 2} \Big(\frac{8 x-3}{2 x-1}-\frac{4 x^{2}+1}{4 x^{2}-1}\Big)=\lim _{x \rightarrow 1 / 2} \Big[\frac{(8 x-3)(2 x+1)-(4 x^{2}+1)}{(4 x^{2}-1)}\Big]$

$ \begin{aligned} & =\lim _{x \rightarrow 1 / 2} \Big[\frac{16 x^{2}+8 x-6 x-3-4 x^{2}-1}{4 x^{2}-1} \Big]\\ & =\lim _{x \rightarrow 1 / 2} \Big[\frac{12 x^{2}+2 x-4}{4 x^{2}-1}\Big] \\ & =\lim _{x \rightarrow 1 / 2} \Big[\frac{2(6 x^{2}+x-2)}{4 x^{2}-1}\Big] \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow 1 / 2} \frac{2(6 x^{2}+4 x-3 x-2)}{4 x^{2}-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[2 x(3 x+2)-1(3 x+2)]}{4 x^{2}-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[(3 x+2)(2 x-1)]}{(2 x)^{2}-(1)^{2}} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)(2 x-1)}{(2 x-1)(2 x+1)} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)}{2 x-1}=\frac{2 \Big(3 \times \frac{1}{2}+2\Big)}{2 \times \frac{1}{2}+1} \\ & =\frac{3}{2}+2=\frac{7}{2} \end{aligned} $

Solution

Given,

$ \lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80 $

$ \begin{aligned} \Rightarrow & n(2)^{n-1}=80 & [\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] \\ \Rightarrow & n(2)^{n-1}=5 \times 16 \\ \Rightarrow & n \times 2^{n-1}=5 \times(2)^{4} \\ \Rightarrow & n \times 2^{n-1}=5 \times(2)^{5-1} \\ \therefore & n=5 \end{aligned} $

Solution

Given, $\quad \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \cdot 3 x}{\frac{\sin 7 x}{7 x} \cdot 7 x}=\frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{\lim _{x \rightarrow 0} \frac{\sin 7 x}{7 x}} \cdot \frac{3 x}{7 x}$

$ =\frac{3}{7} \cdot \frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{\lim _{x \rightarrow 0} \frac{\sin 7 x}{7 x}} $

$ =\frac{3}{7} \quad[\because x \rightarrow 0 \Rightarrow(k x \rightarrow 0) \text {, here } k \text { is real number }] $

Solution

Given, $\quad \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}=\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{[\sin 2(2 x)]^{2}}$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{(2 \sin 2 x \cos 2 x)^{2}} \\ & =\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{4 \sin ^{2} 2 x \cos ^{2} 2 x} & \quad[\because \sin 2 \theta=2 \sin \theta \cos \theta] \\ & =\lim _{x \rightarrow 0} \frac{1}{4 \cos ^{2} 2 x}=\frac{1}{4} & \quad[\because \cos 0=1]\\ \end{aligned} $

Solution

Given, $\quad \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^{2}}=\lim _{x \rightarrow 0} \frac{1-1+2 \sin ^{2} x}{x^{2}} \quad[\because \cos 2 x=1-2 \sin ^{2} x]$

$ \begin{matrix} =\lim _{x \rightarrow 0} \frac{2 \sin ^{2} x}{x^{2}}=2 \lim _{x \rightarrow 0} \frac{\sin ^{2} x}{x^{2}} & \\ =2 \lim _{x \rightarrow 0} \Big(\frac{\sin x}{x}\Big)^{2} & [\because \lim _{x \rightarrow 0} \Big(\frac{\sin x}{x}\Big)^{2}=1] \\ =2 \times 1=2 & \end{matrix} $

Solution

Given, $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}=\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^{3}} \quad[\because \sin 2 x=2 \sin x \cos x]$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^{3}} \\ & =2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \Big(\frac{1-\cos x}{x^{2}}\Big) \end{aligned} $

$ \begin{matrix} =2 \cdot 1 \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} & [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1] \\ \\ =2 \lim _{x \rightarrow 0} \frac{1-1+2 \sin ^{2} \frac{x}{2}}{x^{2}}=2 \lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{4 \times \frac{x^{2}}{4}} & \\ \\ =\frac{2 \cdot 2}{4} \lim _{x \rightarrow 0} \Big(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\Big)^{2}=\lim _{x \rightarrow 0} \Big(\frac{\sin \frac{x^{2}}{2}}{\frac{x}{2}}\Big)^{2}=1 \end{matrix} $

Solution

Given, $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{1-1+2 \sin ^{2} \frac{m x}{2}}{1-1+2 \sin ^{2} \frac{n x}{2}}$

$\Big[\because \quad \cos m x=1-2 \sin ^{2} \frac{m x}{2}$

and $\sin n x=1-2 \sin ^{2} \frac{n x}{2}\Big]$

$=\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{m x}{2}}{\sin ^{2} \frac{n x}{2}}=\lim _{x \rightarrow 0} \frac{\frac{\sin ^{2} \frac{m x}{2}}{(\frac{m x}{2})^{2}} \cdot (\frac{m x}{2})^{2}}{\frac{\sin ^{2} \frac{n x}{2}}{(\frac{n x}{2})^{2}} \cdot (\frac{n x}2)^{2}}\\ \=\frac{\lim _{x \rightarrow 0} \Big(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\Big)^{2}}{\lim _{x \rightarrow 0} \Big(\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\Big)^{2}} \cdot \frac{m^{2} }{n^{2}}$

$[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1]$

$[\because x \rightarrow 0 \Rightarrow k x \rightarrow 0]$

Solution

Given, $\lim _{x \rightarrow \pi / 3} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2} \Big(\frac{\pi}{3}-x\Big)}=\lim _{x \rightarrow \pi / 3} \frac{\sqrt{1-1+2 \sin ^{2} 3 x}}{\sqrt{2} \Big(\frac{\pi}{3}-x\Big)}$

$[\because \cos 2 x=1-2 \sin ^{2} x]$

$=\lim _{x \rightarrow \pi / 3} \frac{\sqrt{2} \sin 3 x}{\sqrt{2} \Big(\frac{\pi}{3}-x\Big)}=\lim _{x \rightarrow \pi / 3} \frac{\sin 3 x}{\frac{\pi}{3}-x}$

$=\lim _{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{\frac{\pi-3 x}{3}} \quad[\because \sin (\pi-\theta)=\sin \theta]$

$=3 \lim _{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{(\pi-3 x)}=3 \times 1 \quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1]$

$=3 \quad [\because x \rightarrow \frac{\pi}{3} \Rightarrow x-\frac{\pi}{3} \rightarrow 0]$

Solution

Given, $\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2} \Big(\sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}\Big)}{\Big(x-\frac{\pi}{4}\Big)}=\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2} \Big(\sin x \cos \frac{\pi}{4}-\cos x \cdot \sin \frac{\pi}{4}\Big)}{\Big(x-\frac{\pi}{4}\Big)}$

$ \begin{matrix} =\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2} \sin x-\frac{\pi}{4}}{x-\frac{\pi}{4}} & \\ \\ =\sqrt{2} \lim _{x \rightarrow \pi / 4} \frac{\sin x-\frac{\pi}{4}}{x-\frac{\pi}{4}}=\sqrt{2} & [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1] \end{matrix} $

$ [\because x \rightarrow \frac{\pi}{4} \Rightarrow x-\frac{\pi}{4} \rightarrow 0] $

Solution

Given, $\lim _{x \rightarrow \pi / 6} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}=\lim _{x \rightarrow \pi / 6} \frac{2 \frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x}{x-\frac{\pi}{6}}$

$ =\lim _{x \rightarrow \pi / 6} \frac{2 \Big(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}\Big)}{\Big(x-\frac{\pi}{6}\Big)}$ $=2 \lim _{x \rightarrow \pi / 6} \frac{\sin \Big(x-\frac{\pi}{6}\Big)}{\Big(x-\frac{\pi}{6}\Big)} $

$ =2 \quad [\because \sin A \cos B-\cos A \sin B=\sin (A-B)] $

$ [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1] $

$ [\because x\rightarrow\frac{\pi}{6} \Rightarrow \Big(x-\frac{\pi}{6}\Big)\rightarrow 0] $

Solution

Given, $\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x+3 x}{2 x} \cdot 2 x}{\frac{2 x+\tan 3 x}{3 x} \cdot 3 x}\\ $

$ =\lim _{x \rightarrow 0} \frac{\Big(\frac{\sin 2 x}{2 x}+\frac{3 x}{2 x}\Big) 2 x}{\Big(\frac{2 x}{3 x}+\frac{\tan 3 x}{3 x}\Big) 3 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{2 x}+\frac{3}{2}}{\frac{2}{3}+\frac{\tan 3 x}{3 x}} \cdot \frac{2}{3} $

$ \begin{aligned} & =\frac{2}{3} \lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{2 x}+\frac{3}{2}}{\frac{2}{3}+\lim _{x \rightarrow 0} \frac{\tan 3 x}{3 x}} \\ & =\frac{2}{3} \Big(\frac{1+\frac{3}{2}}{\frac{2}{3}+1} \Big)\quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1] \\ & =\frac{2}{3} \times \frac{\frac{2}{5}}{\frac{5}{3}}=\frac{2}{3} \times \frac{5}{2} \times \frac{3}{5}=1 \end{aligned} $

Solution

Given, $\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}=\lim _{x \rightarrow 0} \frac{2 \cos \Big(\frac{x+a}{2}\Big) \sin \Big(\frac{x-a}{2}\Big)}{\sqrt{x}-\sqrt{a}}$

$ \begin{aligned} & =\lim _{x \rightarrow a} \frac{2 \cos \Big(\frac{x+a}{2}\Big) \sin \Big(\frac{x-a}{2}\Big)(\sqrt{x+} \sqrt{a})}{(\sqrt{x-} \sqrt{a})(\sqrt{x}+\sqrt{a})} \\ & =\lim _{x \rightarrow 0} \frac{2 \cos \Big(\frac{x+a}{2}\Big) \sin \Big(\frac{x-a}{2}\Big)(\sqrt{x}+\sqrt{a})}{x-a} \\ & =2 \lim _{x \rightarrow a} \cos \Big(\frac{x+a}{2}\Big)(\sqrt{x}+\sqrt{a}) \lim _{x \rightarrow 0} \frac{\sin \Big(\frac{x-a}{2}\Big)}{\frac{x-a}{2}} \\ & =2 \lim _{x \rightarrow 0} \cos \Big(\frac{x+a}{2}\Big)(\sqrt{x}+\sqrt{a}) \cdot \frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin \Big(\frac{x-a}{2}\Big)}{\frac{x-a}{2}} \\ & =2 \cdot \cos \frac{a}{2} \cdot \sqrt{a} \cdot \frac{1}{2} \\ & =\sqrt{a} \cos \frac{a}{2} \end{aligned} $

Solution

Given $\lim _{x \rightarrow \pi / 6} \frac{\cot ^{2} x-3}{cosec x-2}=\lim _{x \rightarrow \pi / 6} \frac{cosec^{2} x-1-3}{cosec x-2} \quad[\because cosec^{2} x=1+\cot ^{2} x]$

$ \begin{aligned} & =\lim _{x \rightarrow \pi / 6} \frac{cosec^{2} x-4}{cosec x-2}=\lim _{x \rightarrow \pi / 6} \frac{(cosec x)^{2}-(2)^{2}}{cosec x-2} \\ & =\lim _{x \rightarrow \pi / 6} \frac{(cosec x+2)(cosec x-2)}{(cosec x-2)}=\lim _{x \rightarrow \pi / 6}(cosec x+2) \\ & =cosec \frac{\pi}{6}+2=2+2=4 \end{aligned} $

Solution

Given, $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+2 \cos ^{2} \frac{x}{2}-1}}{\sin ^{2} x} \quad [\because \cos x=2 \cos ^{2} \frac{x}{2}-1]$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{2 \cos ^{2} \frac{x}{2}}}{\sin ^{2} x} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{2}\Big( 1-\cos \frac{x}{2}\Big)}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{\sqrt{2} \Big(1-1+2 \sin ^{2} \frac{x}{4}\Big)}{\sin ^{2} x} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{2} \Big(2 \sin ^{2} \frac{x}{4}\Big)}{\sin ^{2} x}=\lim _{x \rightarrow 0} 2 \sqrt{2} \frac{\sin ^{2} \frac{x}{4}}{(\frac{x}{4})^{2}} \cdot \frac{\Big(\frac{x}{4}\Big)^{2}}{\sin ^{2} x} \\ & =2 \sqrt{2} \lim _{x \rightarrow 0} \Big(\frac{\sin ^{\frac{x}{4}}}{\frac{x}{4}}\Big)^{2} \cdot \lim _{x \rightarrow 0} \Big(\frac{x}{\sin x}\Big)^{2} \cdot \frac{1}{16} \\ & =2 \sqrt{2} \cdot 1 \cdot 1 \cdot \frac{1}{16}=\frac{1}{4 \sqrt{2}} \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x} & =\lim _{x \rightarrow 0} \frac{\sin 5 x+\sin x-2 \sin 3 x}{x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin 3 x(\cos 2 x-1)}{x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{\frac{1}{3} \times 3 x}(\cos 2 x-1)=6 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}(\cos 2 x-1) \\ & =6 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \cdot \lim _{x \rightarrow 0}(\cos 2 x-1)=6 \times 1 \times 0=0 \end{aligned} $

Solution

Given,

$ \lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}=\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}} $

$ \Rightarrow \quad 4(1)^{4-1}=\lim _{x \rightarrow k} \frac{\frac{x^{3}-k^{3}}{x-k}}{\frac{x^{2}-k^{2}}{x-k}} $

$ \begin{gathered}[ \because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a} \\ =n a^{n-1}] \end{gathered} $

$ \begin{matrix} \Rightarrow & 4=\frac{\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x-k}}{\lim _{x \rightarrow k} \frac{x^{2}-k^{2}}{x-k}} 1 \\ \\ \Rightarrow & 4=\frac{3 k^{2}}{2 k} \Rightarrow 4=\frac{3}{2} k \\ \\ \therefore & k=\frac{4 \times 2}{3}=\frac{8}{3} \end{matrix} $

Differentiate each of the functions w.r.t. $x$ in following questions

Solution

$\quad \frac{d}{d x} \Big(\frac{x^{4}+x^{3}+x^{2}+1}{x}\Big)=\frac{d}{d x} \Big(x^{3}+x^{2}+x+\frac{1}{x}\Big)$

$ \begin{aligned} & =\frac{d}{d x} x^{3}+\frac{d}{d x} x^{2}+\frac{d}{d x} x+\frac{d}{d x} \Big(\frac{1}{x}\Big) \\ & =3 x^{2}+2 x+1+\Big(-\frac{1}{x^{2}}\Big) \\ & =3 x^{2}+2 x+1-\frac{1}{x^{2}} \\ & =\frac{3 x^{4}+2 x^{3}+x^{2}-1}{x^{2}} \end{aligned} $

Solution

Let

$ y=\Big(x+\frac{1}x\Big)^{3} $

$ \begin{aligned} \therefore \quad \frac{d y}{d x}=\frac{d}{d x} \Big(x+\frac{1}x\Big)^{3} & =3 \Big(x+\frac{1}x\Big)^{3-1} \frac{d}{d x} \Big(x+\frac{1}{x}\Big) \\ & =3 \Big(x+\frac{1}x\Big)^{2} \Big(1-\frac{1}{x^{2}}\Big) \\ & =3 x^{2}-\frac{3}{x^{2}}-\frac{3}{x^{4}}+3 \end{aligned} $

Solution

Let

$ \begin{aligned} y & =(3 x+5)(1+\tan x) \\ \frac{d y}{d x} & =\frac{d}{d x}[(3 x+5)(1+\tan x)] \\ & =(3 x+5) \frac{d}{d x}(1+\tan x)+(1+\tan x) \frac{d}{d x}(3 x+5) \quad \text { [by product rule] } \\ & =(3 x+5)(\sec ^{2} x)+(1+\tan x) \cdot 3 \\ & =(3 x+5) \sec ^{2} x+3(1+\tan x) \\ & =3 x \sec ^{2} x+5 \sec ^{2} x+3 \tan x+3 \end{aligned} $

$ \therefore \quad \frac{d y}{d x}=\frac{d}{d x}[(3 x+5)(1+\tan x)] $

Solution

Let

$ \begin{matrix} y =(\sec x-1)(\sec x+1) & \\ y =(\sec ^{2}-1) & {[\because(a+b)(a-b)=a^{2}-b^{2}]} \\ =\tan ^{2} x & \\ \therefore\frac{d y}{d x} =2 \tan x \cdot \frac{d}{d x} \tan x & \\ =2 \tan x \cdot \sec ^{2} x \quad \text { [by chain rule] } \end{matrix} $

Solution

Let $\quad y=\frac{3 x+4}{5 x^{2}-7 x+9}$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{(5 x^{2}-7 x+9) \frac{d}{d x}(3 x+4)-(3 x+4) \frac{d}{d x}(5 x^{2}-7 x+9)}{(5 x^{2}-7 x+9)^{2}} \text { [by quotient rule] } \\ & =\frac{(5 x^{2}-7 x+9) \cdot 3-(3 x+4)(10 x-7)}{(5 x^{2}-7 x+9)^{2}} \\ & =\frac{15 x^{2}-21 x+27-30 x^{2}+21 x-40 x+28}{(5 x^{2}-7 x+9)^{2}} \\ & =\frac{-15 x^{2}-40 x+55}{(5 x^{2}-7 x+9)^{2}} \\ & =\frac{55-15 x^{2}-40 x}{(5 x^{2}-7 x+9)^{2}} \end{aligned} $

Solution

Let

$ \begin{aligned} y & =\frac{x^{5}-\cos x}{\sin x} \\ \frac{d y}{d x} & =\frac{\sin x \frac{d}{d x}(x^{5}-\cos x)-(x^{5}-\cos x) \frac{d}{d x} \sin x}{(\sin x)^{2}} \quad \text { [by quotient rule] } \\ & =\frac{\sin x(5 x^{4}+\sin x)-(x^{5}-\cos x) \cos x}{\sin ^{2} x} \\ & =\frac{5 x^{4} \sin x+\sin ^{2} x-x^{5} \cos x+\cos ^{2} x}{\sin ^{2} x} \\ & =\frac{5 x^{4} \sin x-x^{5} \cos x+\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x} \\ & =\frac{5 x^{4} \sin x-x^{5} \cos x+1}{\sin ^{2} x} \end{aligned} $

Solution

Let

$ \begin{aligned} y & =\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}=\frac{\frac{x^{2}}{\sqrt{2}}}{\sin x} \\ y & =\frac{1}{\sqrt{2}} \cdot \frac{x^{2}}{\sin x} \\ \frac{d y}{d x} & =\frac{1}{\sqrt{2}} \Big[\frac{\sin x \cdot \frac{d}{d x} x^{2}-x^{2} \frac{d}{d x} \sin x}{\sin ^{2} x}\Big] \\ & =\frac{1}{\sqrt{2}} \Big[\frac{\sin x \cdot 2 x-x^{2} \cdot \cos x}{\sin ^{2} x}\Big] \\ & =\frac{1}{\sqrt{2}} \cdot \frac{2 x \sin x-x^{2} \cos x}{\sin ^{2} x} \\ & =\frac{x}{\sqrt{2}}[2 cosec x-x \cot x cosec x] \\ & =\frac{x}{\sqrt{2}} cosec[2-x \cot x] \end{aligned} $ $ \quad [\therefore \quad \frac{d y}{d x}=\frac{1}{\sqrt{2}} \frac{\sin x \cdot \frac{d}{d x} x^{2}-x^{2} \frac{d}{d x} \sin x}{\sin ^{2} x}] $

Solution

Let $y=(a x^{2}+\cot x)(p+q \cos x)$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =(a x^{2}+\cot x) \frac{d}{d x}(p+q \cos x)+(p+q \cos x) \frac{d}{d x}(a x^{2}+\cot x) \quad \text { [by product rule] } \\ & =(a x^{2}+\cot x)(-q \sin x)+(p+q \cos x)(2 a x-cosec^{2} x) \\ & =-q \sin x(a x^{2}+\cot x)+(p+q \cos x)(2 a x-cosec^{2} x) \end{aligned} $

Solution

Let $\quad y=\frac{a+b \sin x}{c+d \cos x}$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{(c+d \cos x) \frac{d}{d x}(a+b \sin x)-(a+b \sin x) \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}} \text { [by quotinet rule] } \\ & =\frac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}} \\ & =\frac{b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}} \\ & =\frac{b \cos x+a d \sin x+b d(\cos ^{2} x+\sin ^{2} x)}{(c+d \cos x)^{2}} \\ & =\frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}} \end{aligned} $

Solution

Let

$ \begin{aligned} & y=(\sin x+\cos x)^{2} \\ & \therefore \quad \frac{d y}{d x}=2(\sin x+\cos x)(\cos x-\sin x) \\ & =2(\cos ^{2} x-\sin ^{2} x) & \text { [by chain rule] } \\ & =2 \cos 2 x & {[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x]} \end{aligned} $

Solution

Let

$ \begin{matrix} & y =(2 x-7)^{2}(3 x+5)^{3} \\ \\ & \frac{d y}{d x} =(2 x-7)^{2} \frac{d}{d x}(3 x+5)^{3}+(3 x+5)^{3} \frac{d}{d x}(2 x-7)^{2} & \text { [by product rule] } \\ \\ & =(2 x-7)^{2}(3)(3 x+5)^{2}(3)+(3 x+5)^{3} 2(2 x-7)(2) \quad \text { [by chain rule] } \\ \\ & =9(2 x-7)^{2}(3 x+5)^{2}+4(3 x+5)^{3}(2 x-7) \\ \\ & =(2 x-7)(3 x+5)^{2}[9(2 x-7)+4(3 x+5)] \\ \\ & =(2 x-7)(3 x+5)^{2}(18 x-63+12 x+20) \\ \\ & =(2 x-7)(3 x+5)^{2}(30 x-43) \end{matrix} $

Solution

Let

$ \begin{matrix} y =x^{2} \sin x+\cos 2 x \\ \\ \frac{d y}{d x} =\frac{d}{d x}(x^{2} \sin x)+\frac{d}{d x} \cos 2 x \\ \\ =x^{2} \cdot \cos x+\sin x 2 x+(-\sin 2 x) \cdot 2 \quad \text { [by product rule] } \\ \\ =x^{2} \cos x+2 x \sin x-2 \sin 2 x \quad \text { [by chain urle] } \end{matrix} $

Solution

Let

$ \begin{matrix} y =\sin ^{3} x \cos ^{3} x \\ \\ \frac{d y}{d x} =\sin ^{3} x \cdot \frac{d}{d x} \cos ^{3} x+\cos ^{3} x \frac{d}{d x} \sin ^{3} x \quad \text { [by product rule] } \\ \\ =\sin ^{3} x \cdot 3 \cos ^{2} x(-\sin x)+\cos ^{3} x \cdot 3 \sin ^{2} x \cos x \quad \text { [by chain rule] } \\ \\ =-3 \cos ^{2} x \sin ^{4} x+3 \sin ^{2} x \cos ^{4} x \\ \\ =3 \sin ^{2} x \cos ^{2} x(\cos ^{2} x-\sin ^{2} x) \\ \\ =3 \sin ^{2} x \cos ^{2} x \cos 2 x \\ \\ =\frac{3}{4}(2 \sin x \cos x)^{2} \cos 2 x \\ \\ =\frac{3}{4} \sin ^{2} 2 x \cos 2 x \end{matrix} $

Solution

Let $ y=\frac{1}{a x^{2}+b x+c}=(a x^{2}+b x+c)^{-1} $

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =-(a x^{2}+b x+c)^{-2}(2 a x+b) \quad \text { [by chain rule] } \\ & =\frac{-(2 a x+b)}{(a x^{2}+b x+c)^{2}} \end{aligned} $

Solution

Let

$ \begin{aligned} f(x) & =\cos (x^{2}+1) \text { and } f(x+h)=\cos \lbrace(x+h)^{2}+1\rbrace \\ \frac{d}{dx}f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\cos \lbrace(x+h)^{2}+1\rbrace -\cos (x^{2}+1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \lbrace\frac{(x+h)^{2}+1+x^{2}+1}{2}\rbrace \sin\lbrace \frac{(x+h)^{2}+1-x^{2}-1}{2}\rbrace}{h} \\ & [\because \cos C-\cos D=-2 \sin \lbrace\frac{C+D}{2}\rbrace \cdot \sin \lbrace\frac{C-D}{2}\rbrace] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\Big[-2 \sin \lbrace \frac{(x+h)^{2}+x^{2}+2}{2}\rbrace \sin \lbrace\frac{(x+h)^{2}-x^{2}}{2}\rbrace\Big] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\Big[-2 \sin \lbrace\frac{(x+h)^{2}+x^{2}+2}{2}\rbrace \sin \lbrace \frac{x^{2}+h^{2}+2 x h-x^{2}}{2}\rbrace\Big] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\Big[-2 \sin \lbrace\frac{(x+h)^{2}+x^{2}+2}{2}\rbrace \sin \lbrace \frac{h^{2}+2 h x}{2}\rbrace\Big] \\ & =-2 \lim _{h \rightarrow 0} \sin \lbrace \frac{(x+h)^{2}+x^{2}+2}{2}\rbrace \lim _{h \rightarrow 0}\lbrace \frac{\sin h \frac{h+2 x}{2}}{h \frac{h+2 x}{2}}\rbrace \times \Big(\frac{h+2 x}{2}\Big) \\ & =-2 \lim _{h \rightarrow 0} \sin \lbrace \frac{(x+h)^{2}+x^{2}+2}{2}\rbrace \lim _{h \rightarrow 0} \Big(\frac{h+2 x}{2}\Big) \quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1] \\ & =-2 x \sin (x^{2}+1) \end{aligned} $

$ \therefore \quad \frac{d}{d x} f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $

Solution

Let $ f(x)=\frac{a x+b}{c x+d} $

$ f(x+h)=\frac{a(x+h)+b}{c(x+h)+d} $

$ \frac{d}{d x} f(x)=\lim _{h \rightarrow 0} \frac{1}{h}[f(x+h)-f(x)] $

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[\frac{a(x+h)+b}{c(x+h)+d}-\frac{a x+b}{c x+d}\Big] $

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[\frac{a x+b+a h}{c(x+h)+d}-\frac{a x+b}{c x+d}\Big] $

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[\frac{(a x+a h+b)(c x+d)-(a x+b{c(x+h)+d}}{{c(x+h)+d}(c x+d)} \Big]$

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[\frac{(a x+a h+b)(c x+d)-(a x+b)(c x+c h+d)}{\lbrace c(x+h)+d)\rbrace(c x+d)} \Big]$

$ a c x^{2}+a c h x+b c x+a d x+a d h+b d $

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[\frac{-{a c x^{2}+a c h x+a d x+b c x+b c h+b d}}{\lbrace c(x+h)+d\rbrace(c x+d)}\Big]$

$ a c x^{2}+a c h x+b c x+a d x+a d h+b d $

$ =\lim _{h \rightarrow 0} \frac{1}{h}\Big[ \frac{-a c x^{2}-a c h x-a d x-b c x-b c h-b d}{\lbrace c(x+h)+d\rbrace(c x+d)} \Big]$

$ =\lim _{h \rightarrow 0} \frac{1}{h} \Big[ \frac{a d h-b c h}{\lbrace c(x+h)+d\rbrace(c x+d)}\Big] $

$ =\lim _{h \rightarrow 0} \frac{a c-b d}{{c(x+h)+d}(c x+d)} $

$ =\frac{a c-b d}{(c x+d)^{2}} $

Solution

Let

Now, $ \begin{aligned} f(x) & =x^{2 / 3} \\ f(x+h) & =(x+h)^{2 / 3} \\ \frac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\Big[(x+h)^{2 / 3}-x^{2 / 3}\Big] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\Big[ x^{2 / 3} \Big(1+\frac{h}{x} \Big)^{2 / 3}-x^{2 / 3}\Big] \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h} \Big[x^{2 / 3} \Big(1+\frac{h}{x} \cdot \frac{2}{3}+\frac{2}{3} \Big(\frac{2}{3}-1\Big) \frac{h^{2}}{x^{2}}+\cdots\Big)-1\Big] \\ & \quad [\because(1+x)^{h}=1+n x+\frac{h(n-1)}{2 !} x^{2}+\cdots] \\ & =\lim _{h \rightarrow 0} \frac{1}{h} \Big[x^{2 / 3} \Big(\frac{2}{3} \cdot \frac{h}{x}-\frac{2}{9} \cdot \frac{h^{2}}{x^{2}}+\cdots\Big)\Big] \\ & =\lim _{h \rightarrow 0} \frac{x^{2 / 3}}{h} \cdot \frac{2}{3} \frac{h}{x} \Big(1-\frac{1}{3} \cdot \frac{h}{x}+\cdots\Big) \\ & =\frac{2}{3} x^{2 / 3-1}=\frac{2}{3} x^{-1 / 3} \end{aligned} $

Alternate Method

Let $ \begin{matrix} f(x) =x^{2 / 3} \\ \\ f(x+h) =(x+h)^{2 / 3} \\ \\ \frac{d}{d x} f(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \\ =\lim _{(h \rightarrow 0)} \Big[\frac{(x+h) 2 / 3-x^{2 / 3}}{h}\Big]=\lim _{(x+h) \rightarrow x} \Big[ \frac{(x+h) 2 / 3-x^{2 / 3}}{(x+h)-x}\Big] \\ \\ =\frac{2}{3}(x)^{2 / 3-1} & \quad [\because \lim _{x \rightarrow a} \frac{x^{x}-a^{n}}{x-a}=n a^{n-1}] \\ \\ =\frac{2}{3} x^{-1 / 3} \end{matrix} $

Solution

Let

$ \begin{aligned} f(x) & =x \cos x \\ \therefore f(x+h) & =(x+h) \cos (x+h) \\ \therefore \frac{d}{dx}f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[(x+h) \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x \cos (x+h)+h \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x\lbrace\cos (x+h)-\cos x\rbrace+h \cos (x+h)] \\ & =\lim _{h \rightarrow 0} \frac{1}{h} \Big[x\lbrace-2 \sin \frac{2 x+h}{2} \sin \frac{h}{2}\rbrace+h \cos (x+h)\Big] \\ & =\lim _{h \rightarrow 0}\Big[-2 x \Big(\sin x+\frac{h}{2}\Big) \frac{\sin \frac{h}{2}}{h}+\cos (x+h)\Big] \\ & =-2 \lim _{h \rightarrow 0} \Big(x \sin x+\frac{h}{2}\Big) \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \frac{1}{2}+\lim _{h \rightarrow 0} \cos (x+h) \\ & =-2 \cdot \frac{1}{2} x \sin x+\cos x \\ & =\cos x-x \sin x \end{aligned} $

Evaluate each of the following limits in following questions

Solution

Given, $\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}$

$ \begin{aligned} & =\lim _{y \rightarrow 0} \frac{\frac{x+y}{\cos (x+y)}-\frac{x}{\cos x}}{y} \\ & =\lim _{y \rightarrow 0} \frac{(x+y) \cos x-x \cos (x+y)}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0} \frac{x \cos x+y \cos x-x \cos (x+y)}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0} \frac{x \cos x-x \cos (x+y)+y \cos x}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0} \frac{x\lbrace\cos x-\cos (x+y)\rbrace+y \cos x}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0} \frac{x\Big[-2 \sin \Big(x+\frac{y}{2}\Big) \sin \Big(\frac{-y}{2}\Big)\Big]+y \cos x}{y \cos x \cos (x+y)} \\ & [\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}] \end{aligned} $

$ \begin{aligned} & =\lim _{y \rightarrow 0} \Big[\frac{x \lbrace 2 \sin (x+\frac{y}{2}) \sin \frac{y}{2}\rbrace +y \cos x}{y \cos x \cos (x+y)}\Big] \\ & =\lim _{y \rightarrow 0} \frac{2 x \sin (x+\frac{y}{2})}{\cos x \cos (x+y)} \cdot \lim _{y \rightarrow 0} \frac{\sin \frac{y}{2}}{\frac{y}{2}} \cdot \frac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y) \\ & \\ & \quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } x \rightarrow 0 \Rightarrow k x \rightarrow 0] \end{aligned} $

$ \begin{aligned} & =\lim _{y \rightarrow 0} \frac{2 x \sin (x+\frac{y}{2})}{\cos x \cos (x+y)} \cdot \frac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y) \\ & =\frac{2 x \sin x}{\cos x \cos x} \cdot \frac{1}{2}+\sec x \\ & =x \tan x \sec x+\sec x \\ & =\sec x(x \tan x+1) \end{aligned} $

Solution

Given, $\lim _{x \rightarrow 0} \frac{[\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha x]}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x \\ $

$ =\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cdot \cos \beta x+\sin 2 \alpha x]}{\cos 2 \beta x-\sin +\cos 2 \alpha x}=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2} \\ $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cos \beta x+\sin 2 \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \quad [\because \cos C-\cos D=2 \sin \frac{C+D}{2} \cdot \sin \frac{D-C}{2}] \\ \\ & =\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cos \beta x+2 \sin \alpha x \cos \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \\ \\ & =\lim _{x \rightarrow 0} \frac{2 \sin \alpha x[\cos \beta x+\cos \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \\ \\ & \begin{matrix} =\lim _{x \rightarrow 0} \frac{\sin \alpha x \Big[2 \cos \frac{\alpha+\beta}{2} x \cos \frac{\alpha-\beta}{2} \Big]\times x}{2 \sin \frac{\alpha+\beta}{2} x \cos \frac{\alpha+\beta}{2} x \cdot 2 \sin \frac{\alpha-\beta}{2} x \cos \frac{\alpha-\beta}{2} x} \\ \\ [\because \cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2} \text { and } \sin 2 \theta=2 \sin \theta \cos \theta] \end{matrix} \\ & =\lim _{x \rightarrow 0} \frac{\sin \alpha x \cdot x}{2 \sin \frac{\alpha+\beta}{2} x \sin \frac{\alpha-\beta}{2} x} \\ & =\frac{1}{2} \lim _{x \rightarrow 0} \frac{\frac{\sin \alpha x}{\alpha x} \cdot x \cdot(\alpha x)}{2 \sin \frac{\frac{\alpha+\beta}{2} x}{\frac{\alpha+\beta}{2} x} \cdot \sin \frac{\frac{\alpha-\beta}{2} x}{\frac{\alpha-\beta}{2} x} \cdot \frac{\alpha+\beta}{2} x \cdot \frac{\alpha-\beta}{2} x} \\ & =\frac{\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin \alpha x}{\alpha x} \cdot \alpha x^{2}}{\lim _{x \rightarrow 0} \sin \frac{\frac{\alpha+\beta}{2} x}{\frac{\alpha+\beta}{2} \lim _{x \rightarrow 0} \frac{\frac{\alpha-\beta}{2} x}{\frac{\alpha-\beta}{2} x} \cdot \frac{\alpha^{2}-\beta^{2}}{4} x^{2}}} \end{aligned} $

$\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ and $x \rightarrow 0 \Rightarrow k x \rightarrow 0$

$ \begin{aligned} & =\frac{1}{2} \cdot \frac{\alpha \cdot 4}{\alpha^{2}-\beta^{2}} \Big[\frac{\lim _{x \rightarrow 0} \frac{\sin \alpha x}{\alpha x}}{\lim _{x \rightarrow 0} \sin \frac{\frac{\alpha+\beta}{2} x}{\frac{\alpha+\beta}{2} x} \lim _{x \rightarrow 0} \sin \frac{\frac{\alpha-\beta}{2} x}{\frac{\alpha-\beta}{2} x}}\Big] \\ \\ & =\frac{1}{2} \cdot \frac{4 \alpha}{\alpha^{2}-\beta^{2}}=\frac{2 \alpha}{\alpha^{2}-\beta^{2}} \end{aligned} $

Solution

Given, $\lim _{x \rightarrow \pi / 4} \frac{\tan ^{3} x-\tan x}{\cos \Big(x+\frac{\pi}{4}\Big)}$

$ \begin{matrix} =\lim _{x \rightarrow \pi / 4} \frac{\tan x(\tan ^{2} x-1)}{\cos \Big(x+\frac{\pi}{4}\Big)}=\lim _{x \rightarrow \pi / 4} \tan x \cdot \lim _{x \rightarrow \pi / 4} \Big(\frac{1-\tan ^{2} x}{\cos \Big(x+\frac{g \pi}{4}\Big)}\Big) \\ =-1 \times \lim _{x \rightarrow \pi / 4} \frac{(1+\tan x)(1-\tan x)}{\cos (x+\frac{\pi}{4})} & {[\because a^{2}-b^{2}=(a+b)(a-b)]} \end{matrix} $

$ =-\lim _{x \rightarrow \pi / 4}(1+\tan x) \lim _{x \rightarrow \pi / 4} \Big[\frac{\cos x-\sin x}{\cos x \cdot \cos \Big(x+\frac{\pi}{4}\Big)}\Big] $

$ \begin{aligned} & =-(1+1) \times \lim _{x \rightarrow \pi / 4} \frac{\sqrt{2} \Big[\frac{1}{\sqrt{2}} \cdot \cos x-\frac{1}{\sqrt{2}} \cdot \sin x\Big]}{\cos x \cdot \cos x+\frac{\pi}{4}}=-2 \sqrt{2} \lim _{x \rightarrow \pi / 4} \Big[\frac{\cos \frac{\pi}{4} \cdot \cos x-\sin \frac{\pi}{4} \cdot \sin x}{\cos x \cdot \cos x+\frac{\pi}{4}}\Big] \\ \\ & {[\because \cos A \cdot \cos B-\sin A \sin B=\cos (A+B)]} \\ \\ & =-2 \sqrt{2} \lim _{x \rightarrow \pi / 4} \frac{\cos \Big(x+\frac{\pi}{4}\Big)}{\cos x \cdot \cos \Big(x+\frac{\pi}{4}\Big)}=-2 \sqrt{2} \times \frac{1}{\frac{1}{\sqrt{2}}}=-2 \sqrt{2} \times \sqrt{2}=-4 \end{aligned} $

Solution

Given, $\quad \lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}\Big(\cos \frac{x}{4}-\sin \frac{x}{4}\Big)}$

$ \begin{aligned} & =\lim _{x \rightarrow \pi} \frac{\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}-2 \cdot \sin \frac{x}{4} \cdot \cos \frac{x}{4}}{\cos \frac{x}{2} \cdot(\cos \frac{x}{4}-\sin \frac{x}{4})} \quad[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \sin 2 \theta=2 \sin \theta \cos \theta] \\ & =\lim _{x \rightarrow \pi} \frac{\Big(\cos \frac{x}{4}-\sin \frac{x}{4}\Big)^{2}}{\Big(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\Big) \Big(\cos\frac{x}{4}-\sin \frac{x}{4}\Big)} \quad[\because \cos ^{2} 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta] \end{aligned} $

$ \begin{aligned} & =\lim _{x \rightarrow \pi} \frac{\cos \frac{x}{4}-\sin \frac{x}{4}}{\Big(\cos \frac{x}{4}+\sin \frac{x}{4}\Big) \Big( \cos \frac{x}{4}-\sin \frac{x}{4} \Big)} \\ & \lim _{x \rightarrow \pi} \frac{1}{\cos \frac{x}{4}+\sin \frac{x}{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{aligned} $

Solution

Given,

$ \lim _{x \rightarrow \pi / 4} \frac{|x-4|}{x-4} $

$ LHL =\lim _{x \rightarrow \frac{\pi^{-}}{4}} \frac{-(x-4)}{x-4} {[\because|x-4|=-(x-4), x<4]} $

$=-1$

$ RHL =\lim _{x \rightarrow \frac{\pi^{+}}{4}} \frac{(x-4)}{x-4}=1 {[\because|x-4|=(x-4), x>4]} $

$\therefore \quad$ LHL $\neq$ RHL

So, limit does not exist.

Solution

Given,

$\quad f(x)=\begin{cases}\frac{k \cos x}{\pi-2 x}, \quad x \neq \frac{\pi}{2}\\ \quad 3, \quad x=\frac{\pi}{2}\end{cases} \\ $

$ LHL=\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \frac{\pi}{2}-h}{\pi-2 \frac{\pi}{2}-h} $

$ =\lim _{h \rightarrow 0} \frac{k \sinh }{\pi-\pi+2 h}=\lim _{h \rightarrow 0} \frac{k \sinh }{2 h} $

$ =\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2} \quad \because \lim _{h \rightarrow 0} \frac{\sin x}{x}=1 $

$ RHL=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi}{2}}+\frac{k \cos \frac{\pi}{2}+h}{\pi-2 \frac{\pi}{2}+h}$

$ =\lim _{h \rightarrow 0} \frac{-k \sin h}{\pi-\pi-2 h}=\lim _{h \rightarrow 0} \frac{k \sinh }{2 h}=\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{2 h}=\frac{k}{2} \text { and } f \frac{\pi}{2}=3 $

$ \text { It is given that, } \quad \lim _{x \rightarrow \pi / 2} f(x)=f \frac{\pi}{2} \Rightarrow \frac{k}{2}=3 $

$ \therefore \quad k=6 $

Solution

Given,

$ \begin{aligned} & f(x)=\begin{cases} x+2, & x \leq-1 \\ c x^{2}, & x>-1 \end{cases} \\ & LHL=\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow–^{-}}(x+2) \\ & =\lim _{h \rightarrow 0}(-1-h+2)=\lim _{h \rightarrow 0}(1-h)=1 \\ & RHL=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} c x^{2}=\lim _{h \rightarrow 0} c(-1+h)^{2} \end{aligned} $

$ \begin{aligned} & \therefore \\ & \text { If } \lim _{x \rightarrow-1} f(x) \text { exist, then } LHL=RHL=c \\ & \therefore \quad c=1 \end{aligned} $

Solution

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

$[\because \sin \theta=\sin (\pi-\theta)]$

$ =-\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{(\pi-x)}=-1 \quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \pi-x \rightarrow 0 \Rightarrow x \rightarrow \pi] $

• Option (a) 1: This is incorrect because the limit $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ does not simplify to 1. The correct simplification involves recognizing that $\sin(\pi - x) = \sin x$ and using the limit property $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$, which results in -1, not 1.

• Option (b) 2: This is incorrect because there is no mathematical operation or simplification in the given limit that would result in a factor of 2. The limit $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ simplifies to -1, not 2.

• Option (d) -2: This is incorrect because the limit $\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ does not involve any factor that would result in -2. The correct simplification results in -1, not -2.

Solution

(a) Given, $\lim _{x \rightarrow 0} \frac{x^{2} \cos x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x^{2} \cos x}{2 \sin ^{2} \frac{x}{2}}$

$[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}]$

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

• Option (b) $\frac{3}{2}$ is incorrect because the limit calculation does not yield $\frac{3}{2}$. The correct limit, as shown, is 2.

• Option (c) $\frac{-3}{2}$ is incorrect because the limit calculation does not yield a negative value. The correct limit is positive and equals 2.

• Option (d) 1 is incorrect because the limit calculation does not yield 1. The correct limit, as shown, is 2.

Solution

(a) Given, $\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}=\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{(1+x)-1}=\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{(1+x)-1}$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1^{n}}{(1+x)-1}=\lim _{(1+x) \rightarrow 1} \frac{(1+x)^{n}-1^{n}}{(1+x)-1} \\ & =n \cdot(1)^{n-1}=n \quad \because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1} \end{aligned} $

• Option (b) 1: The limit evaluates to ( n ), not 1. The expression (\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}) simplifies to ( n ) using the binomial expansion or derivative approach, not to 1.

• Option (c) -n: The limit evaluates to ( n ), not (-n). The expression (\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}) simplifies to ( n ) using the binomial expansion or derivative approach, not to (-n).

• Option (d) 0: The limit evaluates to ( n ), not 0. The expression (\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}) simplifies to ( n ) using the binomial expansion or derivative approach, not to 0.

Solution

(b) Given, $\quad \lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}=\lim _{x \rightarrow 1} \frac{\frac{x^{m}-1}{x-1}}{\frac{x^{n}-1}{x-1}}=\frac{\lim _{x \rightarrow 1} \frac{x^{m}-1^{m}}{x-1}}{\lim _{x \rightarrow 1} \frac{x^{n}-1^{n}}{x-1}}$

$ =\frac{m(1)^{m-1}}{n(1)^{n-1}}=\frac{m}{n} \quad [\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

• Option (a) 1 is incorrect because the limit $\lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}$ depends on the exponents (m) and (n). It simplifies to (\frac{m}{n}), not 1, unless (m = n).

• Option (c) $-\frac{m}{n}$ is incorrect because the limit $\lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}$ results in a positive ratio (\frac{m}{n}) when both (m) and (n) are positive. There is no negative sign involved in the simplification.

• Option (d) $\frac{m^{2}}{n^{2}}$ is incorrect because the limit $\lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}$ simplifies directly to (\frac{m}{n}), not (\frac{m^{2}}{n^{2}}). The exponents do not get squared in the process.

Solution

(a) Given, $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}=\lim _{\theta \rightarrow 0} \frac{2 \sin ^{2} 2 \theta}{2 \sin ^{2} 3 \theta}$

$[\because 1-\cos 2 \theta=2 \sin ^{2} \theta]$

$ \begin{aligned} & =\frac{\lim _{\theta \rightarrow 0} \frac{\sin ^{2} 2 \theta}{(2 \theta)^{2}} \cdot(2 \theta)^{2}}{\lim _{\theta \rightarrow 0} \frac{\sin ^{2} 3 \theta}{(3 \theta)^{2}} \cdot(3 \theta)^{2}}=\frac{4}{9} \cdot \frac{\lim _{\theta \rightarrow 0} \Big(\frac{\sin 2 \theta}{2 \theta}\Big)^{2}}{\lim _{\theta \rightarrow 0} \Big(\frac{\sin 3 \theta}{3 \theta}\Big)^{2}} \quad [\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } x \rightarrow 0 \Rightarrow k x \rightarrow 0] \\ & =\frac{4}{9} \end{aligned} $

• Option (b) $\frac{1}{2}$ is incorrect because the limit calculation shows that the ratio of the squared sine terms results in $\frac{4}{9}$, not $\frac{1}{2}$.

• Option (c) $\frac{-1}{2}$ is incorrect because the limit calculation involves positive squared sine terms, which cannot result in a negative value.

• Option (d) -1 is incorrect because the limit calculation involves positive squared sine terms, and the ratio of these terms results in a positive fraction, not a negative value.

Solution

(c) Given,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{cosec x-\cot x}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}{x}=\lim _{x \rightarrow 0} \frac{1-\cos x}{x \cdot \sin x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2}}=\lim _{x \rightarrow 0} \frac{\tan \frac{x}{2}}{x} \\ & =\lim _{x \rightarrow 0} \frac{\tan \frac{x}{2}}{\frac{x}{2}} \cdot \frac{1}{2}=\frac{1}{2} \end{aligned} $

$ [\because \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1] $

• Option (a) $\frac{-1}{2}$ is incorrect because the limit calculation shows that the expression evaluates to $\frac{1}{2}$, not $\frac{-1}{2}$.
• Option (b) 1 is incorrect because the limit calculation shows that the expression evaluates to $\frac{1}{2}$, not 1.
• Option (d) 1 is incorrect because the limit calculation shows that the expression evaluates to $\frac{1}{2}$, not 1.

Solution

(c) Given, $\quad \lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$

$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}} \cdot \frac{\sqrt{x+1}+\sqrt{1-x}}{\sqrt{x+1}+\sqrt{1-x}} \\ & =\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+1}+\sqrt{1-x})}{(x+1)-(1-x)} \\ & =\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+1}+\sqrt{1-x})}{x+1-1+x} \\ & =\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x} \lim _{x \rightarrow 0}(\sqrt{x+1}+\sqrt{1-x}) \\ & =\frac{1}{2} \cdot 1 \cdot 2=1 \end{aligned} $

• Option (a) 2 is incorrect because the limit calculation shows that the final result is 1, not 2. The steps involve simplifying the expression and using the limit properties, which do not lead to the value 2.

• Option (b) 0 is incorrect because the limit calculation shows that the final result is 1, not 0. The expression simplifies in such a way that the numerator and denominator do not cancel out to zero.

• Option (d) -1 is incorrect because the limit calculation shows that the final result is 1, not -1. The steps involve rationalizing the denominator and using the limit properties, which do not lead to a negative value.

Solution

(d) Given,

$ \begin{aligned} & \lim _{x \rightarrow \pi / 4} \frac{\sec ^{2} x-2}{\tan x-1} \\ = & \lim _{x \rightarrow \pi / 4} \frac{1+\tan ^{2} x-2}{\tan x-1}=\lim _{x \rightarrow \pi / 4} \frac{\tan ^{2} x-1}{\tan x-1} \\ = & \lim _{x \rightarrow \pi / 4} \frac{(\tan x+1)(\tan x-1)}{(\tan x-1)}=\lim _{x \rightarrow \pi / 4}(\tan x+1) \\ = & 2 \end{aligned} $

• Option (a) 3: This option is incorrect because when evaluating the limit, the expression simplifies to (\lim_{x \rightarrow \pi / 4} (\tan x + 1)). Since (\tan(\pi / 4) = 1), the limit evaluates to (1 + 1 = 2), not 3.

• Option (b) 1: This option is incorrect because the simplified expression (\lim_{x \rightarrow \pi / 4} (\tan x + 1)) evaluates to (2) when (\tan(\pi / 4) = 1). Therefore, the limit is not 1.

• Option (c) 0: This option is incorrect because the limit (\lim_{x \rightarrow \pi / 4} (\tan x + 1)) evaluates to (2) when (\tan(\pi / 4) = 1). Hence, the limit is not 0.

Solution

(b) Given, $\quad \lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^{2}+x-3}=\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{(2 x+3)(x-1)}$

$ \begin{aligned} & =\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{(2 x+3)(\sqrt{x}-1)(\sqrt{x}+1)} \\ & =\lim _{x \rightarrow 1} \frac{2 x-3}{(2 x+3)(\sqrt{x}+1)}=\frac{-1}{5 \times 2}=\frac{-1}{10} \end{aligned} $

• Option (a) $\frac{1}{10}$ is incorrect because the correct limit calculation results in a negative value, specifically $\frac{-1}{10}$, not a positive value.

• Option (c) 1 is incorrect because the correct limit calculation does not simplify to 1. The numerator and denominator do not cancel out in a way that would result in the limit being 1.

• Option (d) None of these is incorrect because the correct answer is indeed one of the provided options, specifically option (b) $\frac{-1}{10}$.

Solution

(d) Given,

f(x)= $\begin{cases} \frac{\sin [x]}{[x]}, & {[x] \neq 0 } \\ 0, & {[x]=0 } \end{cases} $

$\therefore$ $ LHL=\lim _{x \rightarrow 0^{-}} f(x) $

$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{\sin [x]}{[x]}=\lim _{h \rightarrow 0} \frac{\sin [0-h]}{[0-h]} \\ & =\lim _{h \rightarrow 0} \frac{-\sin [-h]}{[-h]}=-1 \\ RHL & =\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sin [x]}{[x]} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\sin [0+h]}{[0+h]}=\lim _{h \rightarrow 0} \frac{\sin [h]}{[h]}=1 \end{aligned} $

$\because$ $LHL \neq RHL$

So, limit does not exist.

• Option (a) 1: This option is incorrect because the left-hand limit (LHL) and the right-hand limit (RHL) are not equal. The LHL is -1 and the RHL is 1, so the limit does not exist and cannot be 1.

• Option (b) 0: This option is incorrect because neither the left-hand limit (LHL) nor the right-hand limit (RHL) is 0. The LHL is -1 and the RHL is 1, so the limit does not exist and cannot be 0.

• Option (c) -1: This option is incorrect because the left-hand limit (LHL) is -1, but the right-hand limit (RHL) is 1. Since the LHL and RHL are not equal, the limit does not exist and cannot be -1.

Solution

(c) Given,

$ \begin{aligned} \text { limit } & =\lim _{x \rightarrow 0} \frac{|\sin x|}{x} \\ \therefore \quad LHL=\lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x}=-\lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}=-1 \\ RHL & =\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1 \end{aligned} $

$\because \quad LHL \neq RHL$

So, limit does not exist.

• Option (a) is incorrect because the left-hand limit (LHL) and the right-hand limit (RHL) are not equal. For the limit to exist and be equal to 1, both LHL and RHL must be equal to 1, but in this case, LHL is -1 and RHL is 1.

• Option (b) is incorrect because the limit is not equal to -1. The right-hand limit (RHL) is 1, not -1. For the limit to be -1, both LHL and RHL must be equal to -1, but they are not.

• Option (d) is incorrect because the correct answer is provided in option (c), which states that the limit does not exist due to the inequality of LHL and RHL.

Solution

(d) Given,

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

$ \begin{aligned} \therefore \quad \lim _{x \rightarrow 2^{-}} f(x) & =\lim _{x \rightarrow 2^{-}}(x^{2}-1) \\ & =\lim _{h \rightarrow 0}[(2-h)^{2}-1]=\lim _{h \rightarrow 0}(4+h^{2}-4 h-1) \\ & =\lim _{h \rightarrow 0}(h^{2}-4 h+3)=3 \end{aligned} $

and $\quad \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)$

$ =\lim _{h \rightarrow 0}[2(2+h)+3]=\lim _{h \rightarrow 0}(4+2 h+3)=7 $

So, the quadratic equation whose roots are 3 and 7 is $x^{2}-(3+7) x+3 \times 7=0$ i.e., $x^{2}-10 x+21=0$.

• Option (a) (x^{2}-6x+9=0):

  • The roots of this quadratic equation are 3 and 3 (since ((x-3)^2 = 0)).
  • However, the limits (\lim_{x \rightarrow 2^{-}} f(x)) and (\lim_{x \rightarrow 2^{+}} f(x)) are 3 and 7, respectively. Therefore, the roots should be 3 and 7, not 3 and 3.

• Option (b) (x^{2}-7x+8=0):

  • The roots of this quadratic equation are 1 and 7 (since ((x-1)(x-7) = 0)).
  • The limits (\lim_{x \rightarrow 2^{-}} f(x)) and (\lim_{x \rightarrow 2^{+}} f(x)) are 3 and 7, respectively. Therefore, the roots should be 3 and 7, not 1 and 7.

• Option (c) (x^{2}-14x+49=0):

  • The roots of this quadratic equation are 7 and 7 (since ((x-7)^2 = 0)).
  • The limits (\lim_{x \rightarrow 2^{-}} f(x)) and (\lim_{x \rightarrow 2^{+}} f(x)) are 3 and 7, respectively. Therefore, the roots should be 3 and 7, not 7 and 7.

Solution

(b) Given,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{x \Big[\frac{\tan 2 x}{x}-1\Big]}{x\Big[3-\frac{\sin x}{x}\Big]} \\ & =\frac{\lim _{x \rightarrow 0} 2 \times \frac{\tan 2 x}{2 x}-1}{3-\lim _{x \rightarrow 0} \frac{\sin x}{x}}=\frac{2-1}{3-1}=\frac{1}{2} \end{aligned} $

• Option (a) 2: This option is incorrect because the limit calculation shows that the numerator approaches 1 and the denominator approaches 2, resulting in a limit of (\frac{1}{2}), not 2.

• Option (c) (\frac{-1}{2}): This option is incorrect because the limit calculation shows that both the numerator and the denominator are positive as (x) approaches 0, resulting in a positive limit of (\frac{1}{2}), not a negative value.

• Option (d) (\frac{1}{4}): This option is incorrect because the limit calculation shows that the numerator approaches 1 and the denominator approaches 2, resulting in a limit of (\frac{1}{2}), not (\frac{1}{4}).

Solution

(b) Given, $f(x)=x-[x]$

Now, first we have to check the differentiability of $f(x)$ at $x=\frac{1}{2}$.

$\therefore \quad L f^{\prime} \frac{1}{2}=LHD=\lim _{h \rightarrow 0} \frac{f \Big(\frac{1}{2}-h\Big)-f \Big(\frac{1}{2}\Big)}{-h}$

$ =\lim _{n \rightarrow 0} \frac{\Big(\frac{1}{2}-h\big)-\big(\frac{1}{2}-h\Big)-\frac{1}{2}+\frac{1}{2}}{h}=\lim _{h \rightarrow 0} \frac{\frac{1}{21}-h-0-\frac{1}{2}+0=1}{h} $

and $\quad R f^{\prime} \frac{1}{2}=RHD \lim _{h \rightarrow 0} \frac{f \Big(\frac{1}{2}+h\Big)-f \Big(\frac{1}{2}\Big)}{h}$

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

$\because \quad LHL=RHD$

$\therefore \quad f^{\prime} \frac{1}{2}=1$

• Option (a) $\frac{3}{2}$ is incorrect because the derivative of $f(x) = x - [x]$ at $x = \frac{1}{2}$ is calculated to be 1, not $\frac{3}{2}$. The function $f(x)$ is a piecewise linear function with a slope of 1, so the derivative cannot be $\frac{3}{2}$.

• Option (c) 0 is incorrect because the derivative of $f(x) = x - [x]$ at $x = \frac{1}{2}$ is calculated to be 1, not 0. The function $f(x)$ has a constant slope of 1 between integer points, so the derivative is not zero.

• Option (d) -1 is incorrect because the derivative of $f(x) = x - [x]$ at $x = \frac{1}{2}$ is calculated to be 1, not -1. The function $f(x)$ has a positive slope of 1, so the derivative cannot be negative.

Solution

(d) Given,

$ \begin{aligned} & \text { Now, } \\ & \therefore \quad \frac{d y}{d x}=\frac{1}{2}-\frac{1}{2}=0 \end{aligned} $

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

• Option (a) 1: This is incorrect because the derivative of ( y = \sqrt{x} + \frac{1}{\sqrt{x}} ) is not equal to 1 at ( x = 1 ). The correct derivative calculation shows that the derivative at ( x = 1 ) is 0.

• Option (b) (\frac{1}{2}): This is incorrect because the derivative of ( y = \sqrt{x} + \frac{1}{\sqrt{x}} ) at ( x = 1 ) does not yield (\frac{1}{2}). The correct derivative calculation shows that the derivative at ( x = 1 ) is 0.

• Option (c) (\frac{1}{\sqrt{2}}): This is incorrect because the derivative of ( y = \sqrt{x} + \frac{1}{\sqrt{x}} ) at ( x = 1 ) does not yield (\frac{1}{\sqrt{2}}). The correct derivative calculation shows that the derivative at ( x = 1 ) is 0.

Solution

(a) Given,

$f(x)^{5}=\frac{x-4}{2 \sqrt{x}}$

$ \text { Now, } \quad \begin{aligned} f^{\prime}(x) & =\frac{2 \sqrt{x}-(x-4) \cdot 2 \cdot \frac{1}{2 \sqrt{x}}}{4 x} \\ & =\frac{2 x-(x-4)}{4 x^{3 / 2}}=\frac{2 x-x+4}{4 x^{3 / 2}} \\ & =\frac{x+4}{4 x^{3 / 2}} \\ \therefore \quad f^{\prime}(1) & =\frac{1+4}{4 \times(1)^{3 / 2}}=\frac{5}{4} \end{aligned} $

• Option (b) $\frac{4}{5}$: This option is incorrect because the correct derivative calculation yields $\frac{5}{4}$, not $\frac{4}{5}$. The numerator in the final expression for $f’(1)$ is $5$, and the denominator is $4$, leading to $\frac{5}{4}$.

• Option (c) 1: This option is incorrect because the correct derivative calculation does not simplify to 1. The final expression for $f’(1)$ is $\frac{5}{4}$, not 1.

• Option (d) 0: This option is incorrect because the derivative $f’(1)$ is not zero. The correct calculation shows that $f’(1) = \frac{5}{4}$, indicating a non-zero value.

Solution

(a) Given,

$ \begin{aligned} y & =\frac{1+\frac{1}{x^{2}}}{1-\frac{1}{x^{2}}} \Rightarrow y=\frac{x^{2}+1}{x^{2}-1} \\ \frac{d y}{d x} & =\frac{(x^{2}-1) 2 x-(x^{2}+1)(2 x}{(x^{2}-1)^{2}} \\ \frac{d y}{d x} & =\frac{2 x(x^{2}-1-x^{2}-1)}{(x^{2}-1)^{2}} \\ & =\frac{2 x(-2)}{(x^{2}-1)^{2}}=\frac{-4 x}{(x^{2}-1)^{2}} \end{aligned} $

$ \therefore \quad \frac{d y}{d x}=\frac{(x^{2}-1) 2 x-(x^{2}+1)(2 x)}{(x^{2}-1)^{2}} \quad \text { [by quotient rule] } $

• Option (b) $\frac{-4 x}{x^{2}-1}$ is incorrect because the denominator should be $(x^{2}-1)^{2}$, not $x^{2}-1$. The correct differentiation results in a squared term in the denominator.

• Option (c) $\frac{1-x^{2}}{4 x}$ is incorrect because the numerator and denominator do not match the result of the differentiation. The correct differentiation results in a numerator of $-4x$ and a denominator of $(x^{2}-1)^{2}$.

• Option (d) $\frac{4 x}{x^{2}-1}$ is incorrect because the sign of the numerator is wrong and the denominator should be squared. The correct differentiation results in a negative numerator and a squared term in the denominator.

Solution

(a) Given, $y=\frac{\sin x+\cos x}{\sin x-\cos x}$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \quad \text { [by quotient rule] } \\ & =\frac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x-\cos x)^{2}} \\ & =\frac{-[(\sin x-\cos x)^{2}+(\sin x+\cos x)^{2}]}{(\sin x-\cos x)^{2}} \\ & =\frac{-[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x]}{(\sin x-\cos x)^{2}} \\ & =\frac{-2}{(\sin x-\cos x)^{2}} \\ \therefore \quad \frac{d y}{d x} & =-2 \end{aligned} $

• Option (b) 0: This option is incorrect because the derivative (\frac{d y}{d x}) at (x=0) is calculated to be (-2), not 0. The detailed differentiation process shows that the result is (-2), and there is no step in the calculation that would yield 0.

• Option (c) (\frac{1}{2}): This option is incorrect because the derivative (\frac{d y}{d x}) at (x=0) is (-2), not (\frac{1}{2}). The differentiation process clearly shows that the numerator simplifies to (-2), and the denominator does not affect this result to make it (\frac{1}{2}).

• Option (d) Does not exist: This option is incorrect because the derivative (\frac{d y}{d x}) at (x=0) does exist and is calculated to be (-2). The differentiation process does not encounter any undefined expressions or discontinuities at (x=0).

Solution

(a) Given, $\quad y=\frac{\sin (x+9)}{\cos x}$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{\cos x \cos (x+9)-\sin (x+9)(-\sin x)}{(\cos x)^{2}} \quad \text { [by quotient rule] } \\ & =\frac{\cos x \cos (x+9)+\sin x \sin (x+9)}{\cos ^{2} x} \\ \therefore \quad \frac{d y}{d x} _{x=0} & =\frac{\cos 9}{1} \\ & =\cos 9 \end{aligned} $

• Option (b) $\sin 9$: This option is incorrect because the derivative of the given function at ( x = 0 ) results in ( \cos 9 ), not ( \sin 9 ). The calculation involves the cosine and sine functions evaluated at specific points, leading to the final result of ( \cos 9 ).

• Option (c) 0: This option is incorrect because the derivative of the given function at ( x = 0 ) does not simplify to zero. The correct evaluation of the derivative at ( x = 0 ) yields ( \cos 9 ), which is generally not zero unless ( 9 ) is an odd multiple of ( \frac{\pi}{2} ), which it is not in this context.

• Option (d) 1: This option is incorrect because the derivative of the given function at ( x = 0 ) results in ( \cos 9 ), not 1. The value of ( \cos 9 ) is not equal to 1 unless ( 9 ) is an even multiple of ( \pi ), which it is not in this context.

Solution

(b) Given,

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

$ \begin{aligned} & \therefore \quad f^{\prime}(x)=0+1+2 \times \frac{x}{2}+\ldots+100 \frac{x^{99}}{100} \\ & f^{\prime}(x)=1+x+x^{2}+\ldots+x^{99} \\ & f^{\prime}(1)=1+1+1+\ldots+1 \text { (100 times) } \\ & =100 \end{aligned} $

• Option (a) $\frac{1}{100}$: This option is incorrect because the derivative of the given function at ( x = 1 ) results in a sum of 100 terms, each equal to 1, which sums to 100, not (\frac{1}{100}).

• Option (c) 0: This option is incorrect because the derivative of the given function at ( x = 1 ) is a sum of 100 terms, each equal to 1, which sums to 100, not 0.

• Option (d) Does not exist: This option is incorrect because the function ( f(x) ) is a polynomial, and the derivative of a polynomial always exists. Therefore, ( f’(1) ) exists and is equal to 100.

Solution(d) Given,

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

$ \begin{matrix} \therefore & f^{\prime}(x)=\frac{(x-a) n x^{n-1}-(x^{n}-a^{n})(1)}{(x-a)^{2}} \quad \text { [by quotient rule] } \\ \Rightarrow & f^{\prime}(x)=\frac{n x^{n-1}(x-a)-x^{n}+a^{n}}{(x-a)^{2}} \\ \text { Now, } & f^{\prime}(a)=\frac{n a^{n-1}(0)-a^{n}+a^{n}}{(x-a)^{2}} \\ \Rightarrow & f^{\prime}(a)=\frac{0}{0} \end{matrix} $

So, $f^{\prime}(a)$ does not exist,

Since, $f(x)$ is not defined at $x=a$.

Hence, $f^{\prime}(x)$ at $x=a$ does not exist.

• Option (a) 1: This is incorrect because the derivative ( f’(a) ) does not simplify to 1. The expression for ( f’(x) ) at ( x = a ) results in an indeterminate form ( \frac{0}{0} ), which means the derivative does not exist at that point.

• Option (b) 0: This is incorrect because the derivative ( f’(a) ) does not simplify to 0. The expression for ( f’(x) ) at ( x = a ) results in an indeterminate form ( \frac{0}{0} ), which means the derivative does not exist at that point.

• Option (c) ( \frac{1}{2} ): This is incorrect because the derivative ( f’(a) ) does not simplify to ( \frac{1}{2} ). The expression for ( f’(x) ) at ( x = a ) results in an indeterminate form ( \frac{0}{0} ), which means the derivative does not exist at that point.

Solution

(a) Given,

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

$ \begin{matrix} \therefore f^{\prime}(x) =100 x^{99}+99 x^{98}+\ldots+1+0 \\ =100 x^{99}+99 x^{98}+\ldots+1 \\ \text { Now, }\\ f^{\prime}(1) =100+99+\ldots+1 \\ =\frac{100}{2}[2 \times 100+(100-1)(-1)] & \\ =50[200-99] & \\ =50 \times 101 \\ =5050 \end{matrix} $

• Option (b) 5049: This option is incorrect because the sum of the series (100 + 99 + \ldots + 1) is calculated using the formula for the sum of the first (n) natural numbers, which is (\frac{n(n+1)}{2}). For (n = 100), the sum is (\frac{100 \times 101}{2} = 5050), not 5049.

• Option (c) 5051: This option is incorrect because the sum of the series (100 + 99 + \ldots + 1) is exactly 5050, as derived from the formula (\frac{n(n+1)}{2}). Adding 1 to this sum to get 5051 is incorrect.

• Option (d) 50051: This option is incorrect because it is significantly larger than the correct sum of the series (100 + 99 + \ldots + 1). The correct sum is 5050, and 50051 does not follow from any reasonable miscalculation of the series sum.

Solution

(d) Given, $f(x)=1-x+x^{2}-x^{3}+\ldots-x^{99}+x^{100}$

$ \begin{aligned} f^{\prime}(x) =0-1+2 x-3 x^{2}+\ldots-99 x^{98}+100 x^{99} \\ =-1+2 x-3 x^{2}+\ldots-99 x^{98}+100 x^{99} \\ \therefore \quad f^{\prime}(1) =-1+2-3+\ldots-99+100 \\ =(-1-3-5-\ldots-99)+(2+4+\ldots+100) \quad \because S_{n}=\frac{n}{2}{2 a+(n-1) d} \\ =-\frac{50}{2}[2 \times 1+(50-1) 2]+\frac{50}{2}[2 \times 2+(50-1) 2] \\ =-25[2+49 \times 2]+25[4+49 \times 2] \\ =-25(2+98)+25(4+98) \\ =-25 \times 100+25 \times 102 \\ =-2500+2550 \\ =50 \end{aligned} $

• Option (a) 150: This option is incorrect because the sum of the series of derivatives evaluated at ( x = 1 ) does not result in 150. The correct calculation shows that the sum is 50, not 150.

• Option (b) -50: This option is incorrect because the sum of the series of derivatives evaluated at ( x = 1 ) does not result in -50. The correct calculation shows that the sum is 50, not -50.

• Option (c) -150: This option is incorrect because the sum of the series of derivatives evaluated at ( x = 1 ) does not result in -150. The correct calculation shows that the sum is 50, not -150.

Solution

Given, $f(x)=\frac{\tan x}{x-\pi}=\lim _{x \rightarrow \pi} \frac{\tan x}{x-\pi}=\lim _{\pi \rightarrow x \rightarrow 0} \frac{-\tan (\pi-x)}{-(\pi-x)} \quad [\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1]$

$ =1 $

Solution

Given, $\lim _{x \rightarrow 0} \sin m x \cot \frac{x}{\sqrt{3}}=2$

$ \begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{\sin m x}{m x} \cdot m x \cdot \frac{1}{\tan \frac{x}{\sqrt{3}}}=2 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{\sin m x}{m x} \cdot m x \cdot \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \cdot \frac{1}{\frac{x}{\sqrt{3}}}=2 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{\sin m x}{m x} \cdot \lim _{x \rightarrow 0} \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \cdot \lim _{x \rightarrow 0} \frac{m x}{\frac{x}{\sqrt{3}}}=2 \\ & \begin{matrix} \Rightarrow & \sqrt{3} x=2 \end{matrix} \\ & \therefore \quad m=\frac{2 \sqrt{3}}{3} \end{aligned} $

Solution

Given,

$ \begin{aligned} y & =1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots \\ \frac{d y}{d x} & =0+1+\frac{2 x}{2}+\frac{3 x^{2}}{6}+\frac{4 x^{3}}{4 !} \\ & =1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\ldots \\ & =1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \\ & =y \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 3^{+}} \frac{x}{[x]} & =\lim _{h \rightarrow 0} \frac{(3+h)}{[3+h]} \\ & =\lim _{h \rightarrow 0} \frac{(3+h)}{3}=1 \end{aligned} $