Question

If $f\left(x\right)=\sqrt{1-e^{-x^2}}$, then at $x=0$, $f\left(x\right)$ is

• A. differentiable as well as continuous
• B. continuous but not differentiable
• C. differentiable but not continuous
• D. neither differetiable nor continuous

Answer 1

Answer: (B)

continuous but not differentiable

Here, $f\left(0\right)=0$
$L{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0-h)-f\left(0\right)}{-h}$
$={lim}_{h\to 0}\frac{\sqrt{1-e^{-h^2}}}{-h}$
$={lim}_{h\to 0}\frac{{[1-(1-h^2+\frac{h^4}{2!}-......)]}^{1/2}}{-h}$
$={lim}_{h\to 0}\frac{h{[1-\frac{h^4}{2!}+....]}^{1/2}}{-h}=-1$
$R{f}^{\prime }\left(0\right)={lim}_{h\to 0}\frac{f(0+h)-f\left(0\right)}{h}$
$={lim}_{h\to 0}\frac{\sqrt{1-e^{-h^2}}}{h}$
$={lim}_{h\to 0}\frac{{[1-(1-h^2+\frac{h^4}{2!}-......)]}^{1/2}}{h}=1$

Hence, $f\left(x\right)$ is not differentiable at $x=0$. since
$L{f}^{\prime }\left(0\right)$ and $R{f}^{\prime }\left(0\right)$ are finite, therefore , $f\left(x\right)$ s continuous at $x=0$

Hence, $f\left(x\right)$ is continuous but not differentiable at $x=0$