# Evaluate $\lim \limits_{n\to\infty} ne^{-nx^2}$

From intuition, I'm pretty sure that the limit converges to $0$, as the exponential function would decay towards $0$ more quickly than $n$ would increase towards $\infty$. However, if this is correct, how would I show this rigorously?

I tried using a few limit theorems (e.g. Squeeze Theorem) and played around with manipulating $n = e^{\ln (n)}$ but am still stuck.

• It tends to $0$ if $x \ne 0$, and to $+\infty$ if $x = 0$. – TonyK Feb 18 '16 at 11:19

Squeeze theorem for $\;x\neq 0\;$ , otherwise limit is infinity:

$$0\le\frac n{e^{nx^2}}=\frac n{1+nx^2+\frac{n^2x^4}2+\ldots}\le\frac n{\frac{n^2x^4}2}\longrightarrow0$$

• This is surely the most convincing approach. +1. – Vim Feb 18 '16 at 11:19

Rewrite as $\frac{n}{\exp(n x^2)}$ and use L'Hospital.

We can also use the Stolz–Cesàro theorem. We have that $$\lim_{n\to\infty}\frac{n+1-n}{e^{(n+1)x^2}-e^{nx^2}}=\lim_{n\to\infty}\frac{1}{e^{nx^2}(e^{x^2}-1)}=0$$ since $e^{nx^2}\to\infty$ as $n\to\infty$ if $x\ne0$. By the Stolz–Cesàro theorem, $$\lim_{n\to\infty}\frac{n}{e^{nx^2}}=0$$ if $x\ne0$.

• Very beautiful. +1 – DonAntonio Feb 18 '16 at 11:24

If $x\neq 0$ you can use $\displaystyle e^{-nx^2}=\sum_{k=0}^{\infty} \frac{(-nx^2)^k}{k!}$.

• Why would this be limited to $x \neq 0$? – Jess Feb 18 '16 at 11:20