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
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
$\displaystyle f(r,\theta) = r\sin(\theta)\cos(\theta)$
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.
To prove $\displaystyle \displaystyle \begin{align*} \lim_{(x, y) \to (a, b)} f(x, y) = L \end{align*}$ you need to show $\displaystyle \displaystyle \begin{align*} \sqrt{\left( x - a \right)^2 + \left( y - b \right)^2 } < \delta \implies \left| f(x, y) - L \right| < \epsilon \end{align*}$. So in your case, to prove $\displaystyle \displaystyle \begin{align*} \lim_{(x, y) \to (0, 0)} \frac{x\,y}{\sqrt{x^2 + y^2}} = 0 \end{align*}$ you need to show $\displaystyle \displaystyle \begin{align*} \sqrt{(x - 0)^2 + (y - 0)^2} < \delta \implies \left| \frac{x\,y}{\sqrt{x^2 + y^2}} - 0 \right| < \epsilon \end{align*}$.
This may prove difficult, so instead we will convert to polars. Note that $\displaystyle \displaystyle \begin{align*} x = r\cos{\theta} , y = r\sin{\theta} \end{align*}$, and $\displaystyle \displaystyle \begin{align*} x^2 + y^2 = r^2 \end{align*}$. Then to prove this limit, we would have to show
$\displaystyle \displaystyle \begin{align*} |r| < \delta \implies \left| \frac{r^2\sin{\theta}\cos{\theta}}{r} \right| < \epsilon \end{align*}$.
Working on the second inequality we have
$\displaystyle \displaystyle \begin{align*} \left| \frac{r^2\sin{\theta}\cos{\theta}}{r} \right| &< \epsilon \\ \left| r\sin{\theta}\cos{\theta} \right| &< \epsilon \\ |r| |\sin{\theta}||\cos{\theta}| &< \epsilon \\ |r| &< \epsilon \textrm{ since } |\sin{\theta}| < 1 \textrm{ and } |\cos{\theta}| < 1 \end{align*}$
So if we let $\displaystyle \displaystyle \begin{align*} \delta = \epsilon \end{align*}$ and reverse the process, you will have your proof.