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Prove limit is 0

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    let f(x,y)=(x4+y4)1/3
    a)find f1 for (x,y)[tex]\neq[/tex](0,0)
    b)prove that lim(x,y)[tex]\rightarrow[/tex](0,0)f1(x,y)=0
    c) is f1 contintoius at (0,0)?

    3. The attempt at a solution
    f1=4x³/3(x4+y4)2/3
    along any line y=mx, it is 0, along y=x² I run into a problem, where i get 4x1/3/3(1+x4)2/3, don't know what to do.

    Once i pass this, I have a hunch i need to use squeeze theorm, yet I don't know how.

    EDIT: realized that if I sub in 0 for x in the parabola limit, it goes to 0. I just need help organizing a squeeze theorem equivilent im thinking [tex]\frac{4x^{3}}{3}[/tex]?
     
    Last edited: Dec 6, 2009
  2. jcsd
  3. Dec 6, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Unfortunately, knowing that the limit is 0 for all linear paths and even all parabolic paths is not enough! In order to be certain that the limit exists, and is 0, you would have to show that you get 0 as limit along all paths. And, of course, you can't do that by looking at different kinds of paths. There are simply too many.

    I recommend changing to polar coordinates. That way, the distance to the origin depends on the single variable, r. If the limit, as r goes to 0, does not depend on [itex]\theta[/itex] that is the limit of the function as (x,y) goes to (0,0).
     
  4. Dec 6, 2009 #3
    so you're saying to make r=x²+y²? with x=rcos[tex]\vartheta[/tex] and y=rsin[tex]\vartheta[/tex]? how would I sub that in? or even isolate for x³?
     
  5. Dec 7, 2009 #4
    You just do the sub
    [tex]
    x=r\cos\theta,\ y=r\sin\theta:
    [/tex]

    [tex]
    \frac{4x^3}{3(x^4+y^4)^{2/3}}\\
    =\frac{4r^3\cos^3\theta}{3(r^4(\cos^4\theta+\sin^4\theta))^{2/3}}\\
    =\frac{r^3}{r^{8/3}}\frac{4\cos^3\theta}{3(\cos^4\theta+\sin^4\theta)^{2/3}}\\
    =r^{1/3}\frac{4\cos^3\theta}{3(\cos^4\theta+\sin^4\theta)^{2/3}}\\
    [/tex]

    It almost depends on the angle your approaching in, but yet it doesn't.
     
  6. Dec 7, 2009 #5
    [tex]\int\int_{-\infty}^{\infty} dx,dy=\infty\: dy +C[/tex] is the same as

    [tex]\frac{\infty}{\infty}=\infty.[/tex] which is trivially true.

    [tex]\int\int (x^4+y^4)^\frac{1}{3}=0 \lim\rightarrow{\infty}[/tex]

    Thus:

    [tex] \int\frac{4x^3}{3(x^4+y^4)^{2/3}}\ =0+C \lim\rightarrow\infty [/tex]

    [tex]\int\int f(x,y)={(x4+y4)}^{1/3} dx,dy=0 \lim(x,y)\;\rightarrow{\infty}[/tex]

    [tex](x,y)\neq 0;f1=1[/tex]

    (0,0) is undefined.

    [tex]\lim(x,y)\rightarrow (0,0)f1(x,y)=0[/tex]

    Is an assymptote

    [tex]\frac{d}{dx} f(x,y)={(x4+y4)}^{1/3=[/tex][tex]\rightarrow[/tex]

    [tex]\frac{dx}{dy}\;\rightarrow[/tex]

    [tex]=0 \lim\rightarrow\infty[/tex]

    41e91672f57602d8b653dd5638351dfc.png

    a=0

    u=0

    [tex]x,y\neq{0}[/tex]

    Where the solution is an arbitrary constant or C or c=a.

    see

    http://en.wikipedia.org/wiki/Partial_differential_equation

    becb2a10a4b16cbac0e667d492af8ec9.png

    And

    http://eqworld.ipmnet.ru/en/solutions/npde/npde1401.pdf

    and

    c65dd028c1c9fb80a8288ca893e949da.png
     
    Last edited: Dec 7, 2009
  7. Dec 7, 2009 #6
    Thank you!!!!!
     
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