# Limit of z = f(x,y)

1. Sep 28, 2013

### icesalmon

1. The problem statement, all variables and given/known data
use polar coordinates and L'hopital's rule to find the limit:
Lim (x,y) -> (0,0) of (x2 + y2)*ln(x2+y2)

3. The attempt at a solution
I was told in class we couldn't use l'hopital because of the multivariable thing, and I was also told the coordinate switch from rectangular to polar wasn't possible. I'm not sure if the problem is a trick question, because if I can't use L'hopital I would say the limit doesn't exist at (0,0) but how am i supposed to do this before the ideas of partial derivatives are introduced? How am I supposed to use L'Hopital's rule to help that is.
and when do I use the polar switch? before or after I take L'Hopital?
when I do switch over to polar should (x,y) become (rcos(θ),rsin(θ)) -> (0,0) or do I have to switch over from (0,0) to some other angle? am I picking an angle that simultaneously makes cos and sin zero? I don't think that one is possible, so perhaps I switch to something like (pi/2, pi) or (0,2pi) how do I choose between those? I could see sign errors arising if I make the wrong choice.

2. Sep 28, 2013

### arildno

"and I was also told the coordinate switch from rectangular to polar wasn't possible."
Who told you such a thing??
It is wrong; you can.

3. Sep 28, 2013

### icesalmon

my professor, he probably was referring to the problem at hand, and that it wouldn't help get the correct answer. Maybe he misunderstood what I was asking him. So if I do use polar coordinates
lim (rcos(t), rsin(t)) -> (0,0) [rln(r)] = 0ln(0) still undefined, differentiate wrt r and I get ln(r) + 1 this seems to work since the answer is 1. I need r = 1 so ln(1)= 0 hmm

4. Sep 28, 2013

### brmath

First, notice that when you switched to $(rcos \theta, rsin \theta)$ you put yourself back in one variable, so you could use L'Hospital's Rule.

Now your confusion about the angles. The $\theta$ is whatever is correct for the vector (x,y). For example what angle does the vector (1,1) make with the x-axis? That is your $\theta$ . The vector (2,2) makes the same angle, so what about it differs from (1,1) in polar coordinates? What is $\theta$ for the vector (2,1)?

Now your function $(x^2 + y^2)*log(x^2 + y^2)$ becomes what in polar coordinates? It is the function which is going to 0, not the vector components.

See if you can make some progress now.

5. Sep 29, 2013

### arildno

I believe you must have misunderstood what your professor was saying.

The problem is perfectly well solvable by using first coordinate transformation, and then L'Hopital's rule (for infinity/infinity).

rln(r) may be written as ln(r)/(1/r), upon which you may use L'Hopital