Question

$f\left(x\right)=\{\begin{array}{c}{e}^{-1/x^2},x>0\\ 0,x\le 0\end{array}$, then $f\left(x\right)$ is

• A. Differential at $x=0$
• B. Continuous but not differentiable at $x=0$
• C. Discontinuous at $x=0$
• D. None of the above

Answer 1

The correct option is A Differential at $x=0$

Given function is $f\left(x\right)=\{\begin{array}{c}{e}^{-1/x^2},x>0\\ 0,x\le 0\end{array}$

To check continuity and differentiability of the given function

$\underset{h\to 0}{lim}f\left(x\right)$

$={lim}_{x\to 0}{e}^{-1/x^2}=0$

$\therefore f\left(x\right)$ is continuous at $x=0$

Consider, $\underset{h\to 0^{+}}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{e^{-1/h^2}-0}{h}$

$=\underset{t\to +\infty }{lim}t{e}^{-t^2}$ ..... [Taking $-1/h^2=t⟹t\to \infty ash\to 0^{+}$]

$=\underset{t\to +\infty }{lim}\frac{t}{e^{t^2}}$ which is in the form of $\frac{0}{0}$ if we take it as $\frac{\frac{1}{e^{t^2}}}{\frac{1}{t}}$

$=\underset{t\to +\infty }{lim}\frac{1}{2t{e}^{t^2}}$ ...... [Applying L'Hospital's Rule]

$=0$

Similarly, $\underset{h\to 0^{-}}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$

$=\frac{0-0}{h}=0$

$\therefore \underset{x\to 0^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)$

$\therefore f\left(x\right)$ is differentiable at $x=0$