# Prove the limit exists for xy/(sqrt(x^2+y^2)

## Homework Statement

Find the limit, if it exists, or show that the limit does not exist.
Stewarts Calculus 7th edition 14.2 Q13.

## The Attempt at a Solution

$f(x,y)=\frac{xy}{\sqrt{x^{2}+y^{2}}}$

The limit along any line through (0,0) is 0, as well as along other
paths through (0,0) such as $x=y^{2}$ and $y=x^{2}$

Let $\epsilon>0$ We want to find $\delta>0$ such that

if $0<\sqrt{x^{2}+y^{2}}<\delta$ then $\left|\frac{xy}{\sqrt{x^{2}+y^{2}}}-0\right|<\epsilon$

Edit: Still can't get the latex working

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HallsofIvy
Homework Helper
Change to polar coordinates. That way the distance from (0, 0) depends only on r. If the limit, as r goes to 0, is the same, for all $\theta$, then the limit exists and is that value.

So then $$f(x,y)=\frac{r\cos\theta r\sin\theta}{r^{2}}=r\cos\theta\sin\theta$$

That's pretty cool! So I can just do this for most functions right?

Unfortunately, I have to prove the above using the definition of limits (as that's what we covered) for exams.

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HallsofIvy
Well, the definition says, given $\epsilon> 0$, there exist $\delta>0$ such that if the $\sqrt{x^2+ y^2}= r< \delta$ then $|f(x, y)|< \delta$. In polar coordinates, as you show, that is $|r cos(\theta)sin(\theta)|< \epsilon$ and you know that $cos(\theta)|\le 1$ and $sin(\theta)\le 1$.