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Evaluate $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{(x+y)^2}{x^2+y^2}$

Using polar, we have $x=r\cos(\theta),y=r\sin(\theta)$

Our limit becomes:

$$\lim_{r\to 0}\dfrac{(r\cos(\theta)+r\sin(\theta))^2}{r^2\sin^2(\theta)+r^2\cos(\theta)}=\lim_{r\to 0}\dfrac{r^2\cos^2(\theta)+2r^2\cos(\theta)\sin(\theta)+r^2\sin^2(\theta)}{r^2}$$

Factoring and dividing removes the $r^2$ in the denominator, and we get $1$ as the limit.

However this is not right.

If we consider along the $x-axis$, our limit becomes $1$. If we consider along the line $y=x$, our limit becomes $1/2$, and are clearly not equal.

This means that the limit does not exist but my polar said it does and it equals $1$. Where did I mess up in my polar coordinates?

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    $\begingroup$ It's a quite common mistake to think limits get easier when using polar coordinates, but often (and indeed here) people end up only considering what happens for constant $\theta$, corresponding to straight lines. If you look closely at your calculations you'll probably also find that they are not valid along one of the along one of the axes, as you get get different results. $\endgroup$
    – Henrik supports the community
    Oct 17, 2017 at 21:27
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    $\begingroup$ you get actually $\lim _{ r\rightarrow 0 }{ \frac { { r }^{ 2 }\cos ^{ 2 }{ \theta } +2{ r }^{ 2 }\sin { \theta } \cos { \theta } +{ r }^{ 2 }\sin ^{ 2 }{ \theta } }{ { r }^{ 2 }\cos ^{ 2 }{ \theta } +{ r }^{ 2 }\sin ^{ 2 }{ \theta } } } =\lim _{ r\rightarrow 0 }{ \frac { { r }^{ 2 }+2{ r }^{ 2 }\sin { \theta } \cos { \theta } }{ { r }^{ 2 } } } =1+\sin { 2\theta } $ not $1$ $\endgroup$
    – haqnatural
    Oct 17, 2017 at 21:29
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    $\begingroup$ Note that you're sending $r$ to zero, not $\theta, $ which can vary arbitrarily with $r$ (and how it varies defines the various paths to approach zero). If you fixed $\theta$, (which is taking the limit along a straight line), you'll find, after factoring the $r$, that you get that the limit is $(\cos \theta + \sin \theta)^2,$ not $1$. This clearly depends on $\theta$, and hence the limit does not exist. $\endgroup$
    – stochasticboy321
    Oct 17, 2017 at 21:30
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    $\begingroup$ Also on a less important note, your limit along the line $y=x$ is $2$, not $1/2$. I mention only so it's clear that this accords with the straight line limit $1+\sin(2\theta)$ that somebody derived above. But as Henrik alluded to, even if the limit along all straight lines agreed it would not follow that the limit exists. $\endgroup$
    – spaceisdarkgreen
    Oct 17, 2017 at 22:33
  • $\begingroup$ See also math.stackexchange.com/questions/1307149/… $\endgroup$
    – Arnaud D.
    Aug 27, 2020 at 9:24

2 Answers 2

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Its been extensively discussed and OP is probably a bit exhausted by seeing many view points that some are clear and some are not. Let's give a "nice proof " that the limit does not exist. Just choose two paths: path $1$ is: $x = -y = t$, then the limit is $0$, and path $2$ is: $x = y = t$, then the limit is $2$. Different values of limits show there is no limit at $(0,0)$.

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  • $\begingroup$ This doesn't address at all the actual question that the OP asked: "Where did I mess up in my polar coordinates?" $\endgroup$
    – zipirovich
    Oct 17, 2017 at 23:26
  • $\begingroup$ The problem and solution is clear. I messed up on taking the limit, I took it with respect to $\theta$ instead of $r$. @DeepSea the second half of my question shows how the limit does not exist, $\endgroup$
    – K Split X
    Oct 18, 2017 at 1:48
  • $\begingroup$ @KSplitX: "some one" downvoted me..haha..you see... $\endgroup$
    – DeepSea
    Oct 18, 2017 at 4:59
  • $\begingroup$ Wasn't me, but Ill make it even $\endgroup$
    – K Split X
    Oct 18, 2017 at 11:32
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The above limit did not exists. Indeed for the path $y=-x$ one gets that the limit equals zero, while for the path $y=x$ one gets that

$$\lim_{x\rightarrow 0} \dfrac{(x+x)^2}{x^2+x^2}=2 \neq 0.$$

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