# Can I conclude that $\lim_{x\to0^+}\frac{x^2}{e^{-\frac{1}{x^2}}\cos(\frac{1}{x^2})^2}$ is infinite or it doesn't exists?

$$\lim_{x\to0^+}\frac{x^2}{e^{-\frac{1}{x^2}}\cos(\frac{1}{x^2})^2}$$

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.

• 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? – 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 $$+\infty$$.
• 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. – José Carlos Santos Jan 3 at 9:54
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 $$|\cos (t)| \leq 1$$.
• 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 $\infty$... – User Jan 3 at 10:59
• @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. – Kabo Murphy Jan 3 at 11:48