Answer
$0$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{h \to 0} f'(0)=\lim\limits_{h \to 0 }\dfrac{f(0+h)-f(0)}{h})$
and $\lim\limits_{h \to 0 }\dfrac{1/h}{e^{1/h^2}}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{h \to 0} \dfrac{-1/h^2}{e^{1/h^2}(-2/h^2)}=\lim\limits_{h \to 0} \dfrac{h}{2} e^{-1/h^2}=0$
