# Thread: Two tricky two-variable limits

1. ## Two tricky two-variable limits

I've been working on these two limits all night and can't seem to make any reasonable progress. Any help would be appreciated. Here's the first one:

$\lim_{(x,y) \to (0,0)} \frac{xy^2}{2x-y}$

According to Wolfram Alpha, this limit does not exist. However, I can't seem to find any x,y path that gives any limit other than 0. I tried all lines (x=0, y=0, y=kx), as well as x=y^2 and y=x^2. Any ideas?

And the second one:

$\lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$

Here, I know that

$\lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^2+y^2} = \lim_{(x,y) \to (0,0)} \frac{\frac{1}{x^2+y^2}}{e^{\frac{1}{x^2+y^2}}} = \lim_{t \to \infty} \frac{t}{e^t} = 0$

I can't find a way to use this though.

Thanks!

2. $\displaystyle \lim_{(x,y) \to (0,0)} \frac{xy^2}{2x-y}$

Try approaching on y=x and y=-x.

$
x=sin \; (\frac{1}{2}\alpha)
$

$
y=sin \; \alpha
$

In the second may be polar coordinates be useful.

4. Originally Posted by dwsmith
Try approaching on y=x and y=-x.
I tried y=kx for all x (except 2 of course, as y=2x is not in the domain), it's always 0. But thanks anyway.

Originally Posted by zzzoak
$
x=sin \; (\frac{1}{2}\alpha)
$

$
y=sin \; \alpha
$
Thank you very much, that did it! (I got 4, hope that's correct.)

Originally Posted by zzzoak
In the second may be polar coordinates be useful.
I haven't done a limit in polar coordinates before, but as I understand it, this shold work:

$\displaystyle
\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} = \lim_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4(cos^4 \varphi + sin^4 \varphi)} = \lim_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^2} \cdot \frac{1}{r^2(cos^4 \varphi + sin^4 \varphi)}$

The first fraction approaches 0 (per the formula in my original post). However, it's still $0 \cdot \infty$, so that doesn't really help me. Any ideas on how to continue?

Also, could you point me to a good resource on using polar coordinates with limits? I'm not sure that what I did is really OK -- aren't there any conditions that need to be satisfied before I can say that a limit is the same in polar coordinates? Also, I assumed that I can just ignore the angle $\varphi$, as I'm approaching zero, is that correct?

Thanks again, you've been a great help!

5. You can make substitution
$
t=1/r^2
$

and get

$
\displaystyle {lim_{t \to \infty} \; \frac{t^2}{e^t}\; max( \varphi )
}
$