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how can I prove that: $$ \lim_{x\to 0} \frac{e^{-1/x^2}}{x} = 0 ? $$ I suppose that the exponential "goes" to $0$ faster than linear, but I'm not sure.

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  • $\begingroup$ Try translating it to $y={1\over x}$ and $y\to\pm\infty$. $\endgroup$
    – Heimdall
    Jun 26, 2015 at 14:49

2 Answers 2

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As it is, your limit goes to $\infty$, since $1/x^2$ is very big whenever $x$ approaches zero. A very different story would be if you had to compute $$ \lim_{x\to 0}\frac{e^{-1/x^2}}{x}. $$

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  • $\begingroup$ oh, I made a mistake writing formula, I edited my question. $\endgroup$
    – linofex
    Jun 26, 2015 at 14:49
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Make the change of variable $y = \frac{1}{x^2}$, then $y \to +\infty $, when $x \to 0$ $$ \frac{e^{\frac{-1}{x^2}}}{x} = \sqrt{y}e^{-y} $$

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  • $\begingroup$ I solve the limit, but I would understand how can I say: "this limit is 0, because of.." with the infinitive rules $\endgroup$
    – linofex
    Jun 26, 2015 at 14:54
  • $\begingroup$ Just make sure you remember to convert your equation back into terms of $x$ toward the end. $\endgroup$
    – Pubbie
    Jun 26, 2015 at 15:17

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