# Thread: Two variable limit question, prove with Paths

1. ## Two variable limit question, prove with Paths

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, 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.

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, 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.
Is it this?

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

I'd convert it to polars.

Notice that $x = r\cos{\theta}$

$y = r\sin{\theta}$

and

$x^2 + y^2 = r^2$.

Also note that when $(x, y) \to 0$, then $r \to 0$.

So we have

$\lim_{(x, y) \to (0, 0)}\frac{x^3y^2}{x^2 + y^2} = \lim_{r \to 0}\frac{r^3\cos^3{\theta}r^2\sin^2{\theta}}{r^2}$

$= \lim_{r \to 0}r^3\cos^3{\theta}\sin^2{\theta}$

$= 0$.

3. I already know the limit does not exist, that is not the problem... its just the proving it doesn't, your logic makes sense to prove it exists, however, when starting the problem I was told it did not exist.