Showing that $\lim_{x\to\infty}x^{2}e^{-x^{8}\sin^{2}(x)}$ does not exist. 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.

Take two subsequences
$$x_n = \pi n \implies f(x_n) = x_n^2\to +\infty$$
$$x_n = \pi n + \frac{\pi }{2} \implies f(x_n) = x_n^2 e^{-x_n^8} \to 0 $$
As the subsequential limits are distinct, the limit does not exist.
