How do you find the limit of $ \lim_{y\to0} \frac{x e^{-x^2/y^2}}{y^2}$ I don't know how to solve this limit
$$ \lim_{y\to0} \frac{x e^ { \frac{-x^2}{y^2}}}{y^2}$$
$\frac{1}{e^ { \frac{x^2}{y^2}}} \to 0$
but $\frac{x}{y^2} \to +\infty$
This limit presents the indeterminate form $0 \infty$ ?
 A: $$\lim_{y\to 0}\frac{x}{y^2}e^{-\frac{x^2}{y^2}} = \lim_{n\to +\infty} nx e^{-nx^2}\leq\lim_{n\to +\infty}\frac{nx}{\left(1+\frac{n}{2}x^2\right)^2}=0. $$
A: The limit is a particular case of the limit
$$
\lim_{u \to +\infty} u e^{-\beta u}, \tag{1}
$$
with $\beta >0$. Indeed, just define $u=y^{-2} \to +\infty$ as $y \to 0$. Rewrite (1) as
$$
\lim_{u \to +\infty} \frac{u}{e^{\beta u}}
$$
and apply De l'Hospital's theorem.
A: Here's one sloppy way to work it:
Assume $x>0$, let
$$k= \lim_{y\to0} \frac{xe^{-\frac{x^2}{y^2}}}{y^2}$$
Take natural logarithm of both sides:
$$\ln(k)= \lim_{y\to0} \left[\ln{x}-2\ln{y}-\frac{x^2}{y^2}\right]=-\infty$$
Therefore
$$ k = e^{-\infty} = 0 $$
For $x<0$, substitute $x\to -x$, and you'll end up with:
$$ -k = e^{-\infty} = 0 $$
A: For $x\ne0$, set $x^2/y^2=t$; then, as $y\to0$, we have $t\to\infty$, so the limit becomes
$$
\lim_{t\to\infty}\frac{1}{x}te^{-t}=\frac{1}{x}\lim_{t\to\infty}\frac{t}{e^t}
$$
that's easy to show being $0$. If $x=0$ there's of course nothing to do.
