# Homework Help: Multivariable limit problem

1. Sep 30, 2009

### icosane

1. The problem statement, all variables and given/known data

the limit as x,y go to 0,0 of (x^2*(sin(y))^2) / (x^2 + 2y^2)

3. The attempt at a solution

I can see that the limit goes to 0 when x=y, and when y=0 the limit goes to 1. Is this proof that the limit does not exist? I'm trying to work ahead and learn straight from the book and I'm having difficulty with these types of limits. Any hints on how to approach this type of problem? Thanks.

2. Sep 30, 2009

### HallsofIvy

What you are saying is that there are points (a, a) with a very close to 0 that give a value of the function very close to 0 and that there are points (a, 0) with a very close to 0 that give a value of the function very close to 1. If there were a limit all points close to (0,0) must give a value close to that limit. Clearly, that cannot happen here. Yes, the fact that approaching (0,0) in two different ways gives two different limits means that this limit does not exist.

If you have a problem in which the limit does exist, I recommend changing to polar coordinates. That way the "distance to (0,0)" depends only on r, not $\theta$. If the limit does exist you will get that limit as r goes to 0, irrespective of the value of $\theta$. Notice here you are saying you get one limit as r goes to 0 with $\theta= \pi/4$ and another limit with $\theta= 0$.

3. Sep 30, 2009

### Billy Bob

But note that icosane was wrong about the limit along y=0. When y=0, then the expression is 0/x^2, which is 0 for all nonzero x, and hence its limit is 0 as x approaches 0.

As Halls suggested, use polar coords to help see what is happening.

4. Sep 30, 2009

### icosane

Are polar coordinates necessary the text didn't do that for any examples. Only the squeeze theorem was used. If using polar coordinates is a more powerful way of solving this type of problem I would like to learn that, however. I've only ever been exposed to polar coordinates to change a point such as (1,-1) to polar coordinates. I've also seen a few pretty spiral type things like r = cos 4 theta. How do I go about changing a complex expression like above into polar coordinates? My only guess is to set x = r cos theta, y = r sin theta, but how do I use 2 different theta values? And how is it that r is a variable all of a sudden?

5. Oct 1, 2009

### Billy Bob

Yes, set x = r cos theta, y = r sin theta, but don't use different theta values. Now you have an expression in terms of r and theta instead of x and y. x^2+y^2=r^2 is sometimes helpful. You can use squeeze, and it helps that |sin theta|<= 1 and |cos theta|<= 1 always.

This particular problem is more confusing because there is a sin y already. In this problem, y is approaching 0, so a good "squeeze" to use first is |sin y|<= |y|, and then convert to polar coordinates.