1. Not finding help here? Sign up for a free 30min 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!

Limits of 2 var functions approaching paths of parabolas

  1. Aug 26, 2011 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/F6qLL.jpg
    In the example, it says if the approaching path of the limit is y=x^2 and x=y^2, the limits are 0. I can see the limit is 0 for approaching in the path of x=y^2 since the function becomes f(x,y) becomes f(y^2,y)=3y^5/(y^4+y^2) [itex]\rightarrow[/itex] 0 as (y^2,y) [itex]\rightarrow[/itex] 0.
    But doing the same for the path of y=x^2, I get f(x,x^2)=3x^2*x^2/(x^2+x^4)=3x^4/(x^2+x^4) which [itex]\rightarrow[/itex] 3 as (x,x^2) [itex]\rightarrow[/itex] 0. What am I doing wrong?
     
  2. jcsd
  3. Aug 26, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]\frac {3x^4}{x^2+x^4}=\frac{3x^2}{1+x^2}\rightarrow \frac 0 1[/tex]
     
  4. Aug 26, 2011 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    To display your image, use the 'Insert Image' icon in the 'Go Advanced' input box.
     
  5. Aug 27, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Notice that your example says "we begin to suspect that the limit exists and is 0." (emphasis mine)

    Frankly, I don't know what they are trying to show. The fact that the limit is 0 along some curves may suggest that the limit itself is 0 but it certainly doesn't have to be. I don't see that doing 4 paths rather than 2 tells us anything at all.

    If you convert to polar coordinates, the function becomes
    [tex]\frac{3r^3 cos^2(\theta)sin(\theta)}{r^2}= 3r cos^2(\theta)sin(\theta)[/tex]
    which goes to 0 as r goes to 0 no matter what [itex]\theta[/itex] is. That tells us that
    [tex]\lim_{(x,y)\to (0, 0} \frac{3x^2y}{x^2+ y^2}= 0[/tex]

    Because distance from the point to the origin is given by the single variable, r, as long as the limit, as r goes to 0, is independent of [itex]\theta[/tex], then that limit exists.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limits of 2 var functions approaching paths of parabolas
Loading...