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
On
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 $\frac{z}{e^z} \to 0$.
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 $\lim_{z\to\infty}z/e^z=0$.