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Math Help - [SOLVED] Multivariable limit (x,y)>(0,0)

  1. #1
    Junior Member Revy's Avatar
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    [SOLVED] Multivariable limit (x,y)>(0,0)

    Find limit of \lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^2+y^2}

    Does it go like this?
    \lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho \sin \theta}{\rho^{2}}=\lim_{\rho \rightarrow 0}\rho \cdot \cos^{2}\theta \sin \theta=?
    Last edited by Revy; March 2nd 2010 at 01:32 AM.
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  2. #2
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    Quote Originally Posted by Revy View Post
    Find limit of \lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^2+y^2}

    Does it go like this?
    \lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho \sin \theta}{\rho^{2}}=\lim_{\rho \rightarrow 0}\rho \cdot \cos^{2}\theta \sin \theta=?
    Looks good to me.

    So what happens as \rho \to 0?
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  3. #3
    Junior Member Revy's Avatar
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    Limit is 0
    I'm not sure after all I'm extremely bad at math
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  4. #4
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    Yes.

    Though I found a mistake in your simplification.

    It should be

    \lim_{\rho \to 0}\rho ^2\cos^2{\theta}\sin{\theta}.


    And I'd say you can't be THAT bad at maths if you are doing multivariable calculus...
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  5. #5
    Junior Member Revy's Avatar
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    you did \lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho^{2} \sin \theta}{\rho^{2}}?
    why is it so? After all x=\rho \cdot \cos\theta and y=\rho \cdot \sin\theta

    After trying to understand various problems for 10hours, almost everyone would start to understand something

    In other thread I found that \lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2) => y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.
    In what way does 2 appear?
    Last edited by Revy; March 2nd 2010 at 02:50 AM.
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  6. #6
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    Quote Originally Posted by Revy View Post
    you did \lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho^{2} \sin \theta}{\rho^{2}}?
    why is it so? After all x=\rho \cdot \cos\theta and y=\rho \cdot \sin\theta

    After trying to understand various problems for 10hours, almost everyone would start to understand something

    In other thread I found that \lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2) => y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.
    In what way does 2 appear?
    Haha you are right, for some reason I misread and thought you only need to divide by \rho, not \rho^2.

    Never mind, you got the answer right.


    And in your other case

    \lim_{(x, y) \to (0, 0)}y^2\ln{(x^2 + y^2)} = \lim_{r \to 0}[(r\sin{\theta})^2 \ln{(r^2)}]

     = \lim_{r \to 0}[r^2\sin^2{\theta} \ln{(r^2)}].


    Now remember the logarithm rule \log{(n^p)} = p\log{n}?

    That means you can take the power out as a factor...


    So it becomes

    \lim_{r \to 0}[r^2 \sin^2{\theta} \cdot 2\ln{r}]

     = \lim_{r \to 0} [2r^2 \sin^2{\theta} \ln{r}].


    Notice though that you can't solve this using direct substitution, since \log{0} is undefined. It will tend to the indeterminate 0\cdot (-\infty). This one requires a bit more thought.
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  7. #7
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    The crucial point, for both of these, is that once you have put the problem in polar coordinates, the limit, as r goes to 0, does not depend on \theta! Since r itself determines "closeness" to (0,0), that is what you must have in order that you do not get different limits approaching (0,0) along different paths- in order that the limit exist.
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