Limit of the function $\lim \limits_{(x,y)\to (0,0)} \sin (\frac{x^2}{x+y}) \ (x+y \neq 0)$. Since $\sin x$ is a continuous function at $(0,0)$ it suffices to check if the limit $\lim_{(x,y)\to (0,0)} \frac{x^2}{x+y}$ is finite. 
I seem to be missing the idea in order to show that the limit $\lim_{(x,y)\to (0,0)} \frac{x^2}{x+y}=0$. I tried converting to polar form I get $\lim_{r\to 0} \frac{r \cos^2 \theta}{\cos \theta +\sin \theta}$. It seems that this is not conclusive. Any help?

For  $k\ne 0$ and   $0\ne x\ne k,$ let $y_{(k,x)}=-x+x^2/k$. Then $y_{(k,x)}\ne 0$  and $x^2/(x+y_{(k,x)})=k.$  Obviously  $\lim_{x\to 0}\sin (\;x^2/(x+y_{(k,x)})\;)=\sin k$.
Also we have $y_{(k,x)}\to 0$ as $x\to 0.$
For example with $k=\pi$ we have $\sin (\;x^2/(x+y_{(\pi,x))}\;)\to 0$ as $x\to 0.$ With $k=\pi /2$ we have $\sin (\;x^2/(x+y_{(\pi /2,x)}\;)\to 1$ as $x\to 0.$
