I'm trying to prove the continuity of $\frac{x\sin(xy)}{\sqrt{x^2+y^2}}$ at $(0,0)$. Is the following working correct, and is the way of writing my proof correct?
Let $\delta = \sqrt{\epsilon}$. Then,
$\frac{x\sin(xy)}{\sqrt{x^2+y^2}}$ $\leq$ $\frac{x^2y}{\sqrt{x^2+y^2}}$ $\leq$ $\frac{(x^2+y^2)\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} $ = $(\sqrt{x^2+y^2})^2$ = $\delta^2$ < $\epsilon$
Thank you!
Let $f(x,y)=\frac{x\,\sin(xy)}{\sqrt{x^2+y^2}},\,\,(x,y)\in\mathbb{R}^2\setminus \left\{(0,0)\right\}.$
Let $\epsilon>0.$ As you already wrote,
$$|f(x,y)-0|<x^2+y^2,\,\,(x,y)\neq (0,0)\,\,\,(I).$$
So, if $\delta=\sqrt{\epsilon}$, then for all $(x,y)$ such that $||(x,y)-(0,0)||_{2}<\delta,$ i.e. $x^2+y^2<\delta^2$, according to $(I)$ you get $$|f(x,y)-0|<x^2+y^2<\delta^2=\epsilon.$$
The above proccedure shows that $$\lim_{(x,y)\to (0,0)}f(x,y)=0,$$ so the continuous extension of $f$ to $\mathbb{R}^2$ is $$g(x,y)=\begin{cases} f(x,y),\,\,(x,y)\neq (0,0)\\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(x,y)=(0,0)\end{cases}$$