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

Question

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.

Answer 1

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 $\delta =\epsilon$.