Proving limit of multivariable function exists

  • Thread starter Anakin_k
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
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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
 
  • #3
12
0
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.
 
  • #4
48
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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:
  • #5
Dick
Science Advisor
Homework Helper
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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?
 
  • #6
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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. :)
 
  • #7
Dick
Science Advisor
Homework Helper
26,263
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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.
 
  • #8
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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):

[itex]\lim_{\{x,y\}\to \{1,0\}} \, \frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}[/itex]


If [itex]0<\left\|(x-1)^2+y^2\right\|<\delta[/itex],

then [itex]\left|g(x,y) - 0\right|<\epsilon[/itex]
[itex]\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|[/itex]

What would be a good choice for [itex]\delta[/itex] so that [itex]\ln[x] < \epsilon[/itex]? Plus, it seems like the way that I've worked it out so far makes the IF statement unhelpful to the process.
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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.
 

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