The function $f(x)=e^{-|x|}$ is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these
(a) continuous everywhere but not differentiable at x = 0
Given: $f(x)=e^{-|x|}= \begin{cases}e^{x}, & x \geq 0 \\ e^{-x}, & x<0\end{cases}$
Continuity :
$\lim _{x \rightarrow 0^{-}} f(x)$
$=\lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0} e^{-(0-h)}$
$=\lim _{h \rightarrow 0} e^{h}$
$=1$
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 $f(0)=f(0)=e^{0}=1$
Thus, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f$
Hence, function is continuous at x = 0
Differentiability at x = 0
(LHD at x = 0)
$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$
$=\lim _{h \rightarrow 0} \frac{e^{-(0-h)}-1}{-h}$
$=\lim _{h \rightarrow 0} \frac{e^{h}}{h}$
$=\infty$
Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.
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