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.

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

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

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

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

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