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Limit of f(x,y)

  1. Oct 3, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Let f(x,y) be defined :

    f(x,y) = 0 for all (x,y) unless x4 < y < x2
    f(x,y) = 1 for all (x,y) where x4 < y < x2

    Show that f(x,y) → 0 as (x,y) → 0 on any straight line through (0,0). Determine if lim f(x,y) exists as (x,y) → (0,0).

    2. Relevant equations

    Polar co - ordinates maybe?

    3. The attempt at a solution

    Pretty confused with this one actually. Not sure where to start.

    I want to show f(x,y) → 0 as (x,y) → 0 on any straight line through the origin. So would I pick lets say y=x.

    Then f(x,x) = I'm not sure, having trouble with the inequalities.
     
  2. jcsd
  3. Oct 3, 2012 #2

    jbunniii

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    Is the following true or false? [itex]x^4 < x < x^2[/itex]
     
  4. Oct 3, 2012 #3

    LCKurtz

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    No, you don't want polar coordinates on this. In the xy plane draw your two curves ##y=x^4## and ##y=x^2##. Note in your picture which areas have f(x,y) = 0 and which have f(x,y)=1. Then draw lines ##y=mx## for various ##m##. You should be able to see graphically why f(x,y) → 0 along those lines as (x,y)→ (0,0). Then can you find a path where it doesn't go to zero?
     
  5. Oct 3, 2012 #4

    Zondrina

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    This is false.

    Also @ LC, I used wolfram to plot the curves for me http://www.wolframalpha.com/input/?i=plot+y%3Dx^4+plot+y%3Dx^2

    I understand what you mean geometrically, I can't find any line y=mx where the graph does not go through the origin.

    Its the analytical portion I'm confused about.
     
  6. Oct 3, 2012 #5

    LCKurtz

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    Of course the lines y=mx go through the origin. It is the value of f(x,y) on the line and near (0,0) that you are interested in. Did you observe where in the plane f = 1 and f = 0 on your picture like I suggested?
     
  7. Oct 3, 2012 #6

    SammyS

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    Where does the line, y = mx, intersect the parabola, y = x2 ?
     
  8. Oct 3, 2012 #7

    Zondrina

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    Okay i think i got it.

    So any line y=mx we draw will not be fully contained in the region x^4 < y < x^2.

    So lim f(x,mx) as (x,y) -> (0,0) = 0 for all (x,y).

    Now considering the graph of y=x^3, the path of x^3 is contained within the region x^4 < y < x^2.

    So lim f(x,x^3) as (x,y) -> (0,0) = 1 for all (x,y).

    Since the two limits are not the same the limit doesn't exist?
     
  9. Oct 3, 2012 #8

    LCKurtz

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    That's the idea. You have a bit of work to do to actually prove it. Do you understand why Sammy asked where y=mx intersects y = x2?
     
  10. Oct 3, 2012 #9

    Zondrina

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    They only intersect at the origin (depending on m). Not sure why that's relevant though.
     
  11. Oct 3, 2012 #10

    SammyS

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    Check your algebra. (The only case of one solution is for m=0.)

    [itex]\displaystyle \text{If }x^2=mx\,,\text{ then }\ \ x^2-mx=0 \quad \Rightarrow\quad(x-m)x=0\ .
    [/itex]

    What are the two solutions to that?
     
  12. Oct 3, 2012 #11

    LCKurtz

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    Think about y = mx for m really small and positive, so the line is very very close to horizontal. That line is going to get in between the two curves and f(x,y) will equal 1 in there and be close to the origin. How do you know it doesn't stay between them as x approaches 0? And that gives a problem with the limit being 0 along all straight lines through the origin. That is the crux of this problem.
     
    Last edited: Oct 3, 2012
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