Question

Say True or False:

$e^{-\left|x\right|}$ is differentiable at $x=0$.

Answer 1

$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{e^{-|h|}-1}{h}=\underset{h\to 0}{lim}\frac{e^{-h}-1}{h}=li{m}_{h\to 0}-e^{-|h|}=-1$ ........(Using L'Hospital's Rule)
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{e^{-|h|}-1}{-h}=\underset{h\to 0}{lim}\frac{e^{-h}-1}{-h}=\underset{h\to 0}{lim}{e}^{-|h|}=1$ .........(Using L'Hospital's Rule)
Hence, the function is not differentiable at $x=0$