1. ## Multivariable Limit/Path Question

I don't know how to type with the math code, so this may look weird but bear with me please.

$\displaystyle \lim_{(x, y) \to (0, 0)}\frac{x^3y^2}{x^2 + y^2}$

I have already figured that the limit does not exist, because it comes out to 0/0, however, I have to prove it through paths. I have one path where I changed the y to another x, and got the answer of 0, however, i have been going through multiple paths and can not get anything but 0 to compare it with. Any help is appreciated.

This problem has continued to evade me.

2. Originally Posted by CalcGeek31
I don't know how to type with the math code, so this may look weird but bear with me please.

x^3(y^2)
lim _________
(x,y) -> (0,0) x^2 + y^2

I have already figured that the limit does not exist, because it comes out to 0/0,

== Uuh?? How in the world did you figure that out, anyway?

Because |(x^3)(y^2)/[x^2+y^2]| <= xy^2 --> 0 when (x,y) --> (0,0)...

Tonio

however, I have to prove it through paths. I have one path where I changed the y to another x, and got the answer of 0, however, i have been going through multiple paths and can not get anything but 0 to compare it with. Any help is appreciated.

This problem has continued to evade me.
...

3. Is the ... as in you don't understand what I am asking or?

4. Take another look ... is there because you can't post an empty message...

5. Oh, ok, i mean, that still isn't what Im looking for though...

6. Originally Posted by CalcGeek31
I don't know how to type with the math code, so this may look weird but bear with me please.

$\displaystyle \lim_{(x, y) \to (0, 0)}\frac{x^3y^2}{x^2 + y^2}$

I have already figured that the limit does not exist, because it comes out to 0/0,
Tonio's point, I believe, was that this assumption is wrong and your whole question is wrong!
No, "the limit does not exist, because it comes out to 0/0" is not true. $\displaystyle \frac{x+y}{x+y}$, as (x, y) goes to (0, 0) gives "0/0" but the limit is 1. The fact that "plugging the x, y values in" gives 0/0 only means you have to use some other method to decide.

Surely you saw plenty of examples of limits in one-variable Calculus where you got "0/0" but the limit existed? For example- any derivative!

however, I have to prove it through paths.
Yes, because your method doesn't work!

Unfortunately, paths doesn't help prove a limit does exist.

I have one path where I changed the y to another x, and got the answer of 0, however, i have been going through multiple paths and can not get anything but 0 to compare it with. Any help is appreciated.

This problem has continued to evade me.
You can't prove the limit does not exist, by any method, because the limit does exist.

Unfortunately, you cannot prove that by looking at paths because no matter how many paths you choose some other path might give a different limit. (There are examples where all straight lines through the origin give the same limit but parabolas give a different limit.)

Do this: change to polar coordinates. Let $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$.

That way, r alone measures distance from (0, 0). A point will be "close" to (0, 0) if r is small no matter what [tex]\theta[/math-] is. Take the limit as r goes to 0. If that is independent of $\displaystyle \theta$, then the limit exists and is that number.