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Homework Help: Proving limit of multivariable function exists

  1. Oct 8, 2012 #1
    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. jcsd
  3. Oct 9, 2012 #2
    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
  4. Oct 9, 2012 #3
    When I said approaching from the limit x = y^3, I meant from the path.
  5. Oct 9, 2012 #4

    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?
    Last edited: Oct 9, 2012
  6. Oct 9, 2012 #5


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    It's not always clear why WA reaches the conclusions it does. What do you conclude about WA?
  7. Oct 9, 2012 #6
    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. :)
  8. Oct 9, 2012 #7


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    I agree. Not the first mistake I've seen it make either.
  9. Oct 9, 2012 #8
    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.
  10. Oct 9, 2012 #9


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    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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