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Ok guys I was working on this post but it seems I got it wrong. Since the post has been flagged as duplicate and I don't have enough rep to post comments yet, please help me find where I did something wrong :

For reference, the original post was about proving $\lim_{x\to\infty} f(x) = 0$

Let's note $f(x) = {e^{-1\over x^2}\over x}$ and $g(x) = ln(f(x))= -\left(1\over{x^2.ln(x)} \right)$

We can see that $\lim_{x\to\infty} g(x) = 0$ wich is to say $\lim_{x\to\infty} ln(f(x)) = 0$

Which is $\lim_{x\to\infty} f(x) = 1$

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  • $\begingroup$ What do you mean by "help me find where I did something wrong"? Are you asking about why you couldn't get the correct answer to your math question, or are you asking about why your original post was closed? $\endgroup$ – Ben Grossmann Oct 1 '16 at 20:04
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    $\begingroup$ your $g(x)$ is not correct. you must have $\frac{-1}{x^2}-ln(x)$ $\endgroup$ – hamam_Abdallah Oct 1 '16 at 20:05
  • $\begingroup$ Omnom the original post was not mine, I was trying to help the author and got it wrong. Abdallah you're absolutly right that's where I did a mistake. Thanks for the help guys =) $\endgroup$ – Furrane Oct 1 '16 at 20:17
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Note that for all $x$, $0\le e^{-1/x^2}\le 1$. Thus, given $\epsilon>0$, we have

$$\left|\frac{e^{-1/x^2}}{x}\right|\le \frac1x<\epsilon$$

whenever $x>1/\epsilon$. And we are done.

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  • $\begingroup$ This is very clean proof, congratulation. Although my post was more about what I did wrong in my demonstration ( I misapplied $ln()$ ). $\endgroup$ – Furrane Oct 1 '16 at 21:11

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