Why is $f(x,y)$ continuous if $\lim_{r\to 0}\sin\left(\frac{1}{r^2}\right)$ is not defined

Question

I am currently studying with notes that our professor gave us and there we have to check if a function is continous: Now let $f(x,y) = 0$ for $(x,y) = (0,0)$, $f(x,y) = \sqrt{|xy|}\sin\left(\frac{1}{x^2+y^2}\right)$ for $(x,y) \neq(0,0)$. We learned that you could transform $x$ and $y$ into polar coordinates and if $\lim_{r\to 0} \sqrt{|r^2|}\sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin\left(\frac{1}{(r(cos(\varphi))^2+(r(\sin(\varphi))^2}\right)$ = 0 and then it differentiable in from all sides and thus continuous.

Now my professor wrote that $\lim_{r\to 0} r\sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin(\frac{1}{r^2})$ = 0 since $\lim_{r\to 0} \sqrt{|(\cos(\varphi)(\sin(\varphi)|}\sin(\frac{1}{r^2})$ is bounded but this does not make any sense in my eyes since $\sin\left(\frac{1}{r^2}\right)$ is not defined and thus can't be bounded... Where did I make my mistake?

Thank you

Answer 1

For every real number $x$, $\sin(x)\in[-1,1]$. Therefore$$(\forall r\in\mathbb{R}\setminus\{0\}):\sin\left(\frac1{r^2}\right)\in[-1,1]$$and so, yes, your professor is right.

Answer 2

Recall that $\forall \theta \in \mathbb{R}$ we have that $|\sin \theta| \le 1$ and therefore

$$0\le \left|\sqrt{|xy|} \sin\left(\frac{1}{x^2+y^2}\right)\right|\le \sqrt{|xy|}\to 0$$