Discuss the continuity and differentiability of $f(x)=e^{|x|}$.
Given: $f(x)=e^{|x|}$
$\Rightarrow f(x)= \begin{cases}e^{x}, & x \geq 0 \\ e^{-x}, & x<0\end{cases}$
Continuity:
(LHL at x = 0)
$\lim _{x \rightarrow 0^{-}} f(x)$
$=\lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0} e^{-(0-h)}$
$=\lim _{h \rightarrow 0} e^{h}$
$=1$
$(\mathrm{RHL}$ at $x=0)$
$\lim _{x \rightarrow 0^{+}} f(x)$
$=\lim _{h \rightarrow 0} f(0+h)$
$=\lim _{h \rightarrow 0} e^{(0+h)}$
$=1$
and $f(0)=e^{0}=1$
Thus, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
Hence,function is continuous at x = 0 .
Differentiability at x = 0.
(LHD at $x=0$ )
$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$
$=\lim _{h \rightarrow 0} \frac{e^{-(0-h)}-1}{-h}$
$=\lim _{h \rightarrow 0} \frac{e^{h}-1}{-h}$
$=-1$
$\left[\because \lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1\right]$
$(\mathrm{RHD}$ at $x=0)$
$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}$
$=\lim _{h \rightarrow 0} \frac{e^{(0+h)}-1}{h}$
$=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}$
$=1$ $\left[\because \lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1\right]$
LHD at $(x=0) \neq R H D$ at $(x=0)$
Hence the function is not differentiable at $x=0$.
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