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Thread: 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.

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


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

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

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

    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; Oct 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 $\displaystyle |\ln{x}|<\epsilon$, so:

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

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

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

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

    - 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 $\displaystyle \delta$ 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 $\displaystyle \left|\ln[x]\right|$ is unwise since that means the $\displaystyle 0 < \sqrt{(x-1)^2+y^2}<\delta$ becomes unused in the proof.

    Plus, the $\displaystyle \sqrt{(x-1)^2+y^2}$ is actually $\displaystyle \frac{1}{\sqrt{(x-1)^2+y^2}}$ in the proof. Taking the reciprocal of the $\displaystyle \sqrt{(x-1)^2+y^2}$ would only give a $\displaystyle \geq \frac{1}{\delta}$ and not a $\displaystyle \leq \frac{1}{\delta}$ part. And we don't want to bound from below with $\displaystyle \epsilon$, 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 $\displaystyle \epsilon>0$, set $\displaystyle \delta=\min(1-e^{-\epsilon},e^{\epsilon}-1)$.

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

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

    I think that's a sound argument.

    The $\displaystyle y^2$ 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; Oct 11th 2012 at 02:41 AM.
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