TITLE: Showing that $\lim_{x\to\infty}x^{2}e^{-x^{8}\sin^{2}(x)}$ does not exist.
QUESTION [5 upvotes]: The following function turns up in quantum mechanics as an example of an element of $\mathscr{L}^{2}(\mathbb{R})$ which does not decay to zero at $\pm\infty$:
$$
f(x)
=
x^{2}e^{-x^{8}\sin^{2}(x)}
$$
Intuitively the reason is that $f(x)$ oscillates with a shorter period the larger $x$ becomes and this plays havoc with convergence. I want to show explicitly that this limit does not exist, but it has been a long time since elementary calculus and I haven't been able to get very far. Any help would be appreciated.

REPLY [3 votes]: As showed by Ninad Munshi fine answer limit doesn't exist because we have found two different paths with limit $0$ and $\infty$.
More in general, we can obtain for the limit any other value $L\in(0,\infty)$, notably we have
$$x^{2}e^{-x^{8}\sin^{2}(x)}=e^{2\log x-x^{8}\sin^{2}(x)}$$
and by
$$x_n=2\pi n+\sqrt{\frac{2\log (2\pi n)-\log L}{(2\pi n)^8}} \to \infty$$
we have

$\log x_n \sim \log (2\pi n)$
$x_n^8\sim (2\pi n)^8$
$\sin^{2}(x_n)\sim \frac{2\log (2\pi n)-\log L}{(2\pi n)^8}$

and therefore
$$2\log x_n-x_n^{8}\sin^{2}(x_n) \sim 2\log (2\pi n)-(2\pi n)^8\frac{2\log (2\pi n)-\log L}{(2\pi n)^8}=\log L$$
such that
$$x_n^{2}e^{-x_n^{8}\sin^{2}(x_n)}=e^{2\log x_n-x_n^{8}\sin^{2}(x_n)}\to L$$