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Math Help - epsilon delta proof for 2 variable limit

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    epsilon delta proof for 2 variable limit

    Hey I need to show that the limit(x,y)->(0,0) of xy / sqrt(x^2 + y^2) = 0. I'm not sure how to start manipulating this as I haven't gotten anything useful yet. Some help to get me going would be nice. Thanks
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    Re: epsilon delta proof for 2 variable limit

    I don't know about the epsilon delta stuff. But here is a "conventional" way to show that the limit exists, maybe you can build from this.

    If you change it to polar coordinates, then the function becomes

    f(r,\theta) = r\sin(\theta)\cos(\theta)

    The maximum possible value of sin and cos is 1, but as r --> 0, r causes the product to go to 0, so because the limit holds from all sides, it exists and equals 0.
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    Re: epsilon delta proof for 2 variable limit

    Yea, I found that while searching for a solution, but I need to prove it via the epsilon delta method.
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    Re: epsilon delta proof for 2 variable limit

    Quote Originally Posted by Shanter View Post
    Hey I need to show that the limit(x,y)->(0,0) of xy / sqrt(x^2 + y^2) = 0. I'm not sure how to start manipulating this as I haven't gotten anything useful yet. Some help to get me going would be nice. Thanks
    To prove \displaystyle \begin{align*} \lim_{(x, y) \to (a, b)} f(x, y) = L  \end{align*} you need to show \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*}. So in your case, to prove \displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)} \frac{x\,y}{\sqrt{x^2 + y^2}} = 0 \end{align*} you need to show \displaystyle \begin{align*} \sqrt{(x - 0)^2 + (y - 0)^2} < \delta \implies \left| \frac{x\,y}{\sqrt{x^2 + y^2}} - 0 \right| < \epsilon \end{align*}.

    This may prove difficult, so instead we will convert to polars. Note that \displaystyle \begin{align*} x = r\cos{\theta} , y = r\sin{\theta} \end{align*}, and \displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}. Then to prove this limit, we would have to show

    \displaystyle \begin{align*} |r| < \delta \implies \left| \frac{r^2\sin{\theta}\cos{\theta}}{r} \right| < \epsilon \end{align*}.

    Working on the second inequality we have

    \displaystyle \begin{align*} \left| \frac{r^2\sin{\theta}\cos{\theta}}{r} \right| &< \epsilon \\ \left| r\sin{\theta}\cos{\theta} \right| &< \epsilon \\ |r| |\sin{\theta}||\cos{\theta}| &< \epsilon \\ |r| &< \epsilon \textrm{ since } |\sin{\theta}| < 1 \textrm{ and } |\cos{\theta}| < 1 \end{align*}

    So if we let \displaystyle \begin{align*} \delta = \epsilon \end{align*} and reverse the process, you will have your proof.
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