Originally, i was trying to find the value of $$\lim_{h \to 0} \frac{e^{-\frac{1}{h^2}}}{h}$$ to find out differentiabiltiy of
$f(x) = \begin{cases} e^{-1/x^2} & \text{ if } x \ne 0 \\ 0 & \text{ if } x = 0 \end{cases}$
at 0.
In this case, $$\lim_{h \to 0} \frac{e^{-\frac{1}{h^2}}}{h}=\lim_{x \to \infty} \frac{x}{e^{x^2}}=\lim_{x \to -\infty} \frac{x}{e^{x^2}}$$
So when I applied L'Hospital's rule, then its value was 0.
And I thought that No matter how big $k$ is, $$\lim_{x \to \infty} \frac{x^k}{e^x}$$ will be equal to zero because exponential's increase speed is much faster than polynomial's.
Definitely, we can also apply L'hospital's rule at here, but I think it is not proper qualitative explanation for the fact that exponential is much bigger than polynomial.
Is there any other approach which explains why about this problem?
(In fact, I tried to use $\epsilon-\delta$ but how can i show that there exist some $M$ s.t. for every $M<x$ then $x^k<\epsilon e^x$?)
(And I also tried to use inequality like $e^x>x+1$ for all $x>0$ but it only worked for $k<1$)
