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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

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    $\begingroup$ Hint: $e^{-1/|h|}\le\frac1{1+\frac1{|h|}+\frac1{2|h|^2}+\frac1{6|h|^3}}\le6|h|^3$ $\endgroup$ – robjohn Jan 28 at 13:50
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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 $\lim_{z\to\infty}z/e^z=0$.

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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 $\frac{z}{e^z} \to 0$.

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To use your result, substitute $h\mapsto2h$ and square: $$ \lim_{h\to0^-}\frac{e^{-1/|h|}}{(2h)^2}=\left(\lim_{h\to0^-}\frac{e^{-1/|2h|}}{2h}\right)^2=0 $$

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