# Homework Help: Proving limit of multivariable function exists

1. Oct 8, 2012

### Anakin_k

1. The problem statement, all variables and given/known data
Evaluate or show that the limit DNE.

Limit as (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3.

3. The attempt at a solution
I tried approaching from multiple paths, and it seems that the limit is equal to 0. I used the delta-epsilon method to prove the limit but I've been stuck so far.

Any ideas on how to begin?

Thank you.

Last edited: Oct 9, 2012
2. Oct 9, 2012

### leej72

Approaching from the limit x = y^3, we get

lim = y^3/ 2(y^3)
(x,y) -> (0,0)
= 1/2

Since the value of the limit from the path x = y^3 is not equal to 0 (the value that you have been getting), then the limit does not exist

3. Oct 9, 2012

### leej72

When I said approaching from the limit x = y^3, I meant from the path.

4. Oct 9, 2012

### Anakin_k

Hi,

If we are approaching from the path x = y^3, then

lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3

= lim (x,y) -> (0,0) of (y)*y^2 / y^3+y^3 =

= lim(x,y) -> (0,0) of y^3 / 2y^3. = 1/2

That seems to make sense.

However, if we use the path y=x:
lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3
= lim (x,y) -> (0,0) of x^(7/3) / (x+x^3)
= 0

And then WolframAlpha says 0 too. Why is that?
http://www.wolframalpha.com/input/?i=limit+%28x%2Cy%29+-%3E+%280%2C0%29+of+%28x^1%2F3%29*y^2+%2F+%28x%2By^3%29

Last edited: Oct 9, 2012
5. Oct 9, 2012

### Dick

It's not always clear why WA reaches the conclusions it does. What do you conclude about WA?

6. Oct 9, 2012

### Anakin_k

I updated my post above, to show that I did try some other paths too.

So if both limits are different, then it would mean that the limit does not exist.

I guess that WA isn't perfect after all. :)

7. Oct 9, 2012

### Dick

I agree. Not the first mistake I've seen it make either.

8. Oct 9, 2012

### Anakin_k

Well thanks for clarifying that up you guys.

I have one more similar question with which I need some help. I've gotten a bit further with this one but not enough:

I concluded that the limit equals 0 by trying a few paths (hopefully this time it was right):

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

If $0<\left\|(x-1)^2+y^2\right\|<\delta$,

then $\left|g(x,y) - 0\right|<\epsilon$
$\left|g(x,y)\right| = \left|\frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}\right|= \left|\frac{(x-1)^2}{(x-1)^2+y^2}\right|\left|\ln[x]\right| \leq \left|\frac{(x-1)^2}{(x-1)^2}\right| \left|\ln[x]\right| = \left| \ln[x] \right|$

What would be a good choice for $\delta$ so that $\ln[x] < \epsilon$? Plus, it seems like the way that I've worked it out so far makes the IF statement unhelpful to the process.

9. Oct 9, 2012

### HallsofIvy

You might conclude it but just trying "a few" paths does not prove you will always get 0 along any path. However, in this case the numerator depends only on x so it is easy to see that the fraction approaches 0 as x goes to 1, no matter what y is.