# Solve the following equations

Question:

$f(x)=\left\{\begin{array}{l}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$

Solution:

Checking the right hand and left hand limits for the given function, we have

$\lim _{x \rightarrow 0} f(x)=\frac{e^{1 / x}}{1+e^{1 / x}}$

$=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}=\lim _{h \rightarrow 0} \frac{e^{-1 / h}}{1+e^{-1 / h}}$

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

$=\frac{1}{e^{\infty}-1}=\frac{1}{0-1}=-1$ $\left[\because e^{\infty}=0\right]$

$\lim _{x \rightarrow 0^{+}} f(x)=\frac{e^{1 / x}}{1+e^{1 / x}}$

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

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

$=\frac{1}{e^{-\infty}+1}=\frac{1}{0+1}=1$

$\left[e^{-\infty}=0\right]$

$\lim _{x \rightarrow 0} f(x)=0$

Now, as

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

Thus, f(x) is discontinuous at x = 0.