Question

If $f\left(x\right)=\left\{\begin{array}{ll}{e}^{1/x},\mathrm{if}& x\ne 0\\ 1,\mathrm{if}& x=0\end{array}\right.$ find whether f is continuous at x = 0.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}{e}^{\frac{1}{x}},\mathrm{if}x\ne 0\\ 1,\mathrm{if}x=0\end{array}\right.$

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$
= ​$\underset{h\to 0}{\lim }{e}^{\frac{-1}{h}}=\underset{h\to 0}{\lim }\left(\frac{1}{e^{\frac{1}{h}}}\right)=\frac{1}{\underset{h\to 0}{\lim }{e}^{\frac{1}{h}}}=0$

(RHL at x = 0) = $\underset{\mathrm{x}\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(h\right)$
= ​$\underset{h\to 0}{\lim }{e}^{\frac{1}{h}}=\infty$

Given:
$f\left(0\right)=1$


It is known that for a function $f\left(x\right)$ to be continuous at x = a,
$\underset{\mathrm{x}\to a^{-}}{\lim }f\left(x\right)=\underset{\mathrm{x}\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to 0^{-}}{\lim }f\left(x\right)\ne \underset{\mathrm{x}\to 0^{+}}{\lim }f\left(x\right)$

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