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I Want to prove that $$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ -\frac { 1 }{ { x }^{ 2 } } } }{ x } } =0$$ Is sufficient with right handed limit.

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  • $\begingroup$ Can you prove that $\lim_{x\to \pm\infty} x e^{-x^2}=0$? Well, it is the same. $\endgroup$ Jun 10 '16 at 17:13
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$$\underset{t\to \infty }{\mathop{\lim }} \frac{t}{e^{t^2}}=\underset{t\to \infty }{\mathop{\lim }}\frac{1}{2t\,e^{t^2}}=0$$

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    $\begingroup$ Why was this downvoted? The answer is correct: it is applying l'Hospital. $\endgroup$
    – DonAntonio
    Jun 10 '16 at 15:16
  • $\begingroup$ @Joanpemo No, it's not correct. One has to separate the limits from the left and from the right and do the right substitutions. On the other hand, the OP tells that only the limit from the right is needed, so this answers the partial question. $\endgroup$
    – egreg
    Jun 10 '16 at 15:46
  • $\begingroup$ @egreg Thank you. What left and what right? He's taking the limit when $\;t\to\infty\;$ ... and I think it is the same in the original limit as the "boss" there is $\;e^{-1/t^2}\;$ . $\endgroup$
    – DonAntonio
    Jun 10 '16 at 15:47
  • $\begingroup$ @Joanpemo For $x<0$, substituting $x=1/t$ makes the limit for $t\to-\infty$. If the substitution is $x=-1/t$ a minus sign is missing. $\endgroup$
    – egreg
    Jun 10 '16 at 15:49
  • $\begingroup$ @egreg Thank you, I think you're right about the sign...not that it changes anything basic, but still. $\endgroup$
    – DonAntonio
    Jun 10 '16 at 15:52
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Since $e^y > y$ for all $y \in \mathbb{R}$, we have for $x \neq 0,$ $e^{1/x^2} > 1/x^2 \implies e^{-1/x^2} < x^2,$ and

$$0 < \left|\frac{e^{-1/x^2}}{x}\right| < \frac{x^2}{|x|} = |x|.$$

By the squeeze principle,

$$\lim_{x \to 0}\frac{e^{-1/x^2}}{x} = 0.$$

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A nice case where l'Hôpital just makes things worse, because the derivative of the numerator is $$ \frac{e^{-1/x^2}}{x^3} $$ However, by noting that $$ \lim_{x\to0}\frac{e^{-1/x^2}}{x}= \lim_{x\to0}\frac{e^{-1/x^2}}{x^2}\,x $$ we can as well show that $$ \lim_{x\to0}\frac{e^{-1/x^2}}{x^2} $$ exists and is finite. The function is even, so the limit exists if and only of the limit from the right exists. Thus we can make the substitution $t=1/x^2$ and transform $$ \lim_{x\to0^+}\frac{e^{-1/x^2}}{x^2} = \lim_{t\to\infty}\frac{t}{e^t} $$ and this is computable with l'Hôpital (or other means).

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