$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{{{e^{1/x}}}}{{1 + {e^{1/x}}}},}&{{\rm{ if }}x \ne 0}\\{0,}&{{\rm{ if }}x = 0}\end{array}} \right.$at $x = 0$}

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

At $x = 0,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{1/x}}}}{{1 + {e^{1/x}}}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{1/0 - h}}}}{{1 + {e^{1/0 - h}}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - 1/h}}}}{{1 + {e^{ - 1/h}}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{e^{1/h}}\left( {1 + {e^{ - 1/h}}} \right)}}$

$= \frac{1}{{\frac{1}{0}}} = 0$

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

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

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

$\Rightarrow {\rm{LHL}} \ne {\rm{RHL}}$ at $x = 0$.

Therefore we can say that $f(x)$ is discontinuous at $x = 0$.