Multivariable limit with polar coordinates

Question

Polar coordinates do not reveal the behaviour of $f(x,y)$ when studying

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

In polar coordinates we have

$$ \lim_{r^2 \to \infty} \frac 12 \frac { r^2 \sin (2 \varphi ) }{ e^{ \frac 14 r^4 \sin^2 (2\varphi) } } $$

which goes to zero independently of $\varphi$. However by letting $y = \frac 1x$, we clearly see that the limit cannot exist. Why didn't polar coordinates work out and what other way than letting $y = \frac 1x$ can we show that it does not exist?

Hm, interesting. That substitution is more or less equivalent to $y = 1/x$. I attempted to find an angle $\varphi$ such that the limit would degenerate but was unable to do so. — user124001, Jan 26 '14 at 18:19

score 4 · Answer 1 · answered Jan 09 '15 at 17:07

But the issue is why polar coordinates can lead to error. In fact this is a recurrent misunderstanding. To explain it I turn to limits when $(x,y)\to(0,0)$ (the same thing up to an inversion). The condition

for all $\theta$ there exist $\lim_{\rho\to0}f(\rho\cos\theta,\rho\sin\theta)$ and does not depend on $\theta$,

only says that all limits along lines through the origin exist and coincide. This is not enough for the limit to exist, there could be different limits along paraboles, for instance (hyperbolas in a comment above). What is equivalent to the existence of a limit, say $L$, is

for all $\varepsilon>0$ there is $\delta>0$ such that if $\ 0<\rho<\delta$,
then for all $0\le\theta<2\pi$ we have $\ |f(\rho\cos\theta,\rho\sin\theta)-L|<\varepsilon.$

In other words, what must be independent of $\theta$ is the rate of convergency $\delta$, besides the limit $l$.

score 1 · Answer 2 · answered Jan 27 '14 at 10:02

Consider the two hyperbolas $$\gamma_\pm:\qquad y=\pm{1\over x}\quad(x>0)\ .$$ When $(x,y)\in\gamma_+$ then $f(x,y)={1\over e}$, and when $(x,y)\in\gamma_-$ then $f(x,y)=-{1\over e}$. Since both hyperbolas extends to infinity there are are points $(x,y)$ with $x^2+y^2$ arbitrarily large and $f(x,y)={1\over e}$, and there are other such points with $f(x,y)=-{1\over e}$. It follows that the envisaged limit cannot exist.

Martín-Blas Pérez Pinilla · Answer 3 · 2014-01-27T09:09:06.040

0

Idea: you can study the auxiliar function $z\mapsto z/\exp z^2$. Obviously, $z=xy$.

EDIT: do the limits on the subsets $x=0$ (or $y=0$, or $x=y$) and $y={1\over\sqrt2 \,x}.$

edited Jan 27 '14 at 09:09

answered Jan 26 '14 at 18:58

Martín-Blas Pérez Pinilla

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So we are reducing to a single variable function and now let $u \to \infty$? This doesn't catch the special case where the limit degenerates. – user124001 Jan 26 '14 at 20:17
The next step: possible values of $z=xy$ (and $z/\exp z^2$) when $x^2+y^2=R$? – Martín-Blas Pérez Pinilla Jan 26 '14 at 20:51
Well z can take on arbitrary values since x and y get arbitrarily large (either negative or positive), so what can we really conclude? – user124001 Jan 26 '14 at 21:49
And arbitrary small. – Martín-Blas Pérez Pinilla Jan 26 '14 at 22:05
Well sure, in that case $z \to 0$ and therefore the limit is $0$ since $e^0 = 1$. What am I not realizing here? – user124001 Jan 26 '14 at 22:09
$z=xy$ can take small and large values when $x^2+y^2=R$. And $z/\exp z^2$... (draw the graph) – Martín-Blas Pérez Pinilla Jan 26 '14 at 22:27
Yes that auxiliary function approaches zero as z becomes sufficiently large and goes to zero as z becomes arbitrarily small. Furthermore, it assumes it's maximum value for $\sqrt 2 /2 $. Not realizing it yet... – user124001 Jan 26 '14 at 22:54
When $(x,y)$ varies in $x^2+y^2=R$, $z/\exp z^2$ varies from 0 to $\sqrt2/2$. – Martín-Blas Pérez Pinilla Jan 26 '14 at 23:02
Sure. Having another look, I see that it varies from $- \sqrt 2 /2$ and $\sqrt 2 / 2$. I do not see where this realization is to lead us.. – user124001 Jan 26 '14 at 23:30
The function $\frac{xy}{e^{x^2y^2}}$ varies in an interval without approaching to a supposed limit when $x^2+y^2$ is arbitrarily large. – Martín-Blas Pérez Pinilla Jan 26 '14 at 23:34
Except it does. When $x^2 + y^2$ get really large, it approaches zero. There are multivariable functions that vary in an interval but still have an existing limit, so I do not how see the property of varying is the one from which we can infer the non-existence of the limit. – user124001 Jan 26 '14 at 23:38
Re-read the definition of limit. – Martín-Blas Pérez Pinilla Jan 26 '14 at 23:41
We want for all $\epsilon$ that $|f(x,y) - L| < \epsilon$ that there exists a $\delta$ such that $\sqrt{x^2 + y^2} > \delta$. – user124001 Jan 26 '14 at 23:46
If $\lim$ exists $f$ will vary in smaller and smaller intervals. – Martín-Blas Pérez Pinilla Jan 26 '14 at 23:53

Multivariable limit with polar coordinates

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