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

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

Alternate Method

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

$$\begin{aligned} & \text { Let } & f(x) & =x \cos x \\ & \therefore & f(x+h) & =(x+h) \cos (x+h) \\ & \therefore & \frac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}[(x+h) \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x \cos (x+h)+h \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x\{\cos (x+h)-\cos x\}+h \cos (x+h)] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[x\left\{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}\right\}+h \cos (x+h)\right] \\ & =\lim _{h \rightarrow 0}\left[-2 x \sin \left(x+\frac{h}{2}\right) \frac{\sin \frac{h}{2}}{h}+\cos (x+h)\right] \\ & \qquad\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right] \\ & =-2 \lim _{h \rightarrow 0} x \sin \left(x+\frac{h}{2}\right) \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \frac{1}{2}+\lim _{h \rightarrow 0} \cos (x+h) \\ & =-2 \cdot \frac{1}{2} x \sin x+\cos x \\ & =\cos x-x \sin x \end{aligned}$$

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

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

$$\begin{aligned} {\text { Given, }}\quad & \lim _{x \rightarrow 0} \frac{[\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha x]}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x \\ & \left.=\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cdot \cos \beta x+\sin 2 \alpha x]}{\cos 2 \beta x-\cos 2 \alpha x} \cdot 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\right] \\ & =\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cos \beta x+\sin 2 \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \quad\left[\because \cos C-\cos D=2 \sin \frac{C+D}{2} \cdot \sin \frac{D-C}{2}\right] \\ & =\lim _{x \rightarrow 0} \frac{[2 \sin \alpha x \cos \beta x+2 \sin \alpha x \cos \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin \alpha x[\cos \beta x+\cos \alpha x] x}{2 \sin (\alpha+\beta) x \sin (\alpha-\beta) x} \end{aligned}$$

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

$$=\lim _\limits{x \rightarrow 0} \frac{\sin \alpha x \cdot x}{2 \sin \left(\frac{\alpha+\beta}{2}\right) x \sin \left(\frac{\alpha-\beta}{2}\right) x}$$

$$=\frac{1}{2} \lim _\limits{x \rightarrow 0} \frac{\frac{\sin \alpha x}{\alpha x} \cdot x \cdot(\alpha x)}{2 \sin \frac{\left(\frac{\alpha+\beta}{2}\right) x}{\left(\frac{\alpha+\beta}{2}\right) x} \cdot \sin \frac{\left(\frac{\alpha-\beta}{2}\right) x}{\left(\frac{\alpha-\beta}{2}\right) x} \cdot\left(\frac{\alpha+\beta}{2}\right) x \cdot\left(\frac{\alpha-\beta}{2}\right) x}$$

$$\begin{aligned} & =\frac{\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin \alpha x}{\alpha x} \cdot \alpha x^2}{\lim _\limits{x \rightarrow 0} \sin \frac{\left(\frac{\alpha+\beta}{2}\right) x}{\left(\frac{\alpha+\beta}{2}\right) x} \lim _\limits{x \rightarrow 0} \frac{\left(\frac{\alpha-\beta}{2}\right) x}{\left(\frac{\alpha-\beta}{2}\right) x} \cdot\left(\frac{\alpha^2-\beta^2}{4}\right) x^2} \\ & {\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } x \rightarrow 0 \Rightarrow k x \rightarrow 0\right]} \end{aligned}$$

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

Given, $\lim _\limits{x \rightarrow \pi / 4} \frac{\tan ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}$ $\left[\frac{0}{0}\right.$ form $]$

$$=\lim _\limits{x \rightarrow \pi / 4} \frac{\tan x\left(\tan ^2 x-1\right)}{\cos \left(x+\frac{\pi}{4}\right)}=\lim _\limits{x \rightarrow \pi / 4} \tan x \cdot \lim _\limits{x \rightarrow \pi / 4}\left(\frac{1-\tan ^2 x}{\cos \left(x+\frac{g \pi}{4}\right)}\right)$$

$$=-1 \times \lim _\limits{x \rightarrow \pi / 4} \frac{(1+\tan x)(1-\tan x)}{\cos \left(x+\frac{\pi}{4}\right)} \quad\left[\because a^2-b^2=(a+b)(a-b)\right]$$

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