# How to find limit $\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h^2}$

How can I find limit $$\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h^2}$$

I solve subproblem: $$\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h} = \lim_{h\rightarrow0^-}\frac{1}{e^{1/|h|}\cdot h} =\lim_{y\rightarrow -\infty}\frac{y}{e^{|y|}}=0$$ but I have no idea how to apply that for main target

• Hint: $e^{-1/|h|}\le\frac1{1+\frac1{|h|}+\frac1{2|h|^2}+\frac1{6|h|^3}}\le6|h|^3$ – robjohn Jan 28 at 13:50

Note that, since you have the limit for $$h\to0^-$$, the substitution $$y=1/h=-1/|h|$$ brings the limit into the form $$\lim_{y\to\infty}\frac{y^2}{e^y}$$ Now any limit of the form $$\lim_{y\to\infty}\frac{y^k}{e^y}$$ with $$k>0$$ can be dealt with the substitution $$y=kz$$, so you get $$\lim_{z\to\infty}\frac{k^kz^k}{e^{kz}}=k^k\lim_{z\to\infty}\left(\frac{z}{e^z}\right)^{\!k}$$ Thus you just need to know that $$\lim_{z\to\infty}z/e^z=0$$.
It suffices to prove $$\lim_{y \to +\infty} \frac{y^2}{e^y} = 0$$ using the limit you provided.
Notice that you may write: $$\frac{y^2}{e^y} = \frac{y}{e^{y/2}}\cdot \frac{y}{e^{y/2}} = 4 \cdot \frac{y/2}{e^{y/2}} \cdot \frac{y/2}{e^{y/2}}$$
Now you just have to substitute $$z = y/2$$ and take the limit $$\frac{z}{e^z} \to 0$$.
To use your result, substitute $$h\mapsto2h$$ and square: $$\lim_{h\to0^-}\frac{e^{-1/|h|}}{(2h)^2}=\left(\lim_{h\to0^-}\frac{e^{-1/|2h|}}{2h}\right)^2=0$$