$\lim_{(x,y)\to 0} \frac{x\sin(y)- y\sin(x)}{x^4 + y^4}$ without polar coordinates?

Question

I have the following limit:

$$\lim_{(x,y)\to 0} \frac{x\sin(y)- y\sin(x)}{x^4 + y^4}$$

And I must evaluate it without polar coordinates. I have tried a lot of stuff but nothing works. Can someone give me a hint?

Since $x$ and $y$ are both going to zero, you’re allowed to suppose that they go to zero together, i.e., you can take $x=y$ as a simplification if you would like (or any other strictly increasing function like that). This might make things a good deal easier. — Jack Crawford, Jun 04 '19 at 22:54
@JackCrawford Yes, but this doesn't proves the limit exists nor that it doesn't, no? — Red Banana, Jun 04 '19 at 22:56
Ah, no it does not. I was under the impression you were just trying to evaluate the limit assuming that it already exists. It’s been a while since I’ve done a problem like this, but iirc you just need to prove that it exists along each of the dimensions separately, right? Perhaps try limiting along $(x,0)$ and $(0,y)$ and see that you get the same result? Actually, I’m not sure if this was enough. I remember having to do something with partial derivatives in my relevant undergrad class — checking that the second derivatives in the basis vectors match, or something? — Jack Crawford, Jun 04 '19 at 22:59
I think checking the second derivatives like this guarantee differentiability at the point and hence guarantees continuity at the point so the existence and uniqueness of the limit is guaranteed. Then you may make the $y=x$ substitution I suggested earlier to get the value. — Jack Crawford, Jun 04 '19 at 23:04
I’m not sure if that will work here — just a thought. Sorry I can’t help more. — Jack Crawford, Jun 04 '19 at 23:06
@JackCrawford That also doesn't guarantee the limit exists. Consider $f(x,y)=\frac{2xy}{x^2+y^2},f(0,0)=0$. For $(x,0)$ and $(0,y)$, you get $0$. If you take the path along $x=y$, you get $1$. (That is, there could be a path in which you get a different value). About the derivatives, anything involving derivatives would be a nightmare. Try to ask the second derivatives to Wolfram Alpha, it's even worse. — Red Banana, Jun 04 '19 at 23:12

score 5 · Answer 1 · answered Jun 04 '19 at 23:14

Let $f(x,y)=\dfrac{x\sin(y)-y\sin(x)}{x^4+y^4}$

We have $f(x,x)=0$.

On the other hand we also have

$f(x,2x)=\dfrac{x\sin(2x)-2x\sin(x)}{17x^4}=\dfrac{x(2x-\frac 43x^3+o(x^3))-2x(x-\frac 16x^3+o(x^3))}{17x^4}=\dfrac{-x^4+o(x^4)}{17x^4}$

Thus $\begin{cases}\lim\limits_{x\to 0}f(x,x)=0\\\lim\limits_{x\to 0}f(x,2x)=-\frac 1{17}\end{cases}$

Since they are different, we can conclude that the limit $\lim\limits_{(x,y)\to (0,0)}f(x,y)$ doesn't exists.

DinosaurEgg · Answer 2 · 2019-06-04T23:32:58.670

3

You can easily disprove that the limit exists by considering the one dimensional family of rays $y=\lambda x, \lambda\in\mathbb{R}$ and taking the limit to the origin along them instead, for a given $\lambda$. If the limit exists then it shouldn't depend on $\lambda$.

The limit on the rays boils down to evaluating:

$$L=\lim_{x\to 0}\frac{x\sin(\lambda x)-\lambda x\sin x}{(1+\lambda^4)x^4}$$

which one can evaluate straightforwardly using L'Hopital's rule or a Taylor expansion to yield:

$$L=\frac{\lambda-\lambda^3}{6(1+\lambda^4)}$$

which does depend on $\lambda$ and therefore the limit cannot exist.

edited Jun 04 '19 at 23:32

answered Jun 04 '19 at 23:15

DinosaurEgg

10,775

note that you can extend the method to a function $\lambda(x)=\frac yx$ instead of a fixed one to cover all paths. – zwim Jun 04 '19 at 23:18
On second thought I don't think one even needs to do that, given that $y=\lambda x$ does precisely what we want because it sets $\theta$ to a specific value, namely $\tan\theta=\lambda$ and if the limit exists for all assignments of $\theta$, then we necessarily have covered all the cases except $\theta=\pi/2, 3\pi/2$ which can be covered on a case by case basis. – DinosaurEgg Jun 04 '19 at 23:32
Yes, it was just a side note on your remark in parenthesis, that with some functions, this method can be generalized to a variable lambda and work from there (in these problems, we just need L to be bounded). Here it is not useful, since exhibiting multiple limits on rays from the origin suffice. :) – zwim Jun 04 '19 at 23:37
I realized it's not a correct statement so I edited it out. :P It is definitely an overkill in multivariable calculus, maybe it will be useful in calculus of variations of some sort... – DinosaurEgg Jun 04 '19 at 23:39

score 2 · Accepted Answer · answered Jun 04 '19 at 23:13

The limit does not exist. Take limit along $y=2x$ for example. You get $\lim \frac {2x\sin \, x(\cos \, x -1)} {(1+2^{4})x^{4}}$. Use the fact that the $\frac {\sin\, x} x \to 1$ and use L'Hopital's Rule to see that the limit is $-1/17$. Since the limit along $x=y$ is $0$ it follows that the given limit does not exist.

$\lim_{(x,y)\to 0} \frac{x\sin(y)- y\sin(x)}{x^4 + y^4}$ without polar coordinates?

3 Answers3