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Math Help - Proving the limit of a multivariable function

  1. #1
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    Proving the limit of a multivariable function

    Hi there! I'm newly getting introduced to multivariable functions and their limits. I had an assignment of 6 or 7 of them, and I got the majority except for the last. I've tried a few paths of approach and have concluded that the limit is indeed equal to 0. I just cannot prove it.

    \lim_{\{x,y\}\to \{1,0\}} \, \frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}


    If 0<\left\|(x-1,y)\right\| = \sqrt{(x-1)^2+y^2}<\delta, FIXED AFTER EDIT.

    then \left|g(x,y) - 0\right|<\epsilon
    \left|g(x,y)\right| = \left|\frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}\right|= \newline \left|\frac{(x-1)^2}{(x-1)^2+y^2}\right|\left|\ln[x]\right| \leq \left|\frac{(x-1)^2}{(x-1)^2}\right| \left|\ln[x]\right| = \left| \ln[x] \right|

    What would be a good choice for \delta so that \ln[x] < \epsilon?

    But now that I look at it, it seems like I've done the process a tedious way as I'm not employing the IF statement (simply divided it out).

    Any advice? Thanks in advance.
    Last edited by mrmackey; October 10th 2012 at 09:35 AM.
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  2. #2
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    Re: Proving the limit of a multivariable function

    You want |\ln{x}|<\epsilon, so:

    -\epsilon < \ln{x}<\epsilon

    e^{-\epsilon}<x<e^{\epsilon}

    e^{-\epsilon}-1<x-1<e^{\epsilon}-1.

    So choose \delta=\min(1-e^{-\epsilon},e^{\epsilon}-1).

    - Hollywood
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  3. #3
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    Re: Proving the limit of a multivariable function

    Hi Hollywood,

    Thanks for replying, By the way, I fixed the IF statement in my OP because of a small mistake.

    Secondly, I'm not sure how this choice of \delta helps. In fact, I'm not sure if any choice of it will help. As I mentioned, I think my process to end up at just \left|\ln[x]\right| is unwise since that means the 0 < \sqrt{(x-1)^2+y^2}<\delta becomes unused in the proof.

    Plus, the \sqrt{(x-1)^2+y^2} is actually \frac{1}{\sqrt{(x-1)^2+y^2}} in the proof. Taking the reciprocal of the \sqrt{(x-1)^2+y^2} would only give a \geq \frac{1}{\delta} and not a \leq \frac{1}{\delta} part. And we don't want to bound from below with \epsilon, we want to do so from above.

    Maybe I should restart the process all together? Any advice on how to approach this?
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  4. #4
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    Re: Proving the limit of a multivariable function

    To show that the limit is zero:

    Given \epsilon>0, set \delta=\min(1-e^{-\epsilon},e^{\epsilon}-1).

    Then whenever \sqrt{(x-1)^2+y^2}<\delta, we have |x-1|<\delta, so
    e^{-\epsilon}-1<x-1<e^{\epsilon}-1
    e^{-\epsilon}<x<e^{\epsilon}
    -\epsilon < \ln{x}<\epsilon
    |\ln{x}|<\epsilon

    And as a result,
    \left|g(x,y) - 0\right|=
    \left|g(x,y)\right| = \left|\frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}\right|=
    \left|\frac{(x-1)^2}{(x-1)^2+y^2}\right|\left|\ln[x]\right| \leq
    \left|\frac{(x-1)^2}{(x-1)^2}\right| \left|\ln[x]\right| =
    \left| \ln[x] \right| < \epsilon.

    I think that's a sound argument.

    The y^2 in the denominator always pushes the function closer to zero, so that's why it's not a significant part of the proof.

    - Hollywood
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  5. #5
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    Re: Proving the limit of a multivariable function

    That works perfectly Hollywood! Thanks a bunch for taking the time to help me and for explaining the process (so that I learn for similar problems in the future).

    If someone can mark this as solved, it would be great. Again, thank you!
    Last edited by mrmackey; October 11th 2012 at 02:41 AM.
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