1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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
  6. Oct 9, 2012 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook