# Partial derivatives

1. Apr 21, 2005

### sjsustudent2004

i am trying to solve the following problem:

find the limit of (xy)/((x^2)+(y^2))^(1/2)

as (x,y) approaches (0,0).

i know it's kind of hard to read, but that is xy divided by root(x-squared + y-squared).

the area where i am having a problem is in my arithmatic. how do i multiply the numerator by root(x-squared + y-squared). i know that it should be able to multiply out, and then cancel out the denominator...and the limit should be 0. i just don't know how to show it.

thanks for any help.

2. Apr 21, 2005

### whozum

$$\frac{xy}{\sqrt{x^2+y^2}}$$

Separate the numerator and the denominator into a product of two functions. The limit of the product is the product of the limits.

3. Apr 21, 2005

want the limit of this as x and y approach zero
$$\frac{xy}{\sqrt{x^2+y^2}}$$
perhaps u could try using polar coordinates where
$$x = r \cos(\theta)$$
$$y = r \sin(\theta)$$
$$r^2 = x^2+y^2$$

$$\frac{r^2 \cos(\theta) \sin(\theta)}{r}$$
$$=r \cos(\theta)\sin(\theta)$$
as r - > 0
and your limit is zero (i hope)

4. Apr 21, 2005

### sjsustudent2004

i don't believe we're allowed to take the limit of r. but thanks for trying though =)

(Sorry, I clicked on "edit" when I meant to "quote"!)

Last edited by a moderator: Apr 22, 2005
5. Apr 22, 2005

### HallsofIvy

Staff Emeritus
Why wouldn't you be "allowed to take the limit of r"? Since r measures the distance from (0,0) converting to polar coordinates changes a problem with two variable (x and y) going to 0 to a problem with only one (r) going to 0. As long as the result does NOT depend on θ, it is the limit. If the result does depend on &theta, then the limit does not exist.

6. Apr 22, 2005

i learnt something myself from this thread, i wasnt sure what happened when the limit depended on theta but now i do. thanks for the extra info hallsofivy

7. Apr 22, 2005

### snoble

Halls I think you might be falling for a classic fallacy. Consider the function $$r/\theta$$ where $$0 < \theta \le 2\pi$$. So $$\theta$$ never equals 0 and for each theta the limit is 0 as r approaches 0. But what is the actual definition of limit. For each epsilon there is a delta such that if (x,y) is with delta of (0,0) then f(x,y)<epsilon. So in this case we are saying that there is a delta>0 such that if r<delta then f(x,y)<1. But no matter how small delta is there is always a theta that is smaller than delta/2 so $$\delta/2\theta > 1$$. So there is no limit.

This is a very cooked example and I seem to recall there are ones that appear innocuous. Moral of the story is that if the limit depends on theta then the limit does not exist but the limit may still not exist even if each directional cross section converges.

Vlad's proof is basically right. Just complete it with given epsilon>0 take delta=epsilon. So if r<delta then $$|r \cdot cos(\theta)\cdot sin(\theta)| \le r<\delta$$

Last edited: Apr 22, 2005
8. Apr 23, 2005

### Galileo

Use the squeeze theorem (i.e. find an upper bound)

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq \frac{|xy|}{\sqrt{x^2}}=\frac{|xy|}{|x|}$$

This is usually the standard method of attack. First try a few paths. If they all give the same limit, try to see if the limit exists by finding an upper bound.