Examine the continuity of the function f(x) = […] at x = 0.

The correct answer is Option (2) → Discontinuous at $x = 0$ because the $\text{LHL} = 0$ and $\text{RHL} = 1$. ##

We have,

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

At $x = 0$,

$\text{LHL} = \lim\limits_{x \to 0^-} \frac{e^{1/x}}{1 + e^{1/x}} = \lim\limits_{h \to 0} \frac{e^{1/(0 - h)}}{1 + e^{1/(0 - h)}}$

Put $x = 0 - h$,

$= \lim\limits_{h \to 0} \frac{e^{-1/h}}{1 + e^{-1/h}} = \lim\limits_{h \to 0} \frac{1}{e^{1/h}(1 + e^{-1/h})}$

$= \lim\limits_{h \to 0} \frac{1}{e^{1/h} + 1} = \frac{1}{e^\infty + 1} = \frac{1}{\infty + 1} = \frac{1}{\infty} = 0 \quad [∵e^\infty = \infty]$

$\text{RHL} = \lim\limits_{x \to 0^+} \frac{e^{1/x}}{1 + e^{1/x}}$

Put $x = 0 + h$,

$= \lim\limits_{h \to 0} \frac{e^{1/(0 + h)}}{1 + e^{1/(0 + h)}} = \lim\limits_{h \to 0} \frac{e^{1/h}}{1 + e^{1/h}} = \lim\limits_{h \to 0} \frac{1}{e^{-1/h} + 1} = \frac{1}{e^{-\infty} + 1}$

$= \frac{1}{0 + 1} = 1 \quad [∵e^{-\infty} = 0]$

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

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