Question

If $f\left(x\right)=e^{-1/x^2},x\ne 0$ and $f\left(0\right)=0$ then $f^{{}^{\prime }}\left(0\right)$ is

• A. $0$
• B. $1$
• C. $e$
• D. non existent

Answer 1

The correct option is A $0$

$f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{e^{-1/h^2}-0}{h}=0$
$\underset{h\to 0}{lim}\frac{1}{h{e}^{1/h^2}}=0$
In the denominator the value will be tending to infinity as $e^{\frac{1}{h^2}}$ dominates $h.$
So the correct option is (A)