Two Variable Limit: Solve xy²/(sin(x²+y²))

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Hi,

Homework Statement



lim (xy²)/(sin(x² + y²))
(x,y) -> (0,0)

The Attempt at a Solution


I think the answer is 0, but I don't know how to get rid of the sin (x² + y²). I thought about using taylor series but I'm not sure if that works with two variables.

Can someone help me out?
 
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How about switching to polar coordinates?

r^2 = x^2 + y^2
 
sunjin09 said:
Mod note: help removed[/color]

Maybe someone can justify the tilde somehow ...

I suggest you read this:

https://www.physicsforums.com/showthread.php?t=414380

"On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
"
 
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I think I've solved it.
(xy²)/(sin(x² + y²) = (x² + y²)/(sin(x² + y²) * (xy²/(x²+y²))
And I know both these limits (1 and 0).
 
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DivisionByZro said:
I suggest you read this:

https://www.physicsforums.com/showthread.php?t=414380

"On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
"

sunjin09 said:
I thought his only problem was how to get rid of the sine, which I didn't help with, only helped clarifying his thought in a form that facilitates thinking.

BTW, giving an answer directly is still better than no answer at all, I'll be glad to take any answer to my questions, which more often than not go unanswered.
@sunjin09

By now I see that a PF Moderator has removed your offending post.

DivisionByZro was merely pointing out to you some of the rules for posting help on this Forum.

It's your opinion that "... giving an answer directly is still better than no answer at all, ...", but if you continue to try to exercise that behavior in this Forum, I suspect that the Moderators will take your privileges to post.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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