Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 3} \left(\dfrac{x^{2}-9}{x-3} \right)& =\lim _{x \rightarrow 3} \left(\dfrac{x^{2}-(3)^{2}}{x-3}\right) \\ \\ & =\lim _{x \rightarrow 3} \dfrac{(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} \left(\dfrac{4 x^{2}-1}{2 x-1}\right) & =\lim _{x \rightarrow 1 / 2} \dfrac{(2 x)^{2}-(1)^{2}}{2 x-1} \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{(2 x+1)(2 x-1)}{(2 x-1)}=\lim _{x \rightarrow 1 / 2}(2 x+1) \\ \\ & =2 \times \dfrac{1}{2}+1=1+1=2 \end{aligned} $

Solution

Given,

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

$ \hspace{3.4cm}=\lim _{h \rightarrow 0} \dfrac{(x+h)^{1 / 2}-(x)^{1 / 2}}{(x+h)-x}\quad $ $ \left[\because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] $

$ \hspace{3.4cm}=\dfrac{1}{2} x^{\left({1}/{2}-1\right)}=\dfrac{1}{2} x^{\left({-1}/{2}\right)} \quad$ $ {\left[\because \ \ h \rightarrow 0 \Rightarrow x+h \rightarrow x\right]}$

$ \hspace{3.4cm}=\dfrac{1}{2 \sqrt{x}} $

Solution

Given,

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

$ \hspace{3.7cm}=\dfrac{1}{3} \times 2^{\left({1}/{3} -1 \right)} =\dfrac{1}{3} \times(2)^{\left({-2}/{3} \right)} \qquad\left[ \because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]\quad$

$ \hspace{3.7cm}=\dfrac{1}{3(2)^{2 / 3}} $

Solution

Given,

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

$ \hspace{3.2cm}=\lim _{x \rightarrow 0} \dfrac{\dfrac{(1+x)^{6}-1}{(1+x)-1}}{\dfrac{(1+x)^{2}-1}{(1+x)-1}}$ $\qquad{[\because \ \ x \rightarrow 0 \Rightarrow 1+x \rightarrow 1]}$

$ \hspace{3.2cm}=\dfrac{\lim _{x \rightarrow 0} \dfrac{(1+x)^{6}-(1)^{6}}{(1+x)-1}}{\lim _{x \rightarrow 0} \dfrac{(1+x)^{2}-(1)^{2}}{(1+x)-1}}\quad \left[ \because \ \ \lim _{x \rightarrow a} \dfrac{f(x)}{g(x)}=\dfrac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\right] $

$ \hspace{3.2cm}=\dfrac{6(1)^{6-1}}{2(1)^{2-1}} \qquad\left[\therefore \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] $

$ \hspace{3.2cm}=\dfrac{6 \times 1}{2 \times 1}=\dfrac{6}{2}=3 $

Solution

Given,

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

$ \hspace{4.7cm} =\dfrac{5}{2}(a+2)^{\frac{5}{2}-1} \qquad \left [ \because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] $

$ \hspace{4.7cm}=\dfrac{5}{2}(a+2)^{3 / 2} \qquad {[\because \ \ x \rightarrow a \Rightarrow x+2 \rightarrow a+2]} $

Solution

Given,

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

$\hspace{2.5cm}=\lim _{x \rightarrow 1} \dfrac{(x)^{7 / 2}-1}{\sqrt{x}-1} \cdot \lim _{x \rightarrow 1} \sqrt{x} \qquad[\because \ \ \lim _{x \rightarrow a} f(x) \cdot g(x)=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)]$

$\hspace{2.5cm}=\lim _{x \rightarrow 1} \dfrac{\dfrac{x^{7 / 2}-1}{x-1}}{\dfrac{(x)^{1 / 2}-1}{x-1}} \cdot 1$

$\hspace{2.5cm}=\dfrac{\lim _{x \rightarrow 1} \dfrac{x^{7 / 2}-1}{x-1}}{\lim _{x \rightarrow 1} \dfrac{(x)^{1 / 2}-1}{x-1}} $

$\hspace{2.5cm}=\dfrac{\dfrac{7}{2}(1)^{\frac{7}{2}-1}}{\dfrac{1}{2}(1)^{\frac{1}{2}-1}}=\dfrac{\dfrac{7}{2}}{\dfrac{1}{2}}=\dfrac{7\times 2}{2\times 1}=7 \qquad [ \because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

Solution

Given,

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

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{(\sqrt{3 x-2})^{2}-(\sqrt{x+2})^{2}}\qquad[\because \ \ (a+b)(a-b)=a^{2}-b^{2}] $

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{(3 x-2)-(x+2)} $

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{3 x-2-x-2} $

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x^{2}-4)(\sqrt{3 x-2}+\sqrt{x+2})}{2 x-4} $

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x+2)(x-2)(\sqrt{3 x-2}+\sqrt{x+2})}{2(x-2)} $

$\hspace{4.2cm} =\lim _{x \rightarrow 2} \dfrac{(x+2)(\sqrt{3 x-2}+\sqrt{x+2})}{2} $

$\hspace{4.2cm} =\dfrac{(2+2)(\sqrt{6-2}+\sqrt{2+2})}{2} $

$\hspace{4.2cm} =\dfrac{4(2+2)}{2}=8$

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow \sqrt{2}} \left(\dfrac{x^{4}-4}{x^{2}+3 \sqrt{2} x-8}\right) & =\lim _{x \rightarrow \sqrt{2}} \dfrac{(x^{2})^{2}-(2)^{2}}{x^{2}+3 \sqrt{2} x-8} \\ \\ & =\lim _{x \rightarrow \sqrt{2}} \dfrac{(x^{2}-2)(x^{2}+2)}{x^{2}+4 \sqrt{2} x-\sqrt{2} x-8} \\ \\ & =\lim _{x \rightarrow \sqrt{2}} \dfrac{(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}} \dfrac{(x-\sqrt{2})(x+\sqrt{2})(x^{2}+2)}{(x-\sqrt{2})(x+4 \sqrt{2})} \\ \\ & =\lim _{x \rightarrow \sqrt{2}} \dfrac{(x+\sqrt{2})(x^{2}+2)}{(x+4 \sqrt{2})} \\ \\ & =\dfrac{(\sqrt{2}+\sqrt{2})[(\sqrt{2})^{2}+2]}{(\sqrt{2}+4 \sqrt{2})} \\ \\ & =\dfrac{2 \sqrt{2}(2+2)}{5 \sqrt{2}}=\dfrac{8}{5} \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 1} \dfrac{x^{7}-2 x^{5}+1}{x^{3}-3 x^{2}+2} & = \lim _{x \rightarrow 1} \dfrac{x^{7}-x^{5}-x^{5}+1}{x^{3}-x^{2}-2 x^{2}+2} \\ \\ & = \lim _{x \rightarrow 1} \dfrac{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} \dfrac{\dfrac{x^{5}(x^{2}-1)}{(x-1)}-\dfrac{1(x^{5}-1)}{(x-1)}}{\dfrac{x^{2}(x-1)}{(x-1)}-\dfrac{2(x^{2}-1)}{(x-1)}} \\ \\ & =\dfrac{\lim _{x \rightarrow 1} x^{5}(x+1)-\lim _{x \rightarrow 1} \dfrac{x^{5}-1}{x-1}}{\lim _{x \rightarrow 1} x^{2}-\lim _{x \rightarrow 1}2(x+1)} \\ \\ & =\dfrac{1 \times 2-5 \times(1)^{4}}{1-2 \times 2}=\dfrac{2-5}{1-4} \qquad \left[ \because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] \\ \\ & =\dfrac{-3}{-3}=1 \end{aligned} $

Solution

Given,

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

$\hspace{4.2cm}=\lim _{x \rightarrow 0} \dfrac{(1+x^{3})-(1-x^{3})}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$\hspace{4.2cm}=\lim _{x \rightarrow 0} \dfrac{1+x^{3}-1+x^{3}}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$\hspace{4.2cm}=\lim _{x \rightarrow 0} \dfrac{2 x^{3}}{x^{2}(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$\hspace{4.2cm}=\lim _{x \rightarrow 0} \dfrac{2 x}{(\sqrt{1+x^{3}}+\sqrt{1-x^{3}})}$

$\hspace{4.2cm}=0$

Solution

Given, $\quad $

$ \begin{aligned} \lim _{x \rightarrow-3} \left(\dfrac{x^{3}+27}{x^{5}+243}\right) & =\lim _{x \rightarrow-3} \dfrac{\left(\dfrac{x^{3}+27}{x+3}\right)}{\left(\dfrac{x^{5}+243}{x+3}\right)} \\ \\ & =\lim _{x \rightarrow-3} \dfrac{\left(\dfrac{x^{3}-(-3)^{3}}{x-(-3)}\right)}{\left(\dfrac{x^{5}-(-3)^{5}}{x-(-3)}\right)} \\ \\ & =\dfrac{\lim _{x \rightarrow-3} \left(\dfrac{x^{3}-(-3)^{3}}{x-(-3)}\right)}{\lim _{x \rightarrow-3} \left(\dfrac{x^{5}-(-3)^{5}}{x-(-3)}\right)} \qquad \left[\because \ \ \lim _{x \rightarrow a} \dfrac{f(x)}{g(x)}=\dfrac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}\right] \\ \\ & =\dfrac{3(-3)^{3-1}}{5(-3)^{5-1}}=\dfrac{3}{5} \dfrac{(-3)^{2}}{(-3)^{4}} \qquad \left[\because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]\\ \\ & =\dfrac{3}{5(-3)^{2}}=\dfrac{3}{45}=\dfrac{1}{15} \end{aligned} $

Solution

$\text{Given,}$

$ \begin{aligned} \lim _{x \rightarrow 1 / 2} \left(\dfrac{8 x-3}{2 x-1}-\dfrac{4 x^{2}+1}{4 x^{2}-1}\right) & =\lim _{x \rightarrow 1 / 2} \left[\dfrac{(8 x-3)(2 x+1)-(4 x^{2}+1)}{(4 x^{2}-1)}\right] \\ \\ & =\lim _{x \rightarrow 1 / 2} \left[\dfrac{16 x^{2}+8 x-6 x-3-4 x^{2}-1}{4 x^{2}-1} \right]\\ \\ & =\lim _{x \rightarrow 1 / 2} \left[\dfrac{12 x^{2}+2 x-4}{4 x^{2}-1}\right] \\ \\ & =\lim _{x \rightarrow 1 / 2} \left[\dfrac{2(6 x^{2}+x-2)}{4 x^{2}-1}\right] \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{2(6 x^{2}+4 x-3 x-2)}{4 x^{2}-1} \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{2[2 x(3 x+2)-1(3 x+2)]}{4 x^{2}-1} \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{2[(3 x+2)(2 x-1)]}{(2 x)^{2}-(1)^{2}} \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{2(3 x+2)(2 x-1)}{(2 x-1)(2 x+1)} \\ \\ & =\lim _{x \rightarrow 1 / 2} \dfrac{2(3 x+2)}{2 x + 1}=\dfrac{2 \left(3 \times \dfrac{1}{2}+2\right)}{2 \times \dfrac{1}{2}+1} \\ \\ & =\dfrac{3}{2}+2=\dfrac{7}{2} \end{aligned} $

Solution

Given,

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

$ \begin{aligned} & \Rightarrow n(2)^{n-1}=80 \qquad \left[\because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right] \\ \\ & \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 \quad n=5 \end{aligned} $

Solution

Given,

$\lim _{x \rightarrow 0} \dfrac{\left(\dfrac{\sin 3 x}{3 x}\right) \cdot 3 x}{\left(\dfrac{\sin 7 x}{7 x}\right) \cdot 7 x}=\dfrac{\lim _{x \rightarrow 0} \left(\dfrac{\sin 3 x}{3 x}\right)}{\lim _{x \rightarrow 0} \left(\dfrac{\sin 7 x}{7 x}\right)} \cdot \dfrac{3 x}{7 x}$

$ \hspace{2.8cm}=\dfrac{3}{7} \cdot \dfrac{\lim _{x \rightarrow 0} \left(\dfrac{\sin 3 x}{3 x}\right)}{\lim _{x \rightarrow 0} \left(\dfrac{\sin 7 x}{7 x}\right)} $

$ \hspace{2.8cm}=\dfrac{3}{7} \quad[\because \ \ x \rightarrow 0 \Rightarrow(k x \rightarrow 0) \text {, here } k \text { is real number }] $

Solution

Given,

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

Solution

Given,

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

$\hspace{2.7cm}=\lim _{x \rightarrow 0} \dfrac{2 \sin ^{2} x}{x^{2}}=2 \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{x^{2}} $

$\hspace{2.7cm}=2 \lim _{x \rightarrow 0} \left(\dfrac{\sin x}{x}\right)^{2} \qquad [\because \ \ \lim _{x \rightarrow 0} \left(\dfrac{\sin x}{x}\right)^{2}=1] $

$\hspace{2.7cm}=2 \times 1=2 $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 0} \dfrac{2 \sin x-\sin 2 x}{x^{3}} & =\lim _{x \rightarrow 0} \dfrac{2 \sin x-2 \sin x \cos x}{x^{3}} \qquad[\because \ \ \sin 2 x=2 \sin x \cos x] \\ \\ & =\lim _{x \rightarrow 0} \dfrac{2 \sin x(1-\cos x)}{x^{3}} \\ \\ & =2 \lim _{x \rightarrow 0} \dfrac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \left(\dfrac{1-\cos x}{x^{2}}\right) \\ \\ & =2 \cdot 1 \lim _{x \rightarrow 0} \dfrac{1-\cos x}{x^{2}} \qquad [\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1] \\ \\ & =2 \lim _{x \rightarrow 0} \dfrac{1-1+2 \sin ^{2} \dfrac{x}{2}}{x^{2}}=2 \lim _{x \rightarrow 0} \dfrac{2 \sin ^{2} \dfrac{x}{2}}{4 \times \dfrac{x^{2}}{4}} \\ \\ & =\dfrac{2 \cdot 2}{4} \lim _{x \rightarrow 0} \left(\dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}}\right)^{2}=\lim _{x \rightarrow 0} \left(\dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}}\right)^{2}=1 \end{aligned} $

Solution

Given,

$\lim _{x \rightarrow 0} \dfrac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \dfrac{1-1+2 \sin ^{2} \dfrac{m x}{2}}{1-1+2 \sin ^{2} \dfrac{n x}{2}}\qquad \left[\because \ \ \quad \cos m x=1-2 \sin ^{2} \dfrac{m x}{2} \ \text{and} \ \sin n x=1-2 \sin ^{2} \dfrac{n x}{2}\right]$

$\hspace{3cm}=\lim _{x \rightarrow 0} \dfrac{\sin ^{2} \dfrac{m x}{2}}{\sin ^{2} \dfrac{n x}{2}}$

$\hspace{3cm}=\lim _{x \rightarrow 0} \dfrac{\dfrac{\sin ^{2} \dfrac{m x}{2}}{\left(\dfrac{m x}{2}\right)^{2}} \cdot \left(\dfrac{m x}{2}\right)^{2}}{\dfrac{\sin ^{2} \dfrac{n x}{2}}{\left(\dfrac{n x}{2}\right)^{2}} \cdot \left(\dfrac{n x}2\right)^{2}}$

$\hspace{3cm}=\dfrac{\lim _{x \rightarrow 0} \left(\dfrac{\sin \dfrac{m x}{2}}{\dfrac{m x}{2}}\right)^{2}}{\lim _{x \rightarrow 0} \left(\dfrac{\sin \dfrac{n x}{2}}{\dfrac{n x}{2}}\right)^{2}} \cdot \dfrac{m^{2} }{n^{2}}\qquad$ $\left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1, \quad x \rightarrow 0 \Rightarrow k x \rightarrow 0\right]$

Solution

Given,

$\lim _{x \rightarrow \pi / 3} \dfrac{\sqrt{1-\cos 6 x}}{\sqrt{2} \left(\dfrac{\pi}{3}-x\right)}=\lim _{x \rightarrow \pi / 3} \dfrac{\sqrt{1-1+2 \sin ^{2} 3 x}}{\sqrt{2} \left(\dfrac{\pi}{3}-x\right)}\qquad$ $[\because \ \ \cos 2 x=1-2 \sin ^{2} x]$

$\hspace{3.5cm}=\lim _{x \rightarrow \pi / 3} \dfrac{\sqrt{2} \sin 3 x}{\sqrt{2} \left(\dfrac{\pi}{3}-x\right)}$

$\hspace{3.5cm}=\lim _{x \rightarrow \pi / 3} \dfrac{\sin 3 x}{\dfrac{\pi}{3}-x}$

$\hspace{3.5cm}=\lim _{x \rightarrow \pi / 3} \left(\dfrac{\sin (\pi-3 x)}{\dfrac{\pi-3 x}{3}}\right) \qquad[\because \ \ \sin (\pi-\theta)=\sin \theta]$

$\hspace{3.5cm}=3 \lim _{x \rightarrow \pi / 3} \dfrac{\sin (\pi-3 x)}{(\pi-3 x)}=3 \times 1 \qquad [\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1]$

$\hspace{3.5cm}=3$

Solution

Given,

$\lim _{x \rightarrow \pi / 4} \dfrac{\sqrt{2} \left(\sin x \cdot \dfrac{1}{\sqrt{2}}-\cos x \cdot \dfrac{1}{\sqrt{2}}\right)}{\left(x-\dfrac{\pi}{4}\right)}=\lim _{x \rightarrow \pi / 4} \dfrac{\sqrt{2} \left(\sin x \cos \dfrac{\pi}{4}-\cos x \cdot \sin \dfrac{\pi}{4}\right)}{\left(x-\dfrac{\pi}{4}\right)}$

$\hspace{6cm}=\lim _{x \rightarrow \pi / 4} \dfrac{\sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)}{x-\dfrac{\pi}{4}} $

$\hspace{6cm}=\sqrt{2} \lim _{x \rightarrow \pi / 4} \dfrac{\sin\left(x-\dfrac{\pi}{4}\right)}{x-\dfrac{\pi}{4}}=\sqrt{2} \qquad \left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1\right] $

Solution

Given,

$\lim _{x \rightarrow \pi / 6} \dfrac{\sqrt{3} \sin x-\cos x}{x-\dfrac{\pi}{6}}=\lim _{x \rightarrow \pi / 6} \left(\dfrac{2 \dfrac{\sqrt{3}}{2} \sin x-\dfrac{1}{2} \cos x}{x-\dfrac{\pi}{6}}\right)$

$\hspace{4cm}=\lim _{x \rightarrow \pi / 6} \dfrac{2 \left(\sin x \cos \dfrac{\pi}{6}-\cos x \sin \dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6}\right)}$

$\hspace{4cm}=2 \lim _{x \rightarrow \pi / 6} \dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6}\right)} \qquad \left[ \because \ \ \quad \sin A \cos B-\cos A \sin B=\sin (A-B), \quad \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1\right] $

$\hspace{4cm}=2 $

Solution

Given,

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

$\hspace{3.1cm}=\lim _{x \rightarrow 0} \dfrac{\left(\dfrac{\sin 2 x}{2 x}+\dfrac{3 x}{2 x}\right) 2 x}{\left(\dfrac{2 x}{3 x}+\dfrac{\tan 3 x}{3 x}\right) 3 x}$

$\hspace{3.1cm}=\lim _{x \rightarrow 0} \dfrac{\dfrac{\sin 2 x}{2 x}+\dfrac{3}{2}}{\dfrac{2}{3}+\dfrac{\tan 3 x}{3 x}} \cdot \dfrac{2}{3}$

$\hspace{3.1cm}=\dfrac{2}{3} \lim _{x \rightarrow 0} \dfrac{\dfrac{\sin 2 x}{2 x}+\dfrac{3}{2}}{\dfrac{2}{3}+\lim _{x \rightarrow 0} \dfrac{\tan 3 x}{3 x}} $

$\hspace{3.1cm}=\dfrac{2}{3} \left(\dfrac{1+\dfrac{3}{2}}{\dfrac{2}{3}+1} \right)\qquad \left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1 \text { and } \lim _{x \rightarrow 0} \dfrac{\tan x}{x}=1\right] $

$\hspace{3.1cm}=\dfrac{2}{3} \times \dfrac{\dfrac{5}{2}}{\dfrac{5}{3}}=\dfrac{2}{3} \times \dfrac{5}{2} \times \dfrac{3}{5}=1$

Solution

Given,

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

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow \pi / 6} \left(\dfrac{\cot ^{2} x-3}{ \ cosec \ x-2}\right) & =\lim _{x \rightarrow \pi / 6} \dfrac{ \ cosec \ ^{2} x-1-3}{ \ cosec \ x-2} \qquad[\because \ \ \ cosec \ ^{2} x=1+\cot ^{2} x] \\ \\ & =\lim _{x \rightarrow \pi / 6} \dfrac{ \ cosec \ ^{2} x-4}{ \ cosec \ x-2} \\ \\ & =\lim _{x \rightarrow \pi / 6} \dfrac{( \ cosec \ x)^{2}-(2)^{2}}{ \ cosec \ x-2} \\ \\ & =\lim _{x \rightarrow \pi / 6} \dfrac{( \ cosec \ x+2)( \ cosec \ x-2)}{( \ cosec \ x-2)} \\ \\ & =\lim _{x \rightarrow \pi / 6}( \ cosec \ x+2) \\ \\ & = \ cosec \ \dfrac{\pi}{6}+2=2+2=4 \end{aligned} $

Solution

$\text{Given,}$

$ \begin{aligned} \lim _{x \rightarrow 0} \dfrac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x} & = \lim _{x \rightarrow 0} \dfrac{\sqrt{2}-\sqrt{1+2 \cos ^{2} \dfrac{x}{2}-1}}{\sin ^{2} x} \quad \left[\because \ \ \cos x=2 \cos ^{2} \dfrac{x}{2}-1\right] \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\sqrt{2}-\sqrt{2 \cos ^{2} \dfrac{x}{2}}}{\sin ^{2} x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\sqrt{2}\left( 1-\cos \dfrac{x}{2}\right)}{\sin ^{2} x}=\lim _{x \rightarrow 0} \dfrac{\sqrt{2} \left(1-1+2 \sin ^{2} \dfrac{x}{4}\right)}{\sin ^{2} x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\sqrt{2} \left(2 \sin ^{2} \dfrac{x}{4}\right)}{\sin ^{2} x}=\lim _{x \rightarrow 0} 2 \sqrt{2} \dfrac{\sin ^{2} \dfrac{x}{4}}{\left(\dfrac{x}{4}\right)^{2}} \cdot \dfrac{\left(\dfrac{x}{4}\right)^{2}}{\sin ^{2} x} \\ \\ & =2 \sqrt{2} \lim _{x \rightarrow 0} \left(\dfrac{\sin {\dfrac{x}{4}}}{\dfrac{x}{4}}\right)^{2} \cdot \lim _{x \rightarrow 0} \left(\dfrac{x}{\sin x}\right)^{2} \cdot \dfrac{1}{16} \\ \\ & =2 \sqrt{2} \cdot 1 \cdot 1 \cdot \dfrac{1}{16}=\dfrac{1}{4 \sqrt{2}} \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow 0} \dfrac{\sin x-2 \sin 3 x+\sin 5 x}{x} & =\lim _{x \rightarrow 0} \dfrac{\sin 5 x+\sin x-2 \sin 3 x}{x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{x}=\lim _{x \rightarrow 0} \dfrac{2 \sin 3 x(\cos 2 x-1)}{x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{2 \sin 3 x}{\dfrac{1}{3} \times 3 x}(\cos 2 x-1)=6 \lim _{x \rightarrow 0} \dfrac{\sin 3 x}{3 x}(\cos 2 x-1) \\ \\ & =6 \lim _{x \rightarrow 0} \dfrac{\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} \dfrac{x^{4}-1}{x-1}=\lim _{x \rightarrow k} \left(\dfrac{x^{3}-k^{3}}{x^{2}-k^{2}}\right) $

$ \Rightarrow \quad 4(1)^{4-1}=\lim _{x \rightarrow k} \dfrac{\left(\dfrac{x^{3}-k^{3}}{x-k}\right)}{\left(\dfrac{x^{2}-k^{2}}{x-k}\right)}\qquad\left[\because \ \ \lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a} =n a^{n-1}\right]$

$ \begin{aligned} \Rightarrow \quad & 4=\dfrac{\lim _{x \rightarrow k} \left(\dfrac{x^{3}-k^{3}}{x-k}\right)}{\lim _{x \rightarrow k} \left(\dfrac{x^{2}-k^{2}}{x-k}\right)} \\ \\ \Rightarrow \quad & 4=\dfrac{3 k^{2}}{2 k} \\ \\ \Rightarrow \quad & 4=\dfrac{3}{2} k \\ \\ \therefore\quad & k=\dfrac{4 \times 2}{3}=\dfrac{8}{3} \end{aligned} $

Solution

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

Solution

Let

$ y=\left(x+\dfrac{1}x\right)^{3} $

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

Solution

Let

$ \begin{aligned} y & =(3 x+5)(1+\tan x) \\ \\ \dfrac{d y}{d x} & =\dfrac{d}{d x}[(3 x+5)(1+\tan x)] \\ \\ & =(3 x+5) \dfrac{d}{d x}(1+\tan x)+(1+\tan x) \dfrac{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 \dfrac{d y}{d x}=3 x \sec ^{2} x+5 \sec ^{2} x+3 \tan x+3 $

Solution

$ \begin{aligned} \text{Let,} \quad y & =(\sec x-1)(\sec x+1) \\ \\ y & =(\sec ^{2}-1) \qquad \left[\because \ (a+b)(a-b)=a^{2}-b^{2}\right] \\ \\ & =\tan ^{2} x \\ \\ \therefore \quad \dfrac{d y}{d x} & =2 \tan x \cdot \dfrac{d}{d x} \tan x \\ \\ & =2 \tan x \cdot \sec ^{2} x \qquad\left[\text {by chain rule}\right] \end{aligned} $

Solution

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

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

Solution

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

Solution

$\text{Let,}$

$ \begin{aligned} y & =\dfrac{x^{2} \cos \dfrac{\pi}{4}}{\sin x}=\dfrac{\dfrac{x^{2}}{\sqrt{2}}}{\sin x} \\ \\ y & =\dfrac{1}{\sqrt{2}} \cdot \dfrac{x^{2}}{\sin x} \\ \\ \dfrac{d y}{d x} & =\dfrac{1}{\sqrt{2}} \left[\dfrac{\sin x \cdot \dfrac{d}{d x} x^{2}-x^{2} \dfrac{d}{d x} \sin x}{\sin ^{2} x}\right] \\ \\ & =\dfrac{1}{\sqrt{2}} \left[\dfrac{\sin x \cdot 2 x-x^{2} \cdot \cos x}{\sin ^{2} x}\right] \\ \\ & =\dfrac{1}{\sqrt{2}} \cdot \dfrac{2 x \sin x-x^{2} \cos x}{\sin ^{2} x} \\ \\ & =\dfrac{x}{\sqrt{2}}[2 \ cosec \ x-x \cot x \ cosec \ x] \\ \\ & =\dfrac{x}{\sqrt{2}} \ cosec \ [2-x \cot x] \end{aligned} $

Solution

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

$ \begin{aligned} \therefore \quad \dfrac{d y}{d x} & =(a x^{2}+\cot x) \dfrac{d}{d x}(p+q \cos x)+(p+q \cos x) \dfrac{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

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

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

Solution

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

Solution

$ \begin{aligned} \text{Let,} \quad y & =(2 x-7)^{2}(3 x+5)^{3} \\ \\ \dfrac{d y}{d x} & =(2 x-7)^{2} \dfrac{d}{d x}(3 x+5)^{3}+(3 x+5)^{3} \dfrac{d}{d x}(2 x-7)^{2} \qquad \text { [by product rule] } \\ \\ & =(2 x-7)^{2}(3)(3 x+5)^{2}(3)+(3 x+5)^{3} 2(2 x-7)(2) \qquad \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{aligned} $

Solution

$ \begin{aligned} \text{Let,} \quad y & =x^{2} \sin x+\cos 2 x \\ \\ \dfrac{d y}{d x} & =\dfrac{d}{d x}(x^{2} \sin x)+\dfrac{d}{d x} \cos 2 x \\ \\ & =x^{2} \cdot \cos x+\sin x 2 x+(-\sin 2 x) \cdot 2 \qquad \text { [by product rule] } \\ \\ & =x^{2} \cos x+2 x \sin x-2 \sin 2 x \qquad \text { [by chain urle] } \end{aligned} $

Solution

$ \begin{aligned} \text{Let,} \quad y & =\sin ^{3} x \cos ^{3} x \\ \\ \dfrac{d y}{d x} & =\sin ^{3} x \cdot \dfrac{d}{d x} \cos ^{3} x+\cos ^{3} x \dfrac{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 \\ \\ & =\dfrac{3}{4}(2 \sin x \cos x)^{2} \cos 2 x \\ \\ & =\dfrac{3}{4} \sin ^{2} 2 x \cos 2 x \end{aligned} $

Solution

$ \text{Let,} \quad y=\dfrac{1}{a x^{2}+b x+c}=(a x^{2}+b x+c)^{-1} $

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

Solution

$ \begin{aligned} \text{Let,} \quad f(x) & =\cos (x^{2}+1) \text { and } f(x+h)=\cos \left{(x+h)^{2}+1\right} \\ \\ \dfrac{d}{dx}f(x) & =\lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ \\ & =\lim _{h \rightarrow 0} \dfrac{\cos \left{(x+h)^{2}+1\right} -\cos (x^{2}+1)}{h} \\ \\ & =\lim _{h \rightarrow 0} \dfrac{-2 \sin \left{\dfrac{(x+h)^{2}+1+x^{2}+1}{2}\right} \sin\left{ \dfrac{(x+h)^{2}+1-x^{2}-1}{2}\right}}{h} \\ \\ & \qquad \left[\because \ \ \cos C-\cos D=-2 \sin \left{\dfrac{C+D}{2}\right} \cdot \sin \left{\dfrac{C-D}{2}\right}\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left{ \dfrac{(x+h)^{2}+x^{2}+2}{2}\right} \sin \left{\dfrac{(x+h)^{2}-x^{2}}{2}\right}\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left{\dfrac{(x+h)^{2}+x^{2}+2}{2}\right} \sin \left{ \dfrac{x^{2}+h^{2}+2 x h-x^{2}}{2}\right}\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left{\dfrac{(x+h)^{2}+x^{2}+2}{2}\right} \sin \left{ \dfrac{h^{2}+2 h x}{2}\right}\right] \\ \\ & =-2 \lim _{h \rightarrow 0} \sin \left{ \dfrac{(x+h)^{2}+x^{2}+2}{2}\right} \lim _{h \rightarrow 0}\left{ \dfrac{\sin h \dfrac{h+2 x}{2}}{h \dfrac{h+2 x}{2}}\right} \times \left(\dfrac{h+2 x}{2}\right) \\ \\ & =-2 \lim _{h \rightarrow 0} \sin \left{ \dfrac{(x+h)^{2}+x^{2}+2}{2}\right} \lim _{h \rightarrow 0} \left(\dfrac{h+2 x}{2}\right) \quad [\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1] \\ \\ & =-2 x \sin (x^{2}+1) \end{aligned} $

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

Solution

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

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

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

$ \hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{a(x+h)+b}{c(x+h)+d}-\dfrac{a x+b}{c x+d}\right] $

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{a x+b+a h}{c(x+h)+d}-\dfrac{a x+b}{c x+d}\right] $

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{(a x+a h+b)(c x+d)-(a x+b{c(x+h)+d}}{{c(x+h)+d}(c x+d)} \right]$

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{(a x+a h+b)(c x+d)-(a x+b)(c x+c h+d)}{\left{ c(x+h)+d)\right}(c x+d)} \right]$

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[\dfrac{{a c x^{2}+a c h x+b c x+a d x+a d h+b d}-{a c x^{2}+a c h x+a d x+b c x+b c h+b d}}{\left{ c(x+h)+d\right}(c x+d)}\right]$

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[ \dfrac{a c x^{2}+a c h x+b c x+a d x+a d h+b d-a c x^{2}-a c h x-a d x-b c x-b c h-b d}{\left{ c(x+h)+d\right}(c x+d)} \right]$

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{1}{h} \left[ \dfrac{a d h-b c h}{\left{ c(x+h)+d\right}(c x+d)}\right] $

$\hspace{1.4cm}=\lim _{h \rightarrow 0} \dfrac{a c-b d}{{c(x+h)+d}(c x+d)} $

$\hspace{1.4cm}=\dfrac{a c-b d}{(c x+d)^{2}} $

Solution

$ \begin{aligned} \text{Let,} \ \ f(x) & =x^{2 / 3} \\ \\ f(x+h) & =(x+h)^{2 / 3} \\ \\ \dfrac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}\left[(x+h)^{2 / 3}-x^{2 / 3}\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}\left[ x^{2 / 3} \left(1+\dfrac{h}{x} \right)^{2 / 3}-x^{2 / 3}\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h} \left[x^{2 / 3} \left{\left(1+\dfrac{h}{x} \cdot \dfrac{2}{3}+\dfrac{2}{3} \left(\dfrac{2}{3}-1\right) \dfrac{h^{2}}{x^{2}}+\cdots\right)-1\right}\right] \qquad \left[\because \ \ \ (1+x)^{h} =1+n x+\dfrac{h(n-1)}{2 !} x^{2}+\cdots\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h} \left[x^{2 / 3} \left(\dfrac{2}{3} \cdot \dfrac{h}{x}-\dfrac{2}{9} \cdot \dfrac{h^{2}}{x^{2}}+\cdots\right)\right] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{x^{2 / 3}}{h} \cdot \dfrac{2}{3} \dfrac{h}{x} \left(1-\dfrac{1}{3} \cdot \dfrac{h}{x}+\cdots\right) \\ \\ & =\dfrac{2}{3} x^{(2 / 3-1)}=\dfrac{2}{3} x^{(-1 / 3)} \end{aligned} $

Alternate Method

$ \begin{aligned} \text{Let,} \ \ f(x) & = x^{2 / 3} \\ \\ f(x+h) & =(x+h)^{2 / 3} \\ \\ \dfrac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ \\ & =\lim _{(h \rightarrow 0)} \left[\dfrac{(x+h)^ {2 / 3}-x^{2 / 3}}{h}\right]=\lim _{(x+h) \rightarrow x} \left[ \dfrac{(x+h)^ {2 / 3}-x^{2 / 3}}{(x+h)-x}\right] \\ \\ & =\dfrac{2}{3}(x)^{2 / 3-1} \qquad \left[\because \ \ \lim _{x \rightarrow a} \dfrac{x^{x}-a^{n}}{x-a}=n a^{n-1}\right] \\ \\ & =\dfrac{2}{3} x^{-1 / 3} \end{aligned} $

Solution

$ \begin{aligned} \text {Let,} \ \ f(x) & =x \cos x \\ \\ \therefore \ \ f(x+h) & =(x+h) \cos (x+h) \\ \\ \therefore \ \ \dfrac{d}{dx}f(x) & =\lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}[(x+h) \cos (x+h)-x \cos x] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}[x \cos (x+h)+h \cos (x+h)-x \cos x] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h}[x\left{\cos (x+h)-\cos x\right}+h \cos (x+h)] \\ \\ & =\lim _{h \rightarrow 0} \dfrac{1}{h} \left[x\left{-2 \sin \dfrac{2 x+h}{2} \sin \dfrac{h}{2}\right}+h \cos (x+h)\right] \\ \\ & =\lim _{h \rightarrow 0}\left[-2 x \sin \left(x+\dfrac{h}{2}\right) \dfrac{\sin \dfrac{h}{2}}{h}+\cos (x+h)\right] \\ \\ & =-2 \lim _{h \rightarrow 0} x \sin \left(x+\dfrac{h}{2}\right) \lim _{h \rightarrow 0} \dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}} \cdot \dfrac{1}{2}+\lim _{h \rightarrow 0} \cos (x+h) \\ \\ & =-2 \cdot \dfrac{1}{2} x \sin x+\cos x \\ \\ & =\cos x-x \sin x \end{aligned} $

Evaluate each of the following limits in following questions

Solution

$\text{Given,}$

$\lim _{y \rightarrow 0} \dfrac{(x+y) \sec (x+y)-x \sec x}{y}$

$ \begin{aligned} & =\lim _{y \rightarrow 0} \dfrac{\dfrac{x+y}{\cos (x+y)}-\dfrac{x}{\cos x}}{y} \\ \\ & =\lim _{y \rightarrow 0} \dfrac{(x+y) \cos x-x \cos (x+y)}{y \cos x \cos (x+y)} \\ \\ & =\lim _{y \rightarrow 0} \dfrac{x \cos x+y \cos x-x \cos (x+y)}{y \cos x \cos (x+y)} \\ \\ & =\lim _{y \rightarrow 0} \dfrac{x \cos x-x \cos (x+y)+y \cos x}{y \cos x \cos (x+y)} \\ \\ & =\lim _{y \rightarrow 0} \dfrac{x\left{\cos x-\cos (x+y)\right}+y \cos x}{y \cos x \cos (x+y)} \\ \\ & =\lim _{y \rightarrow 0} \dfrac{x\left[-2 \sin \left(x+\dfrac{y}{2}\right) \sin \left(\dfrac{-y}{2}\right)\right]+y \cos x}{y \cos x \cos (x+y)} \qquad \left[\because \ \ \cos C-\cos D=-2 \sin \dfrac{C+D}{2} \cdot \sin \dfrac{C-D}{2}\right] \end{aligned} $

$ \begin{aligned} & =\lim _{y \rightarrow 0} \left[\dfrac{x \left{ 2 \sin (x+\dfrac{y}{2}) \sin \dfrac{y}{2}\right} +y \cos x}{y \cos x \cos (x+y)}\right] \\ \\ & =\lim _{y \rightarrow 0} \dfrac{2 x \sin (x+\dfrac{y}{2})}{\cos x \cos (x+y)} \cdot \lim _{y \rightarrow 0} \dfrac{\sin \dfrac{y}{2}}{\dfrac{y}{2}} \cdot \dfrac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y)\qquad \left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1 \text { and } x \rightarrow 0 \Rightarrow k x \rightarrow 0\right] \\ \\ & =\lim _{y \rightarrow 0} \dfrac{2 x \sin (x+\dfrac{y}{2})}{\cos x \cos (x+y)} \cdot \dfrac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y) \\ \\ & =\dfrac{2 x \sin x}{\cos x \cos x} \cdot \dfrac{1}{2}+\sec x \\ \\ & =x \tan x \sec x+\sec x \\ \\ & =\sec x(x \tan x+1) \end{aligned} $

Solution

$\text{Given,}$

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

$ =\lim _{x \rightarrow 0} \dfrac{[2 \sin \alpha x \cdot \cos \beta x+\sin 2 \alpha x]}{\cos 2 \beta x-\cos 2 \alpha x}=2 \sin \left(\dfrac{C+D}{2}\right) \cos \left(\dfrac{C-D}{2}\right) \\ \\ $

$ \begin{aligned} & =\lim _{x \rightarrow 0} \dfrac{[2 \sin \alpha x \cos \beta x+\sin 2 \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \qquad \left[\because \ \ \cos C-\cos D=2 \sin \left(\dfrac{C+D}{2}\right) \cdot \sin \left(\dfrac{D-C}{2}\right)\right] \\ \\ & =\lim _{x \rightarrow 0} \dfrac{[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} \dfrac{2 \sin \alpha x[\cos \beta x+\cos \alpha x] x}{2 \sin (\alpha+\beta) x \cdot\sin (\alpha-\beta) x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\sin \alpha x \left[2 \cos \left(\dfrac{\alpha+\beta}{2}\right) x \cos \left(\dfrac{\alpha-\beta}{2}\right) \right]\times x}{2 \sin \left(\dfrac{\alpha+\beta}{2}\right) x \cos \left(\dfrac{\alpha+\beta}{2}\right) x \cdot 2 \sin \left(\dfrac{\alpha-\beta}{2}\right) x \cos \left(\dfrac{\alpha-\beta}{2}\right) x} \\ \\ & \left[\because \ \ \cos C+\cos D=2 \cos \left(\dfrac{C+D}{2}\right) \cos \left(\dfrac{C-D}{2}\right) \text { and } \sin 2 \theta=2 \sin \theta \cos \theta\right] \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\sin \alpha x \cdot x}{2 \sin \left(\dfrac{\alpha+\beta}{2}\right) x \sin \left(\dfrac{\alpha-\beta}{2}\right) x} \\ \\ & =\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{\dfrac{\sin \alpha x}{\alpha x} \cdot x \cdot(\alpha x)}{ \dfrac{2 \sin\left(\dfrac{\alpha+\beta}{2}\right) x}{\left(\dfrac{\alpha+\beta}{2}\right) x} \cdot \dfrac{\sin\left(\dfrac{\alpha-\beta}{2}\right) x}{\sin\left(\dfrac{\alpha-\beta}{2}\right) x} \cdot \left(\dfrac{\alpha+\beta}{2}\right) x \cdot \left(\dfrac{\alpha-\beta}{2}\right) x} \\ \end{aligned} $

$ \begin{aligned} & =\dfrac{1}{2} \cdot \dfrac{\alpha \cdot 4}{\alpha^{2}-\beta^{2}} \left[\dfrac{\lim _{x \rightarrow 0} \left(\dfrac{\sin \alpha x}{\alpha x}\right)}{\lim _{x \rightarrow 0} \sin \dfrac{\left(\dfrac{\alpha+\beta}{2}\right) x}{\left(\dfrac{\alpha+\beta}{2}\right) x} \lim _{x \rightarrow 0} \sin \dfrac{\left(\dfrac{\alpha-\beta}{2}\right) x}{\left(\dfrac{\alpha-\beta}{2}\right) x}}\right] \\ \\ & =\dfrac{1}{2} \cdot \left(\dfrac{4 \alpha}{\alpha^{2}-\beta^{2}}\right)=\dfrac{2 \alpha}{\alpha^{2}-\beta^{2}}\qquad\left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1\right] \end{aligned} $

Solution

Given,

$ \begin{aligned} \lim _{x \rightarrow \pi / 4} \dfrac{\tan ^{3} x-\tan x}{\cos \left(x+\dfrac{\pi}{4}\right)} & = \lim _{x \rightarrow \pi / 4} \dfrac{\tan x(\tan ^{2} x-1)}{\cos \left(x+\dfrac{\pi}{4}\right)} \\ \\ & =\lim _{x \rightarrow \pi / 4} \tan x \cdot \lim _{x \rightarrow \pi / 4} \left(\dfrac{1-\tan ^{2} x}{\cos \left(x+\dfrac{ \pi}{4}\right)}\right) \\ \\ & =-1 \times \lim _{x \rightarrow \pi / 4} \dfrac{(1+\tan x)(1-\tan x)}{\cos (x+\dfrac{\pi}{4})} {\qquad[\because \ \ a^{2}-b^{2}=(a+b)(a-b)]} \\ \\ & =-\lim _{x \rightarrow \pi / 4}(1+\tan x) \lim _{x \rightarrow \pi / 4} \left[\dfrac{\cos x-\sin x}{\cos x \cdot \cos \left(x+\dfrac{\pi}{4}\right)}\right] \\ \\ & =-(1+1) \times \lim _{x \rightarrow \pi / 4} \dfrac{\sqrt{2} \left[\dfrac{1}{\sqrt{2}} \cdot \cos x-\dfrac{1}{\sqrt{2}} \cdot \sin x\right]}{\cos x \cdot \cos \left(x+\dfrac{\pi}{4}\right)} \\ \\ & =-2 \sqrt{2} \lim _{x \rightarrow \pi / 4} \left[\dfrac{\cos \dfrac{\pi}{4} \cdot \cos x-\sin \dfrac{\pi}{4} \cdot \sin x}{\cos x \cdot \cos \left(x+\dfrac{\pi}{4}\right)}\right] \qquad {[\because \ \ \cos A \cdot \cos B-\sin A \sin B=\cos (A+B)]} \\ \\ & =-2 \sqrt{2} \lim _{x \rightarrow \pi / 4} \dfrac{\cos \left(x+\dfrac{\pi}{4}\right)}{\cos x \cdot \cos \left(x+\dfrac{\pi}{4}\right)} \\ \\ &=-2 \sqrt{2} \times \dfrac{1}{\dfrac{1}{\sqrt{2}}}=-2 \sqrt{2} \times \sqrt{2}=-4 \end{aligned} $

Solution

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

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

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

Solution

$\text{Given,}$

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

$ \text{LHL} =\lim _{x \rightarrow {\pi^{-}}/{4}} \dfrac{-(x-4)}{x-4}=-1\qquad {\left[\because \ \ |x-4|=-(x-4), x < 4\right]} $

$ \text{RHL} =\lim _{x \rightarrow {\pi^{+}}/{4}} \dfrac{(x-4)}{x-4}=1 \qquad {\left[\because \ \ |x-4|=(x-4), x > 4\right]} $

$\therefore \ \ \text{LHL} \neq \text{RHL}$

So, limit does not exist.

Solution

Given,

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

$ \text{LHL}=\lim _{x \rightarrow \ {\pi^{-}}/{2}} \dfrac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \dfrac{k \cos \left(\dfrac{\pi}{2}-h\right)}{\pi-2 \left(\dfrac{\pi}{2}-h\right)} $

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

$ \qquad=\dfrac{k}{2} \lim _{h \rightarrow 0} \dfrac{\sin h}{h}=\dfrac{k}{2} \cdot 1=\dfrac{k}{2} \qquad \left[ \because \ \ \lim _{h \rightarrow 0} \dfrac{\sin x}{x}=1 \right]$

$ \text{RHL}=\lim _{x \rightarrow {\pi^{+}}/{2}} \dfrac{k \cos x}{\pi-2 x}=\lim _{{ \ h}\to{0}}\dfrac{k \cos \left(\dfrac{\pi}{2}+h\right)}{\pi-2 \left(\dfrac{\pi}{2}+h\right)}$

$ \qquad=\lim _{h \rightarrow 0} \left(\dfrac{-k \sin h}{\pi-\pi-2 h}\right)=\lim _{h \rightarrow 0} \dfrac{k \sin h }{2 h}$

$\qquad=\dfrac{k}{2} \lim _{h \rightarrow 0} \dfrac{\sin h}{ h}=\dfrac{k}{2} \text { and } f \left(\dfrac{\pi}{2}\right)=3 $

$ \text { It is given that, } $

$ \lim _{x \rightarrow \pi / 2} f(x)=f \left(\dfrac{\pi}{2}\right) $

$\Rightarrow \dfrac{k}{2}=3 $

$ \therefore \ \ k=6 $

Solution

Given,

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

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

Solution

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

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

$ =-\lim _{x \rightarrow \pi} \dfrac{\sin (\pi-x)}{(\pi-x)}=-1 \qquad \left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1 \right]$

Solution

(a) Given, $\lim _{x \rightarrow 0} \dfrac{x^{2} \cos x}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{x^{2} \cos x}{2 \sin ^{2} \dfrac{x}{2}}\qquad\left[\because \ \ 1-\cos x=2 \sin ^{2} \dfrac{x}{2}\right]$

$=2 \lim _{x \rightarrow 0} \dfrac{\dfrac{x^2}{2}}{\sin ^2 \dfrac{x}{2}} \cdot \lim _{x \rightarrow 0} \cos x=2 \cdot 1=2\qquad\left[\because \ \ \lim _{x \rightarrow 0} \dfrac{\sin x}{x}=1 \right]$

Solution

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

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

Solution

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

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

Solution

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

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

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

Solution

(c) Given,

$ \begin{aligned} \lim _{x \rightarrow 0} \dfrac{ \ cosec \ x-\cot x}{x} & =\lim _{x \rightarrow 0} \dfrac{\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}}{x}=\lim _{x \rightarrow 0} \dfrac{1-\cos x}{x \cdot \sin x} \\ \\ & =\lim _{x \rightarrow 0} \dfrac{2 \sin ^{2} \dfrac{x}{2}}{x \cdot 2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}}=\lim _{x \rightarrow 0} \dfrac{\tan \dfrac{x}{2}}{x} \qquad \left[\because \cos x = 1 - 2 \sin^2 \dfrac{x}{2}\right] \\ \\ & =\lim _{x \rightarrow 0} \dfrac{\tan \dfrac{x}{2}}{\dfrac{x}{2}} \cdot \dfrac{1}{2}=\dfrac{1}{2}\qquad \left[\because \ \ \lim _{\theta \rightarrow 0} \dfrac{\tan \theta}{\theta}=1\right] \end{aligned} $

Solution

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

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

Solution

(d) $\text{Given,}$

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

Solution

(b) $\text{Given, }$

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

Solution

(d) Given,

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

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

$\because \ $ $ \ \mathrm{LHL \neq RHL} \ $ So, limit does not exist.

Solution

(c) Given,

$ \begin{aligned} \text { limit } & =\lim _{x \rightarrow 0} \dfrac{|\sin x|}{x} \\ \\ \therefore \ \ \text{LHL} & =\lim _{x \rightarrow 0^{-}} \dfrac{-\sin x}{x}=-\lim _{x \rightarrow 0^{-}} \dfrac{\sin x}{x}=-1 \\ \\ \text{RHL} & =\lim _{x \rightarrow 0^{+}} \dfrac{\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 \ \ \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 $\quad \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)$

$ \hspace{2.9cm}=\lim _{h \rightarrow 0}[2(2+h)+3]$

$\hspace{2.9cm}=\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} \dfrac{\tan 2 x-x}{3 x-\sin x} & =\lim _{x \rightarrow 0} \dfrac{x \left[\dfrac{\tan 2 x}{x}-1\right]}{x\left[3-\dfrac{\sin x}{x}\right]} \\ \\ & =\dfrac{\lim _{x \rightarrow 0} 2 \times \dfrac{\tan 2 x}{2 x}-1}{3-\lim _{x \rightarrow 0} \dfrac{\sin x}{x}} \\ \\ &=\dfrac{2-1}{3-1}=\dfrac{1}{2} \end{aligned} $

Solution

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

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

$\therefore \ \ L f^{\prime} \left(\dfrac{1}{2}\right)=\text{LHD}$

$\hspace{1.5cm}=\lim _{h \rightarrow 0} \dfrac{f \left(\dfrac{1}{2}-h\right)-f \left(\dfrac{1}{2}\right)}{-h}$

$\hspace{1.5cm} =\lim _{n \rightarrow 0} \dfrac{\left(\dfrac{1}{2}-h\right)-\left(\dfrac{1}{2}-h\right)-\dfrac{1}{2}+\dfrac{1}{2}}{h}$

$\hspace{1.5cm}=\lim _{h \rightarrow 0} \dfrac{\dfrac{1}{21}-h-0-\dfrac{1}{2}+0=1}{h} $

and $\quad R f^{\prime} \left(\dfrac{1}{2}\right)=\text{RHD} $

$\hspace{2.5cm} = \lim _{h \rightarrow 0} \dfrac{f \left(\dfrac{1}{2}+h\right)-f \left(\dfrac{1}{2}\right)}{h}$

$\hspace{2.5cm} =\lim _{h \rightarrow 0} \dfrac{\dfrac{1}{2}+h-\dfrac{1}{2}+h-\dfrac{1}{2}+\dfrac{1}{2}}{h}$

$\hspace{2.5cm} =\lim _{h \rightarrow 0} \dfrac{\dfrac{1}{2}+h-0-\dfrac{1}{2}+0}{h}=1 $

$\because \ \ \quad \text{LHD}=\text{RHD}$

$\therefore \ \ f^{\prime} \left(\dfrac{1}{2}\right)=1$

Solution

(d) We have,

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

On taking differentiating both sides w.r.t x , we get

$ \dfrac{d y}{d x}=\dfrac{1}{2 \sqrt{x}}-\dfrac{1}{2(x)^{{3}/{2}}} $

Since, $\mathrm{x}=1$ Therefore,

$ \begin{aligned} & \left.\dfrac{d y}{d x}\right|{x=1}=\dfrac{1}{2 \sqrt{1}}-\dfrac{1}{2(1)^{{3}/{2}}} \\ \\ & \left.\dfrac{d y}{d x}\right|{x=1}=\dfrac{1}{2}-\dfrac{1}{2} \\ \\ & \left.\dfrac{d y}{d x}\right|_{x=1}=0 \end{aligned} $

Hence, this is the answer.

Solution

(a) $\text{Given,}$

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

Solution

(a) Given,

$ \begin{aligned} y & =\dfrac{1+\dfrac{1}{x^{2}}}{1-\dfrac{1}{x^{2}}} \Rightarrow y=\dfrac{x^{2}+1}{x^{2}-1} \\ \\ \dfrac{d y}{d x} & =\dfrac{(x^{2}-1) 2 x-(x^{2}+1)(2 x)}{(x^{2}-1)^{2}}\qquad \left[\text { by quotient rule }\right] \\ \\ \dfrac{d y}{d x} & =\dfrac{2 x(x^{2}-1-x^{2}-1)}{(x^{2}-1)^{2}} \\ \\ & =\dfrac{2 x(-2)}{(x^{2}-1)^{2}}=\dfrac{-4 x}{(x^{2}-1)^{2}} \end{aligned} $

$ \therefore \ \ \dfrac{d y}{d x}= \dfrac{-4 x}{(x^{2}-1)^{2}} $

Solution

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

$ \begin{aligned} \therefore \ \ \dfrac{d y}{d x} & =\dfrac{(\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] } \\ \\ & =\dfrac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x-\cos x)^{2}} \\ \\ & =\dfrac{-[(\sin x-\cos x)^{2}+(\sin x+\cos x)^{2}]}{(\sin x-\cos x)^{2}} \\ \\ & =\dfrac{-[\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}} \\ \\ & =\dfrac{-2}{(\sin x-\cos x)^{2}} \\ \\ \therefore \ \ \dfrac{d y}{d x} & =-2 \end{aligned} $

Solution

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

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

Solution

(b) Given,

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

$ \begin{aligned} & \therefore \ \ f^{\prime}(x)=0+1+2 \times \dfrac{x}{2}+\ldots+100 \dfrac{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)=\dfrac{x^{n}-a^{n}}{x-a} $

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

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{aligned} \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 \\ \\ & =\dfrac{100}{2}[2 \times 100+(100-1)(-1)] & \\ \\ & =50[200-99] \\ \\ & =50 \times 101 \\ \\ & =5050 \end{aligned} $

Solution

(d) Given,

$ \begin{aligned} f(x) & =1-x+x^{2}-x^{3}+\ldots-x^{99}+x^{100} \\ \\ 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 \ \ f^{\prime}(1) & =-1+2-3+\ldots-99+100 \\ \\ & =(-1-3-5-\ldots-99)+(2+4+\ldots+100) \\ \\ & =-\dfrac{50}{2}[2 \times 1+(50-1) 2]+\dfrac{50}{2}[2 \times 2+(50-1) 2] \qquad \left[\because \ \ S_{n}=\dfrac{n}{2}{2 a+(n-1) d}\right] \\ \\ & =-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,

$ \begin{aligned} f(x)=\dfrac{\tan x}{x-\pi} & =\lim _{x \rightarrow \pi} \dfrac{\tan x}{x-\pi} \\ \\ & =\lim _{\pi \rightarrow x \rightarrow 0} \dfrac{-\tan (\pi-x)}{-(\pi-x)}=1 \qquad \left[ \ \because \ \ \ \lim _{x \rightarrow 0} \dfrac{\tan x}{x}=1\right] \end{aligned} $

Solution

Given, $\lim _{x \rightarrow 0} \sin m x \cot \left(\dfrac{x}{\sqrt{3}}\right)=2$

$ \begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow 0} \left(\dfrac{\sin m x}{m x}\right) \cdot m x \cdot \dfrac{1}{\tan \left(\dfrac{x}{\sqrt{3}}\right)}=2 \\ \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \left(\dfrac{\sin m x}{m x}\right) \cdot m x \cdot \dfrac{\dfrac{x}{\sqrt{3}}}{\tan \left(\dfrac{x}{\sqrt{3}}\right)} \cdot \dfrac{1}{\dfrac{x}{\sqrt{3}}}=2 \\ \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \left(\dfrac{\sin m x}{m x}\right) \cdot \lim _{x \rightarrow 0} \dfrac{\dfrac{x}{\sqrt{3}}}{\tan \left(\dfrac{x}{\sqrt{3}}\right)} \cdot \lim _{x \rightarrow 0} \left(\dfrac{m x}{\dfrac{x}{\sqrt{3}}}\right)=2 \\ \\ & \begin{aligned} \Rightarrow \quad \sqrt{3} \ m=2 \end{aligned} \\ \\ & \therefore \ \ m=\dfrac{2 \sqrt{3}}{3} \end{aligned} $

Solution

Given,

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

Solution

Given,

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