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Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$ exists. If it does, then calculate its value.

My attempt: $$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$

But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$ Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$

So the left hand and right hand limits cannot be equal, and therefore the limit does not exist.

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    $\begingroup$ You should invest the time to learn how to typeset your mathematics with LaTeX. See this and then this. $\endgroup$ – Mike Pierce May 31 '15 at 23:11
  • $\begingroup$ I've formatted your question into MathJax. I recommend looking at the edit so you can see what changes. You'll see that your own math syntax is easily converted to MathJax. (for example (x+y)^2/(x^2+y^2) becomes \frac {(x+y)^2}{x^2+y^2}. $\endgroup$ – GPerez May 31 '15 at 23:17
  • $\begingroup$ wolframalpha.com/input/… Wolfram says that the limit doens't exists, so take two paths to (0,0) and find that function goes to two different values in differents paths. $\endgroup$ – Luis Felipe May 31 '15 at 23:19
  • $\begingroup$ Remember that in $2D$, there are many paths you can take for limits as opposed to $1D$. $\endgroup$ – MathNewbie May 31 '15 at 23:26
  • $\begingroup$ Thanks! I didn't know how to get the formatting, will have a look into those links and try to format it properly myself next time. $\endgroup$ – mathstack May 31 '15 at 23:31
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Consider $$f(x,y)=\frac{(x+y)^2}{x^2+y^2} .$$ If you take the path $(0,y)$, then: $$\displaystyle\lim_{y\to0} f(0,y) =\lim_{y\to0} \frac{y^2}{y^2}=1$$

If you take the path $(x,x)$, then: $$ \lim_{x\to0}f(x,x) =\lim_{x\to0} \frac{(2x)^2}{2x^2}=2$$

So, the limit doesn't exist.

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  • $\begingroup$ Why downvote? Notice me first for improve my answer. $\endgroup$ – Luis Felipe May 31 '15 at 23:46
  • $\begingroup$ Done (although I did not vote). (While we are at it, you might also ask about the upvotes...) $\endgroup$ – Did Jun 1 '15 at 13:42
  • $\begingroup$ One guy was giving down votes in every question and answer, nevermind $\endgroup$ – Luis Felipe Jun 1 '15 at 17:45
  • $\begingroup$ This might be, but what do you think of the upvotes to the previous version of your answer? $\endgroup$ – Did Jun 1 '15 at 18:06
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    $\begingroup$ I am not alluding to $\LaTeX$ formatting, and I have the feeling that you know it... Never mind. $\endgroup$ – Did Jun 1 '15 at 18:18
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Although the question has been answered aptly, it might be instructive to see the application of polar coordinate transformation in the analysis of limits of the type herein.

To that end, we transform the problem to polar coordinates. Letting $x=\rho \cos\phi$ and $y=\rho \sin \phi$, we have

$$\frac{(x+y)^2}{x^2+y^2}=1+\sin 2\phi$$

Obviously, the limit

$$\lim_{\rho \to 0} (1+\sin 2\phi)$$

is highly dependent on the migration of $\phi$ as $\rho \to 0$. Thus, the limit does not exist.

Previous answers have as paths $1)$ $\phi =\pi/2$ for which the limit is $1$ and $2)$ $\phi = \pi/4$ for which the limit is $2$.

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  • $\begingroup$ I think our answers are off by an arctan or tan depending on one's point of view. $\endgroup$ – Baby Dragon Jun 1 '15 at 0:00
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    $\begingroup$ @BabyDragon Yes, true if $k$ is allowed to vary with $x$ and $y$. In this answer, we note that $\phi$ is viewed as $\phi (\rho)$. Thus, non-linear paths are included for consideration. $\endgroup$ – Mark Viola Jun 1 '15 at 0:05
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Here's another way $$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2} $$ Using polar coordinates, we have $$ \lim\limits_{r\to 0^+} \frac{\left(r\cos\phi+r\sin\phi\right)^2}{r^2\cos^2\phi + r^2\sin^2\phi} $$ $$ = \lim\limits_{r\to 0^+} \frac{r^2\left(\cos\phi+\sin\phi\right)^2}{r^2\left(\cos^2\phi + \sin^2\phi\right)} $$ $$ = \lim\limits_{r\to 0^+} \frac{\left(\cos\phi+\sin\phi\right)^2}{\cos^2\phi + \sin^2\phi} $$ $$ = \lim\limits_{r\to 0^+} \left(\cos\phi+\sin\phi\right)^2 $$ $$ = \sin(2\phi)+1 $$ This limit is clearly dependent on $\phi$. Therefore $$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2}\ \mbox{does not exist} $$

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A trick I like to use for these sorts of problems is to set $x=kt$, and $y=t$, where I may vary $k$ at will to possibly be able to produce a variety of values. Then if $f(x,y)$ is a function of two variables, and I am interested in $$lim_{(x,y)\to(0,0)}f(x,y) , $$ I need to look at $$lim_{t\to 0}f(kt , t).$$ Let us set (this actually gives almost all the lines through the origin). In your particular problem, you get that $$f(x,y) = \frac{(x+y)^2}{x^2+y^2}.$$ Now plugging in $x=kt$, and $y=t$, we get, $$\frac{t^2(k+1)^2}{t^2(k^2+1)}=\\ \frac{(k+1)^2}{(k^2+1)}.$$ Therefore you may make the limit approach any number in the range of $$k\mapsto \frac{(k+1)^2}{(k^2+1)}.$$

This corresponds to taking the limit by approaching the value (0,0) along (almost) any ray that contains $(0,0)$. One can often quickly see that a limit does not exist very quickly using this trick. But it does have a gotcha: If all of the rays do in fact give the same answer, it is possible that there is some non-straight line path that will give a different answer. Here What is $\lim_{(x,y)\to(0,0)} (x^3+y^3)/(x^2-y^2)$? , I believe is an example of a function where it behaves on the rays where the limit still fails to exist.

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  • $\begingroup$ Why downvote? can you tell him what are wrong before give a downvote? I'm giving a upvote to counteract $\endgroup$ – Luis Felipe May 31 '15 at 23:47
  • $\begingroup$ @LuisFelipeVillavicencioLopez I do not particularly care about the down vote per se. I would not mind addressing any questions about the answer though. $\endgroup$ – Baby Dragon May 31 '15 at 23:49
  • $\begingroup$ I think people here already forgotten when initiated in Calculus and had doubts, consider such questions as silly and qualify negative, I do not like. $\endgroup$ – Luis Felipe May 31 '15 at 23:52
  • $\begingroup$ @LuisFelipeVillavicencioLopez You may be right, you may be wrong:). This answer is not the easiest answer to read. Of course such sentiments would be appropriate on MathOverflow. $\endgroup$ – Baby Dragon May 31 '15 at 23:58
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    $\begingroup$ Upvotes "to counteract downvotes" (or the other way around) should be avoided. $\endgroup$ – Did Jun 1 '15 at 13:43
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Your final answer is correct, but the way there is just slightly misleaded. The main problem is that you seperate your limit into a sum of limits. The first, $$\lim_{(x,y)\to(0,0)} \frac{x^2+y^2}{x^2+y^2}$$ converges and is equal to $1$, as you note. The second though, $$\lim_{(x,y)\to(0,0)} \frac {2xy}{x^2+y^2}$$ diverges.

The first problem (but probably not the most important) is that it doesn't diverge for the reasons you state (if you look closely, your argument actually concludes that the above limit is $0$). A proof of its divergence would be to consider the limit through points $(x,x)$ vs. through $(0,y)$, as suggested in another answer.

The more important problem though, is that this reasoning cannot conclude that the original limit doesn't exist! The equality (compactly stated) $$\lim f+g = \lim f + \lim g$$ holds only if both the limits of $f$ and of $g$ exist. In that case we can conclude that the original limit exists and is equal to blah blah. However there is nothing to be said in the case that either limit doesn't exist.

You could modify your method by keeping the expansions inside the limit operand: $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2} = \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{x^2+y^2} + \frac{2xy}{x^2+y^2}$$

and conclude by the "path-dependence" argument.

Just to make clear what I mean, here's an example where the same argument would conclude that a very evidently convergent limit, doesn't exist! $$0 = \lim_{x\to +\infty} \sin x-\sin x \stackrel{?}{=}\lim_{x\to +\infty}\sin x - \lim_{x\to +\infty}\sin x$$

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If we approach $y=mx $ then $\lim_{(x,y)\to (0,0)}\frac{(x+y)^2}{x^2+y^2} =\lim_{x\to 0}\frac{(x+xm)^2}{x^2+m^2x^2}=\lim_{x\to 0}\frac{(1+m)^2}{1+m^2}=\frac{(1+m)^2}{1+m^2} $ which is depends on $m$ so limit is not unique . So limit does not exist .

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