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

[mrcleanhands]

Homework Statement

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

Homework Equations

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

 

[HallsofIvy]

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.

 

[mrcleanhands]

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.

 

[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.