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Math Help - Limit problem??

  1. #1
    Junior Member
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    Question Limit problem??

    How do I evaluate this without getting zero on the denominator making it undefined??

    Question is..

    limit x2/sqrt(x2 + y2) = 0
    (x,y) => (0,0)

    x2=x squared
    sqrt= square root

    Also Im a newbie, how can I use proper maths symbols here??

    Cheers

    Nappy
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    Have you tried using polar co-ordinates?

     <br />
x= r\cos(\theta), y= r\sin(\theta)<br />
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by Nappy View Post
    How do I evaluate this without getting zero on the denominator making it undefined??

    Question is..

    limit x2/sqrt(x2 + y2) = 0
    (x,y) => (0,0)

    x2=x squared
    sqrt= square root

    Also Im a newbie, how can I use proper maths symbols here??

    Cheers

    Nappy
    Polar coordinates is good choice here.
    Or, use the 2-path rule on the two paths: y=x and y=0.
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  4. #4
    Master Of Puppets
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    Quote Originally Posted by General View Post
    Or, use the 2-path rule on the two paths: y=x and y=0.
    Not sure I have seen this method, could you explain further?
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  5. #5
    MHF Contributor chisigma's Avatar
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    In such cases the following change of variables ever works...

    x= \rho\cdot \cos \theta

    y= \rho\cdot \sin \theta (1)

    Using (1) we obtain...

    \lim_{(x,y) \rightarrow (0,0)} \frac{x^{2}}{\sqrt{x^{2}+y^{2}}}= \lim_{\rho \rightarrow 0} \frac{\rho^{2}\cdot \cos^{2} \theta}{\rho}= \lim_{\rho \rightarrow 0} \rho \cdot \cos^{2} \theta (2)

    Kind regards

    \chi \sigma
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  6. #6
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    Question

    Thanks guys,

    If I was asked to prove the above directly from the definition of a limit would that answer be sufficient???
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