Continuity of a function with 2 variables Im stuck in this exercise:
Study the continuity of the next function:
$$f(x,y) = \begin{cases} \frac{x\sin(x^2+y^2)}{x^2+y^2}&\text{si } (x,y)\not=(0,0)\\ 0 &\text{si }(x,y) =(0,0). \end{cases}$$
I have gone this far:
$x=r\cos\theta $
$y=r\sin\theta $
so
$$\frac{x\sin(x^2+y^2)}{x^2+y^2}=\frac{r\cos\theta·\sin(r^2\cos^2\theta+r^2\sin^2\theta)}{r^2\cos^2\theta+r^2\sin^2\theta}=\frac{r\cos\theta·\sin(r^2)}{r^2}=\frac{\cos\theta·\sin(r^2)}{r}$$
Then the limit when $r\to0$
$$\lim_{r\rightarrow0}\frac{\cos\theta·\sin(r^2)}{r}=\lim_{r\rightarrow0}\frac{\cos\theta·r^2}{r}=\lim_{r\rightarrow0}{\cos\theta·r}=0 \text{ for all }\theta \in [0,2\pi)$$
From here I don't know what I have to do in order to say if its continuous

You don't need to change to polar coordinates. Notice that at any point $(x,y)\ne (0,0)$ the function is continuous since it is a quotient of continuous functions. So, you just need to check the limit as $(x,y)$ goes to $(0,0)$. Noting that as $(x,y)$ goes to $(0,0)$ so does $x^2+y^2$, and since $\lim_{t\to 0}\sin(t)/t=1$ you may conclude that the limit of the function as $(x,y)\to (0,0)$ is equal to $0$ (note the multiplication by $x$ in the numerator). Since $f(0,0)=0$ be definition it follows that the function is continuous everywhere. 
