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Math Help - Multivariable Limit Problem

  1. #1
    Junior Member
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    Multivariable Limit Problem

    I am trying to show that the following function is differentiable at (0,0):

    f (x,y) :
    ((x^4 + y^4) / (x^2 + y^2)) + 1 if (x,y) =/= (0,0)
    1 if (x, y) = (0,0)

    I have found the Linear Approximation, L at (0 0) = 1
    and that R at (0 0) = f (x, y) - L = (x^4 + y^4) / (x^2 + y^2)
    and finally, that R / ||(x,y) - (0,0)|| = ((x^4 + y^4) / (x^2 + y^2)) / sqrt(x^2 + y^2)

    Now, I need to show that the limit as (x,y) -> (0,0) of R / ||(x,y) - (0,0)|| = 0

    Which brings me to my actual problem...I cannot figure out how to go about solving this limit. If anyone can point me in the right direction I would appreciate it.
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  2. #2
    MHF Contributor

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    Try putting it in polar coordinates. x^2+ y^2= r^2, of course, and x^4+ y^4= r^4 cos^4(\theta)+ r^4 sin^4(\theta) so \frac{x^4+ y^4}{x^2+ y^2}= r^2\frac{1}{cos^4(\theta)+ sin^4(\theta)} and so your "R/||(x,y- (00)||" is \frac{r^2}{r}\frac{1}{cos^4(\theta)+ sin^4(\theta)}= r\frac{1}{cos^4(\theta)+ sin^4(\theta)}.

    The advantage of polar coordinates here is that the "closeness" to (0, 0) is measured by r alone. No matter what \theta is, that goes to 0 as r goes to 0.
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  3. #3
    Junior Member
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    I appreciate the help, but I was hoping to figure out how to solve this limit using Squeeze Theorem or Continuity rather than polar coordinates.
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