How to show $f'(0)$ exists for $f(x) = e^{-1\over x^2} \sin\left({1\over x}\right)$ for $x\neq 0$ and $f(0)=0$?

Question

My attempt at the question

We know the derivative $f'(x)$ exists if the following limit exists $$ \lim_{h\to 0}\left(\frac {f(x+h)-f(x)}{h}\right)$$

Plugging in the values we have $$ f'(0)=\lim_{h\to 0}\left( \frac {e^{-1\over h^2} \sin({1\over h})}{h}\right)$$

So now all we have to show is that the limit exists.

But I cannot proceed further I cannot verify that left hand limit is same as right hand limit.

I tried using expansion of exponential and sine function but I could not simplify further. I even tried to convert it to some form so as to apply L'Hôpital's rule but that did not work out either.

Can anyone show how can we solve this?

Answer 1

Hint

$$\left|\frac {e^{-1\over h^2}\sin({1\over h})}{h})\right| \leq\left|\frac {e^{-1\over h^2}}{h}\right|$$

$$\lim_{h \to 0} \frac {e^{-1\over h^2}}{|h|}$$ can be calculated with the substritution $x= \frac{1}{|h|}$.