Prove $\lim_{x \to 0} \frac{1}{x^{2k}}e^{-\frac{1}{x}} = 0$ (IMPROVED, first attempt added).

Question

Fix $k \in \mathbb{N}_{>0}$. How to prove that $$ \lim_{x \to 0} \frac{1}{x^{2k}}e^{-\frac{1}{x}} = 0? $$ Attempt so far:

We use the suggestion of Mark Viola in the comment to obtain $$ e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2x} + \frac{1}{3!x^3} + \ldots + \frac{1}{(2k+1)!x^{2k+1}} + O(|1/x^{2k+1}|). $$ This then gives \begin{align*} \frac{1}{x^{2k}}e^{-\frac{1}{x}} &= \frac{1}{x^{2k}} \cdot \frac{1}{\left(1 + \frac{1}{x} + \frac{1}{2x} + \frac{1}{3!x^3} + \ldots + \frac{1}{(2k+1)!x^{2k+1}} + O(|1/x^{2k+1}|) \right)} \\ & = \frac{1}{x^{2k}+x^{2k-1} + \frac{1}{2}x^{2k-2} + \ldots + \frac{1}{(2k+1)!} + \frac{1}{x(2k+1)!} + O(|x^{2k}/x^{2k+1}|)} \end{align*} It seems like I'm on a dead end here since all terms in the denominator go to $0$ so the term goes to $\infty$.

Answer 1

First of all, I am not so sure that your limit is correct as stated. The limit only works if you approach $0$ from the right hand side. Performing a substitution $x=1/t$, the limit becomes $\lim_{t \to \infty} \frac{t^{2k}}{e^{t}} $. Numerator is of polynomial nature, denominator is e-power. You can finish the rest?

Answer 2

To imranfat's answer we can add the observation that, since $\int_0^\infty t^n e^{-t}=n!$ is finite for $n\ge 0$, your function $\to 0$ as $t\to\infty$. But that's a one-sided $x\to 0^+$ limit; for $x\to 0^-$, the function diverges.