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?

We can see there's a problem with your method if 
$\;\cos\theta+\sin\theta=0\iff\tan\theta=-1\iff \theta=-\frac\pi4\;$ , in the trigonometric circle.
We can try for example:
$$\begin{align*}&y=x:\implies \frac{x^2}{x+y}=\frac{x^2}{2x}=\frac x2\xrightarrow[(x,y)\to(0,0)]{}0\\{}\\
&y=x^2-x:\implies\frac{x^2}{x+y}=\frac{x^2}{x+x^2-x}=\frac {x^2}{x^2}=1\xrightarrow[(x,y)\to(0,0)]{}1\end{align*}$$
and thus the limit doesn't exist.
