# Limit DNE

1. Jan 29, 2008

### sinClair

1. The problem statement, all variables and given/known data
For a problem I came down to having to show that the limit of (-2xy)/sqrt(x^2+y^2) does not exist as (x,y)->(0,0)

2. Relevant equations

3. The attempt at a solution
I tried taking iterated limits and showing they are not equal but I still get 0. I also tried approaching (0,0) from various directions i.e. setting x=y and x=y^2 but that has not helped either. Can anyone help me out? Thanks.

2. Jan 30, 2008

### Gib Z

PS. When you have an expression symmetrical in x and y, that is replacing every x by y and every y by x, gives the original expression, interchanging limits, taking x then y, then y then x, will always give the same result.

3. Jan 30, 2008

### sinClair

Agh, I actually posted the wrong limit. I actually meant (-2xy^2)/sqrt(x^2+y^2).

4. Jan 30, 2008

### HallsofIvy

Staff Emeritus
Gib_Z, even the original would go to 0 on the line y= -x so that does not help.

sinClair, are you sure you have the problem right? In fact, there is no way to prove the limit does not exist for the very good reason that it does exist! Notice that the "total power" in the numerator is greater than in the denominator: in the first fraction given, the power in the numerator is 2, in the denominator 1. In the second, with y2 instead of y, the power in the numerator is 3, in the denominator 1. That is enough to guarentee that the limit does exist and is 0.

One way to handle limits in 2 variables, especially when the limit is taken at (0,0), is to change to polar coordinates. In polar coordinates the single variable r measures the distance from (0,0). The limit exists if and only if the limit as r goes to infinity is independent of $\theta$. Since $x= r cos(\theta)$ and $y= r sin(\theta)$, $\sqrt{x^2+ y^2}= r$ and your limit becomes
$$\frac{-2r^3 sin^2(\theta)cos(\theta)}{r}= -2r^2 sin^2(\theta)cos(\theta)[/itex] and that goes to 0 no matter what $\theta$ is. 5. Jan 30, 2008 ### sinClair So it turns out I was responsible for some bad arithmetic and got the limit expression wrong. I was really worried when you replied, Ivy! The actual limit is (-2xy^2)/(x^2+y^2)^(3/2) as (x,y)->(0,0). Thanks for the response Ivy. I was not aware of that method of handling limits. But do you mean that "as r goes to 0 is independent of theta"? 6. Jan 30, 2008 ### HallsofIvy Staff Emeritus I mean that the limit depends only on r, not on $\theta$! Now you have [tex]\frac{-2xy^2}{(x^2+ y^2)^{3/2}}$$

In polar coordinates that is
$$\frac{-2r^3 cos(\theta)sin^2(\theta)}{r^3}= -2 cos(\theta)sin^2(\theta)$$

That has no "r" in it. It's limit depends upon $\theta$- no matter how close to (0,0) you are, different values of $\theta$ will give different values to the function so there is no limit.

Of course, now that we finally have the correct formula, what you were doing will work:

If y= x,
$$f(x,y)= f(x,x)= \frac{-2x^3}{(2^{3/2}x^3}= -2^{-1/3}$$
while if y= -x, x= -y and
$$f(x,y)= f(-y,y)= \frac{2y^3}{(2^{3/2}y^3}= 2^{-1/3}$$

Since those are different, the limit does not exist.

7. Jan 30, 2008

Ok, got it.