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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 $0 \infty$ ?

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4 Answers 4

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$$\lim_{y\to 0}\frac{x}{y^2}e^{-\frac{x^2}{y^2}} = \lim_{n\to +\infty} nx e^{-nx^2}\leq\lim_{n\to +\infty}\frac{nx}{\left(1+\frac{n}{2}x^2\right)^2}=0. $$

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  • $\begingroup$ Couldn't one just argument, that $ \limes n to \inf n * e^(-n) = 0 $ ? I think you are using the same argumentation to create the denominator of your last term before 0, however I don't see, why you picked exactly that term estimation as an upper limit. $\endgroup$
    – Imago
    Jan 4, 2015 at 11:52
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    $\begingroup$ I used $e^{nx^2}=\left(e^{nx^2/2}\right)^2\geq(1+nx^2/2)^2.$ $\endgroup$ Jan 4, 2015 at 11:54
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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.

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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: $$ -k = e^{-\infty} = 0 $$

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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.

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