We have, $f(x)=\left\{\begin{array}{cc}3 x+5, & \text { if } x \geq 2 \\ x^2, & \text { if } x<2\end{array}\right.$ at $x=2$.

At $x=2$, $$\mathrm{LHL}=\lim _{x \rightarrow 2^{-}}(x)^2$$

$=\lim _\limits{h \rightarrow 0}(2-h)^2=\lim _\limits{h \rightarrow 0}\left(4+h^2-4 h\right)=4$

and

$$\begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow 2^{+}}(3 x+5) \\ & =\lim _{h \rightarrow 0}[3(2+h)+5]=11 \end{aligned}$$

Since, $\mathrm{LHL} \neq \mathrm{RHL}$ at $x=2$

So, $f(x)$ is discontinuous at $x=2$.

We have, $f(x)=\left\{\begin{array}{ll}\frac{1-\cos 2 x}{x^2}, & \text { if } x \neq 0 \\ 5, & \text { if } x=0\end{array}\right.$ at $x=0$.

$$\begin{aligned} \text { At } x=0,\quad\mathrm{LHL} & =\lim _{x \rightarrow 0^{-}} \frac{1-\cos 2 x}{x^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 2(0-h)}{(0-h)^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 2 h}{h^2} \quad [\because \cos (-\theta)=\cos \theta]\\ & =\lim _{h \rightarrow 0} \frac{1-1+2 \sin ^2 h}{h^2} \quad \left[\because \cos 2 \theta=1-2 \sin ^2 \theta\right]\\ & =\lim _{h \rightarrow 0} \frac{2(\sin h)^2}{(h)^2} \quad \left[\because \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right]\\ & =2 \\ \text { RHL } & =\lim _{x \rightarrow 0^{+}} \frac{1-\cos 2 x}{x^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 2(0+h)}{(0+h)^2} \\ & =\lim _{h \rightarrow 0} \frac{2 \sin ^2 h}{h^2}=2 \quad \left[\because \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right]\\ \text{and}\quad f(0) & =5 \\ \text{Since,}\quad \text { LHL } & =\text { RHL } \neq f(0) \end{aligned}$$

Hence, $f(x)$ is not continuous at $x=0$.

We have, $f(x)=\left\{\begin{array}{ll}\frac{2 x^2-3 x-2}{x-2}, & \text { if } x \neq 2 \\ 5, & \text { if } x=2\end{array}\right.$ at $x=2$.

$$\begin{aligned} \text { At } x=2, \quad \mathrm{LHL} & =\lim _{x \rightarrow 2^{-}} \frac{2 x^2-3 x-2}{x-2} \\ & =\lim _{h \rightarrow 0} \frac{2(2-h)^2-3(2-h)-2}{(2-h)-2} \\ & =\lim _{h \rightarrow 0} \frac{8+2 h^2-8 h-6+3 h-2}{-h} \\ & =\lim _{h \rightarrow 0} \frac{2 h^2-5 h}{-h}=\lim _{h \rightarrow 0} \frac{h(2 h-5)}{-h}=5 \\ \mathrm{RHL} & =\lim _{x \rightarrow 2^{+}} \frac{2 x^2-3 x-2}{x-2} \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{2(2+h)^2-3(2+h)-2}{(2+h)-2} \\ & =\lim _{h \rightarrow 0} \frac{8+2 h^2+8 h-6-3 h-2}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 h^2+5 h}{h}=\lim _{h \rightarrow 0} \frac{h(2 h+5)}{h}=5 \\ & \text { and } \\ & f(2)=5 \\ & \therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(2) \end{aligned}$$

So, $f(x)$ is continuous at $x=2$.

We have, $$f(x)=\left\{\begin{array}{ll} \frac{|x-4|}{2(x-4)}, & \text { if } x \neq 4 \\ 0, & \text { if } x=4 \end{array} \text { at } x=4 .\right.$$

At $x=4$,

$$\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 4^{-}} \frac{|x-4|}{2(x-4)} \\ & =\lim _{h \rightarrow 0} \frac{|4-h-4|}{2[(4-h)-4]}=\lim _{h \rightarrow 0} \frac{|0-h|}{(8-2 h-8)} \\ & =\lim _{h \rightarrow 0} \frac{h}{-2 h}=\frac{-1}{2} \quad \text { and } \quad f(4)=0 \neq \mathrm{LHL} \end{aligned}$$

So, $f(x)$ is discontinuous at $x=4$.

We have, $f(x)=\left\{\begin{array}{ll}|x| \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$ at $x=0$.

At $x=0$,

$$\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 0^{-}}|x| \cos \frac{1}{x}=\lim _{h \rightarrow 0}|0-h| \cos \frac{1}{0-h} \\ & =\lim _{h \rightarrow 0} h \cos \left(\frac{-1}{h}\right) \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0 \end{aligned}$$

$$\mathrm{RHL}=\lim _\limits{x \rightarrow 0^{+}}|x| \cos \frac{1}{x}$$

$$=\lim _\limits{h \rightarrow 0}|0+h| \cos \frac{1}{(0+h)}$$

$$=\lim _\limits{h \rightarrow 0} h \cos \frac{1}{h}$$

$=0 \times[$ an oscillating number between $-1$ and $1$]$=0$

and $$f(0)=0$$

Since, $$\mathrm{LHL}=\mathrm{RHL}=f(0)$$

So, $f(x)$ is continuous at $x=0$.