5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

Clearly, $f(x)$ is continuous and differentiable for all non zero $x.$

Now $\mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0} {e^x} = 1$,

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x){e^{ - x}} = 1$

Also, $f(0) = {e^0} = 1$. So, $f(x)$ is continuous for all $x$.

($LHD $ at $x = 0)$ $ = {\left( {\frac{d}{{dx}}({e^x})} \right)_{x = 0}} = 1$

( $RHD $  at $x = 0)$  $ = {\left( {\frac{d}{{dx}}({e^{-x}})} \right)_{x = 0}} = 1$

So, $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is not differentiable at $L\,f'\,(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$.

Hence $f(x) = {e^{ - \,|\,x\,|}}$ is everywhere continuous but not differentiable at $x = 0$.