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!

Homework Help: Finding limits in R3

  1. Sep 21, 2010 #1
    find the limit if it exists or show that the limit does not exist for:

    limit(x,y) ---> ((x)(y^2))/((x^2)+(y^4)) as (x,y) ----> (0,0)

    I took the limit approaching the origin along the x axis and y axis.
    So, took the limit as (x,y) ---> (x,0) and came up with 0.
    Then, I took the limit as (x,y) ---> (0,y) and came up with 0 again.

    Is 0 my limit or am I missing something?
  2. jcsd
  3. Sep 21, 2010 #2
    No, you are not missing anything. Your answer is correct.
  4. Sep 21, 2010 #3
    WOOOO thanks
  5. Sep 22, 2010 #4


    User Avatar
    Science Advisor

    No, unfortunately you are missing a lot! Once you move beyond one dimension, the limit as you approach the "target" point, from any direction must be the same. Most Calculus texts have examples of that in 2 dimensions but the same is true for 3 dimensions. Just getting the same result approaching (0, 0, 0) along the coordinate axes, or even along every straight line, does NOT imply you will get the same thing along any curve.

    (Hold on, I just realized that, although you titled this "Finding limits is R3", it is really a problem in 2 dimensions, not 3. Oh, well, same thing is true.)

    If, for example, you take the limit as (x, y) goes to (0, 0) along the parabola [tex]x=y^2[/tex], the limit becomes
    [tex]\lim_{y\to 0}\frac{(y^2)(y^2)}{(y^2)^2+ y^4)}= \lim_{y\to 0}\frac{y^4}{2y^4}= \lim_{y\to 0}\frac{1}{2}= \frac{1}{2}[/tex].

    Since that is different from the limit as you approach (0, 0) along the axes (or, in fact, along any straight line), the limit does not exist!
  6. Sep 22, 2010 #5
    thanks!! I talked to my teacher about this and she agrees with you HallsOfIvy. I see it now
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook