Let $f : \mathbb{R}\setminus\{0\}\to \mathbb{R}$ be given by

$$f(x) = \dfrac{2}{x^3}\exp\left(\frac{-1}{x^2}\right).$$

It is not hard to see that $\lim_{x\to 0}f(x)=0$. In truth, if $x\to 0$, we know that $-1/x^2\to -\infty$ so that $\exp(-1/x^2)\to 0$. Now although $2/x^3\to \infty$, since it is the exponential it goes to zero faster than $2/x^3$ goes to infinity and everything works.

Now this is not rigorous, but is the underlying idea of this limit. I'm trying to prove this limit rigorously, but I'm not getting any idea on how to do this. I think that using the definition will be quite complicated, so I believe there could be another way.

How can I show this limit is zero rigorously?

  • $\begingroup$ @GitGud, It was all a typo. The limit is when $x\to 0$, not $x\to \infty$. In that case $-1/x^2\to -\infty$ and $\exp(-1/x^2)\to 0$. If it were $x\to \infty$ as I wrote first then surely we would have $-1/x^2\to 0$ and hence $\exp(-1/x^2)\to \infty$. $\endgroup$ – user1620696 Nov 3 '15 at 12:26
  • $\begingroup$ Using definition shouldn't be too hard (relative to other 'rigorous' methods). What do you know about $\exp$? The Taylor expansion is useful. How are you defining $\exp$? What inequalities regarding $\exp$ have been proven? $\endgroup$ – Git Gud Nov 3 '15 at 12:27
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    $\begingroup$ Possible duplicate of Prove that $x^m/e^x\to0$ when $x\to+\infty$, for every $m$ $\endgroup$ – skyking Nov 3 '15 at 12:44

From the series expansion of $e^x$, we know that if $y \ge 0$, then $e^y \ge 1 + y + \frac12 y^2 > \frac12 y^2$. Putting $y=1/x^2$, we get $e^{1/x^2} > 1/(2x^4)$ for all non-zero $x \in \mathbb R$. Hence $e^{-1/x^2} < 2x^4$.

You can take it from here.


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