Find $\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{x+y}$ 
Find $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{x+y}$$ exist or DNE.

$f(x,y)$ along the lines $y=mx$
\begin{align}
\lim_{(x,mx)\rightarrow (0,0)}\frac{\sin (mx^2)}{x+mx}&=\lim_{x\rightarrow 0}\left(\frac{\sin (mx^2)}{x}\frac{1}{1+m}\right)\\
&=0
\end{align}
By applying L'Hospital's Rule, we can show this limit is $0$ except when $m=-1$ . But the answer says limit DNE. Maybe I have to choose different path. Then I think whyn't choose $y=mx^n$ and again find limit $0$. I don't use polar or spherical coordinates because here is no $x^2+y^2$ terms. Eventually I can't use $\lim_{p\rightarrow a}f(g(p))=f(\lim_{p\rightarrow a}g(p))($usage of continuiuty$)$ also. Is there another approach for this problem$?$ 
 A: No, the limit doesn't exist. See what happens when $(x,y)$ is of the form $\left(-\frac1n+\frac1{n^2},\frac1n\right)$ ($n\in\mathbb N$).
A: Your argument fails because you are only exploring a prticular family of paths wich is not sufficent to show that the limit exists.
To proceed, we have that
$$\frac{\sin (xy)}{x+y}=\frac{\sin (xy)}{xy}\frac{xy}{x+y}$$
and $\frac{\sin (xy)}{xy} \to 1$ but $\frac{xy}{x+y}$ has no limit indeed

*

*for $x=0 $
$$\frac{xy}{x+y}=0$$


*for $x=t$ and $y=-t+t^2$ with $t\to 0$
$$\frac{xy}{x+y}=\frac{-t^2+t^3}{t-t+t^2}=-1+t \to -1$$

Edit
Unfortunately there are not general rules to find critical paths and we need to proceed case by case.
In that case it easy to find the path for which the limit is equal to $0$.
For the other path, a good strategy which often works, is to select at first a path such that the denominator is equal to zero that is $x=-y$ in this case but since $f(x,y)$ is not defined at those points we add to $y$ an extra smaller term that is $t^2$. In some cases the first guess doesn't work and we need to use a different extra term for $y$.
A: HINT  Your argument fails when $m=-1$.
A: Hint
For $xy \ne 0$, rewrite
$$\frac{\sin (xy)}{x+y}
=\frac{\sin (xy)}{xy}\frac{xy}{x+y}
$$
and since
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{xy}=1$$
the problem reduces to studying
$$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x+y}$$
Detailed information about this limit can be found at Does $\lim \frac{xy}{x+y}$ exist at (0,0)?
