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Math Help - Find limits

  1. #1
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    Thumbs up Find limits

    \lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)

    How to find the above limits?
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    \lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)

    How to find the above limits?
    using polar coordinates we have y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r. since (x,y) \to (0,0), we have r \to 0+ and thus, as a result, \ln r < 0. hence 2r^2 \ln r < 2r^2 \sin^2 \theta \ln r < 0.

    finally, since \lim_{r\to0+} 2r^2 \ln r = 0, the squeeze theorem shows that the required limit is 0.
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