Solve this

Given:

$f(x)= \begin{cases}e^{\frac{1}{x}}, & \text { if } x \neq 0 \\ 1, & \text { if } x=0\end{cases}$

We observe

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$=\lim _{h \rightarrow 0} e^{\frac{-1}{h}}=\lim _{h \rightarrow 0}\left(\frac{1}{e^{\frac{1}{h}}}\right)=\frac{1}{\lim _{h \rightarrow 0} e^{\frac{1}{h}}}=0$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0} e^{\frac{1}{h}}=\infty$

Given:

$f(0)=1$

It is known that for a function $f(x)$ to be continuous at $x=a$,

$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$

But here,

$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$

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