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Math Help - prove f approaching to a limit

  1. #1
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    prove f approaching to a limit

    Let f(x,y) = xy( \dfrac{x^2-y^2}{x^2+y^2}). show that |f(x,y)|\le\dfrac{1}{2}(x^2+y^2), and hence prove that f(x,y)approaches a limit as (x,y)-->(0,0)
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  2. #2
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    Notice that ~|2xy|\le x^2+y^2.
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  3. #3
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    I'm not exactly sure how we'd be able to use that, Plato. This particular question is tricky, and our textbook doesn't provide very good examples (or rather ANY examples) on how to do this.

    If anyone could provide some more details, that'd be great.
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  4. #4
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    Quote Originally Posted by Runty View Post
    I'm not exactly sure how we'd be able to use that, Plato. This particular question is tricky, and our textbook doesn't provide very good examples (or rather ANY examples) on how to do this.
    Is it clear to you that |x^2-y^2|\le |x^2+y^2|?

    Then it follows at once  \left| {xy\dfrac{{x^2  - y^2 }}<br />
{{x^2  + y^2 }}} \right| \leqslant \left| {xy} \right| \leqslant  {\dfrac{{x^2  + y^2 }}{2}}

    There nothing complicated about it.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Is it clear to you that |x^2-y^2|\le |x^2+y^2|?

    Then it follows at once  \left| {xy\dfrac{{x^2  - y^2 }}<br />
{{x^2  + y^2 }}} \right| \leqslant \left| {xy} \right| \leqslant  {\dfrac{{x^2  + y^2 }}{2}}

    There nothing complicated about it.
    Wow, I am slow. Then again, proof-work has never been my forte in mathematics (I'm more of a number cruncher).

    One thing, though: wouldn't it be ~2|xy|\le x^2+y^2 instead of ~|2xy|\le x^2+y^2?

    I say this because of the inequality ~2|ab|\le a^2+b^2.
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  6. #6
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    Quote Originally Posted by Runty View Post
    One thing, though: wouldn't it be ~2|xy|\le x^2+y^2 instead of ~|2xy|\le x^2+y^2?.
    They are exactly the same |Ax|=|A||x|.

    So |2x|=2|x|.
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  7. #7
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    I found another potential answer to the question through searching online. This answer uses polar coordinates, though I'm not sure how accurate it is.

    Let x=r\cos t and y=r\sin t.
    \left| {xy\dfrac{{x^2 - y^2 }}<br />
{{x^2 + y^2 }}} \right|=\dfrac{r^4|\sin t\cos t(\cos^2t-\sin^2t)|}{r^2}=\dfrac{r^2|\sin2t\cos2t|}{2}<\dfra  c{r^2}{2}<\epsilon if
    r<\sqrt{2\epsilon} is independent of the angle t, so the limit is 0.

    The person who posted this didn't give all the steps I could've wanted, so expect it to have some errors.
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  8. #8
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    Quote Originally Posted by Plato View Post
    They are exactly the same |Ax|=|A||x|.

    So |2x|=2|x|.
    so can we just conclude that since 1/2 (x^2+y^2) is approaching to 0 when (x,y) approaches (0,0), so f(x,y) is also approaching to a limit which is less or equal to 0?
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