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Give me a clue how to find the limit as x and y approach zero of $(x^2+y^2)*\sin(1/xy)$...I thought about multiplying up and down with $xy$ but that didn't give me anything....

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Squeezing principle. Sine of anything is bounded. – jim Mar 19 at 9:03
@Aaron to prove the limit existence, you should not take a particular case of $x$ and $y$. – Sami Ben Romdhane Mar 19 at 9:05
There is the unpleasantness that if for example we travel towards $(0,0)$ on the $x$-axis, our function is not defined. – André Nicolas Mar 19 at 9:21
When you get an answer that is helpful, you may choose to accept exactly one answer: to accept an answer, click on the $\checkmark$ to the left of the answer you'd like to accept. – amWhy Mar 19 at 18:20

2 Answers

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Hint $$|\sin(1/xy)|\leq 1$$ and $$x^2+y^2\to0.$$

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$a=\frac{1}{xy}$

$ \ Lim_ {a \to \infty} \sin a=[1,-1]=z$, A real value.

$ \ Lim_ {x,y \to 0} \ (x^2+y^2) \sin a=0.$

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