# Proving limit of multivariable function exists

## Homework Statement

Evaluate or show that the limit DNE.

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

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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

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

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

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

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Dick
Homework Helper
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.

But then why does WolframAlpha say otherwise?
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

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

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

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. :)

Dick
Homework Helper
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. :)

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

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.

HallsofIvy