My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.

I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.

WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.

I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.

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    $\begingroup$ someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic? $\endgroup$ – User Jan 3 at 11:02

The limit$$\lim_{x\to0^+}\frac{x^2}{\exp\left(-\frac1{x^2}\right)}$$is indeed $+\infty$. Since $\dfrac1{\cos^2\left(\frac1{x^2}\right)}\geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+\infty$.

  • $\begingroup$ thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them? $\endgroup$ – User Jan 3 at 9:53
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    $\begingroup$ That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,\varepsilon)$, then it doesn't exist. $\endgroup$ – José Carlos Santos Jan 3 at 9:54
  • $\begingroup$ ok, will check. thanks! $\endgroup$ – User Jan 3 at 9:55
  • $\begingroup$ @user376343 yes it's $\endgroup$ – User Jan 3 at 9:55

The function is not finite valued. If you allow the value $\infty$ then the limit is $\infty$. Use L'Hôpital's Rule twice to show that $\frac {e^{y}} {y^{2}} \to \infty$ as $y \to \infty$. Replace $y$ by $\frac 1 {x^{2}}$ and use the fact that $|\cos (t)| \leq 1$.

  • $\begingroup$ Thanks! value of $\infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $\infty$... $\endgroup$ – User Jan 3 at 10:59
  • $\begingroup$ @User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined. $\endgroup$ – Kabo Murphy Jan 3 at 11:48

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