# $\lim_{h\to0}\frac 1{he^{\frac 1{h^2}}}=0$

I would like to prove that $$\displaystyle\lim_{h\to0}\frac 1{he^{\frac 1{h^2}}}=0.$$

But L'Hopital's Rule doesn't seem to help, nor does the Sandwich Theorem. Any suggestions?

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substitution $u = 1/h$. – Yimin Mar 5 '13 at 20:45
There are in principle two cases to examine. If $h\lt 0$, set $x=1/|h|$. – André Nicolas Mar 5 '13 at 20:52
@Yimin Thanks! [][][] – Ryan Mar 5 '13 at 20:52
@AndréNicolas Oh yes, thanks for pointing out! – Ryan Mar 5 '13 at 20:54

$\dfrac 1{he^{\frac 1{h^2}}}=\dfrac {\frac{1}{h}}{e^{\frac 1{h^2}}}\xrightarrow[h\to0^+]{}\dfrac{+\infty}{+\infty}$ L'Hôpital $\Rightarrow$

$\displaystyle \lim_{h\to 0^+}\dfrac 1{he^{\frac 1{h^2}}}=\lim_{h\to0^+}\dfrac {-\frac{1}{h^2}}{-\dfrac{2e^{\frac 1{h^2}}}{h^3}}=\lim_{h\to0^+}\dfrac {h}{2e^{\frac 1{h^2}}}=0.$

Do the same for the $\lim_{h\to0^-}$

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I think $\lim_{x \to 0+}$ should be replaced by $\lim_{h \to 0+}$ . – user52976 Apr 12 '13 at 5:01

Set $x= 1/h$:

$$\lim_{h \to 0} {1 \over h e^{1/h^2}} = \lim_{x \to \infty } {x \over e^{x^2}} = 0$$

as the limit of a continuous function $f$ obeys the rule $\lim_{x \to a}f(x) = f(a)$

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Substitute, $h=\frac{1}{t}$,
then $$\lim_{h\to 0^+}\frac{1}{he^{1/{h^2}}}=\lim_{t\to\infty}\frac{t}{e^{t^2}}=0$$
and similarly, $$\lim_{h\to 0^-}\frac{1}{he^{1/{h^2}}}=\lim_{t\to -\infty}\frac{t}{e^{t^2}}=0$$
Since , $$\lim_{h\to 0^+}\frac{1}{he^{1/{h^2}}}=\lim_{h\to 0^-}\frac{1}{he^{1/{h^2}}}$$
therefore, $\lim_{h\to 0}\frac{1}{he^{1/{h^2}}}$ exists and equal to $0$