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] $

Solution

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

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

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

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} $

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}] $

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} $

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] $

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} $

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} $

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} $

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.

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.

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., $x^{2}-10 x+21=0$.

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} $

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 $

$\because \quad LHL=RHD$

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

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}} $

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} $

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] } $

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} $

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} $

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} $

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.

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} $

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} $

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} $