# How do I prove the limit $\frac{\sin(xy)}{\sqrt{x^2 + y^2}}$ as (x, y) approaches (0, 0) using $\epsilon - \delta$

I know that I can convert this limit to polar coordinates and solve the limit, but I want to see how I would do it using the $$\epsilon - \delta$$ definition of a limit.

This is my work so far:

We know that $$\left|{\frac{\sin(xy)}{\sqrt{x^2 + y^2}}} - 0\right| < \epsilon$$ and $$\left| \sqrt{x^2 + y^2} \right| < \delta$$

Then, \begin{align} \left|\sin(xy)\right| &< \epsilon \left|\sqrt{x^2 + y^2}\right| \\ \frac{\left|\sin(xy)\right|}{\epsilon} &< \left|\sqrt{x^2 + y^2}\right| \end{align}

I am stuck here, as normally I would get an expression that matches $$\delta$$, but here the signs are switched.

Let $$(x,y)\not =(0,0)$$.

$$|\sin (xy)| \le |xy|;$$ $$x^2+y^2 \ge |xy|$$;

$$0\le |\dfrac{\sin (xy)}{\sqrt{x^2+y^2}}| \le\dfrac{|xy|}{\sqrt{x^2+y^2}}\le$$

$$\dfrac{x^2+y^2}{\sqrt{x^2+y^2}}= \sqrt{x^2+y^2}.$$

Choose $$\delta =\epsilon$$.