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Math Help - Multivariable Limit, Proof

  1. #1
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    Multivariable Limit, Proof

    Hi guys,

    Having a little trouble with a question for this week's homework;

    The question is,

    Apply the definition of the limit to show that

    lim (x, y) -> (0, 1) for ((x^2)y(y-1)^2)/(x^2+(y-1)^2) = 0

    Usual me would just sub in values to prove but obviously I can't, as 0/0 would occur. I've tried looking up a few different techniques to tackle it ("squeezing" method, delta epsilon) but I can't get my head around applying them to this question.

    Can anyone give me a hint?

    Also, sorry for the equation format, is there a tutorial on how to set it up all pretty for the future? I'm new

    Thanks!
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  2. #2
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    Re: Multivariable Limit, Proof

    As for the typesetting, we use LaTeX. There is a LaTeX subforum here were you can get some assistance.

    As for the question, the definition of a multivariable limit is that for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists a $\displaystyle \begin{align*} \delta > 0 \end{align*}$ such that $\displaystyle \begin{align*} \sqrt{ \left( x - a \right) ^2 + \left( y - b \right) ^2 } < \delta \implies \left| f(x,y) - L \right| < \epsilon \end{align*}$. Then $\displaystyle \begin{align*} \lim_{(x,y) \to (a,b)} f(x,y) = L \end{align*}$.

    So in this case $\displaystyle \begin{align*} f(x,y) = \frac{x^2\,y\,\left( y - 1 \right) ^2 }{x^2 + \left( y-1 \right) ^2 }, a = 0, b = 1, L = 0 \end{align*}$. Go from here.
    Thanks from topsquark
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  3. #3
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    Re: Multivariable Limit, Proof

    Thanks mate, you're a diamond! I'll write it up and post it back a wee later

    And thanks for the LaTeX hint, I'll do some research.
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  4. #4
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    Re: Multivariable Limit, Proof

    Alright I gave it a try and just couldn't come up with something I'm happy with. I did heaps of practice on other multivariable limits, but this one confuses me for some reason... I've gotten up to

    $\displaystyle \begin{align*} \sqrt{ \left( x \right) ^2 + \left( y - 1 \right) ^2 } < \delta \implies \left| \frac{x^2y(y-1)^2}{x^2+(y-1)^2} \right| < \epsilon \end{align*}$

    which then implies $\displaystyle \begin{align*} x < \delta \end{align*}$ and $\displaystyle \begin{align*} (y-1) < \delta\end{align*}$

    Now, do I sub in delta for x and (y-1)? Assuming I do, and I guess if (y-1) < delta then y < delta + 1, then I get

    $\displaystyle \begin{align*} \left| \frac{\delta^3+\delta^2}{2} \right| \end{align*}$

    What does this mean? What am I doing?! Any further advice would be appreciated

    Thank you!
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