# How to show f′(0) exists for f(x)=e^{\frac{-1}{x^2}} \sin(\frac 1x) for

How to show f′(0) exists for $f\left(x\right)={e}^{\frac{-1}{{x}^{2}}}\mathrm{sin}\left(\frac{1}{x}\right)$ for $x\ne 0$ and f(0)=0?
You can still ask an expert for help

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Hattie Schaeffer
Hint
$|\frac{{e}^{\frac{-1}{{h}^{2}}}\mathrm{sin}\left(\frac{1}{h}\right)}{h}|\le |\frac{{e}^{-\frac{1}{{h}^{2}}}}{h}|$
$\underset{h\to 0}{lim}\frac{{e}^{-\frac{1}{{h}^{2}}}}{|h|}$
can be calculated with the substritution $x=\frac{1}{|h|}$