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

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

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

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

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

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

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 $]$

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

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

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

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

$$ \begin{aligned} = & \lim _{x \rightarrow \pi} \frac{\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)} \quad\left[\because a^2-b^2=(a+b)(a-b)]\right.\\ & \lim _{x \rightarrow \pi} \frac{1}{\cos \frac{x}{4}+\sin \frac{x}{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{aligned}$$

$$\begin{array}{rlr} \text{Given,}\quad & \lim _\limits{x \rightarrow \pi / 4} \frac{|x-4|}{x-4} & \\ \mathrm{LHL} & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{4}} \frac{-(x-4)}{x-4} & {[\because|x-4|=-(x-4), x<4]} \\ & =-1 & \\ \mathrm{RHL} & =\lim _\limits{x \rightarrow \frac{\pi^{+}}{4}} \frac{(x-4)}{x-4}=1 & {[\because|x-4|=(x-4), x>4]} \\ \therefore\quad \mathrm{LHL} & \ne \mathrm{RHL} & \end{array}$$

So, limit does not exist.

$$\begin{aligned} \text{Given,}\quad f(x) & =\left\{\begin{array}{cc} \frac{k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 3, & x=\frac{\pi}{2} \end{array}\right. \\ \therefore \quad\mathrm{LHL} & =\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)} \\ & =\lim _{h \rightarrow 0} \frac{k \sin h}{\pi-\pi+2 h}=\lim _{h \rightarrow 0} \frac{k \sin h}{2 h} \\ & =\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\frac{k}{2} \cdot 1=\frac{k}{2} \quad \left[\because \lim _{h \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}$$

$$\begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi}{2}}+\frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)} \\ & =\lim _{h \rightarrow 0} \frac{-k \sin h}{\pi-\pi-2 h}=\lim _{h \rightarrow 0} \frac{k \sin h}{2 h}=\frac{k}{2} \lim _{h \rightarrow 0} \frac{\sin h}{2 h}=\frac{k}{2} \text { and } f\left(\frac{\pi}{2}\right)=3 \end{aligned}$$

$$\begin{aligned} &\text { It is given that, }\\ &\lim _{x \rightarrow \pi / 2} f(x)=f\left(\frac{\pi}{2}\right) \Rightarrow \frac{k}{2}=3\\ &\therefore \quad k=6 \end{aligned}$$