The right hand and left hand limit of the function are respectively.
\(f(x)=\left\{\begin{array}{cc}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.\)

We have, \(f(x)=\left\{\begin{array}{cc}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text { if } x \neq 0 \\ 0, & x=0\end{array}\right.\)
\(\begin{aligned}
\text { Right hand limit } &=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\
&=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1} \\
&=\lim _{h \rightarrow 0} \frac{1-e^{-1 / h}}{1+e^{-1 / h}} \\
& \equiv \frac{1-0}{1+0} \equiv 1 \\
\text { Left hand limit }=& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \\
=& \lim _{h \rightarrow 0} f(-h)
\end{aligned}\)
\(=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1}\)
\(=\lim _{h \rightarrow 0}\left(\frac{e^{\frac{1}{1 / h}}-1}{\frac{1}{e^{1 / h}}+1}\right)=\frac{0-1}{0+1}=-1\)