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We want to find

$$ \large \lim_{x^2 + y^2 \to \infty } \frac {xy}{ e^{x^2y^2} }$$

It looks like it goes to zero but if we let $y = \frac 1x$then the limit is equal to $\frac 1e$ i.e. the function becomes constant when the domain is restricted to the curve $y = 1/x$. By changing to polar coordinates, $x = r \cos \theta, y = r \sin \theta$ we get

$$ \large \lim_{r^2 \to \infty} \frac { r^2 \sin \theta \cos \theta }{e^{ r^4 \sin^2 \theta \cos^2 \theta }} = \frac 12 \frac { r^2 \sin 2\theta }{e^{ \frac 14 r^4 \sin^2 2\theta }} $$

Now one would say that since $e^{r^4}$ grows faster than $r^2$, the limit goes to zero. But as clearly demonstrated earlier, the limit does not exist. So, how can we now show from expression above that this limit is not necessarily zero?

pedalshifter
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1 Answers1

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I wouldn't use polar coordinate. Just use the curve $y=x$. Then the expression becomes

$\dfrac{x^2}{e^{x^4}}$. This limit is clearly $0$, while the limit along the curve $y=\dfrac{1}{x}$ is non zero, as you have observed. So limit doesn't exist.

voldemort
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