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 $x=0$.

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

At $x=a$,

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

and $f(a)=0$

$$\therefore$$ $\mathrm{LHL = RHL}=f(a)$

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

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

$$\begin{aligned} &\text { At } x=0 \text {, }\\ &\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 0^{-}} \frac{e^{1 / x}}{1+e^{1 / x}}=\lim _{h \rightarrow 0} \frac{e^{1 / 0-h}}{1+e^{1 / 0-h}} \\ & =\lim _{h \rightarrow 0} \frac{e^{-1 / h}}{1+e^{-1 / h}}=\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}\left(1+e^{-1 / h}\right)} \\ & =\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}+1}=\frac{1}{e^{\infty}+1}=\frac{1}{\infty+1} \quad \quad\left[\because e^{\infty}=\infty\right] \\ & =\frac{1}{\frac{1}{0}}=0 \\ \mathrm{RHL} & =\lim _{x \rightarrow 0^{+}} \frac{e^{1 / x}}{1+e^{1 / x}} \\ & =\lim _{h \rightarrow 0} \frac{e^{1 / 0+h}}{1+e^{1 / 0+h}}=\lim _{h \rightarrow 0} \frac{e^{1 / h}}{1+e^{1 / h}} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{e^{-1 / h}+1}=\frac{1}{e^{-\infty}+1} \\ & =\frac{1}{0+1}=1 \end{aligned}\\ &\left[\because e^{-\infty}=0\right] \end{aligned}$$

Hence, $\quad \mathrm{LHL} \neq \mathrm{RHL}$ at $x=0$.

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