How do I find $\lim\limits_{(x,y)\to0}(x^2+y^2)\sin\left(\frac{1}{xy}\right)$

Question

As the title suggested ,I am kinda stuck at this limit.I tried the following :

-We know that
$$-1\le \sin\left(\frac{1}{xy}\right)\le1$$

when $xy!=0$.

-From here I tried to use a squeeze rule so I multiplied by $x^2+y^2$ thus having $$(x^2+y^2)\le(x^2+y^2) \sin\left(\frac{1}{xy}\right)\le(x^2+y^2)$$

But limit of $\lim\limits_{(x,y)\to 0}x^2+y^2=0$ ; the same goes for its counterpart. So if the two limits are $0$ then our desired limit is also $0$.

Is this wrong? If so could you please give me an alternate solution?

This is correct and there is no need for any alternate solution. — Did, May 07 '17 at 14:42
As I am new to this website ,I didn't know about these rules of formatting my question. Will use in my future posts. — alex2597, May 07 '17 at 14:53

score 0 · Answer 1 · answered Jun 04 '22 at 17:03

0

Your proof and answer is correct. But, even though this is probably just a typo, on the step where you multiplied all parts of the inequality by $x^2+y^2$, you forgot to put a negative sign in front of the first part. You put $(x^2+y^2)$, but for the first part, it should be $-x^2-y^2$, or $-(x^2+y^2)$.

answered Jun 04 '22 at 17:03

ArthD21

382

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – Dennis Gulko Jun 04 '22 at 18:33
Okay, thank you, I will do that next time! – ArthD21 Jun 04 '22 at 21:33

How do I find $\lim\limits_{(x,y)\to0}(x^2+y^2)\sin\left(\frac{1}{xy}\right)$

1 Answers1