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$?$
Best Answer
No, the limit doesn't exist. See what happens when $(x,y)$ is of the form $\left(-\frac1n+\frac1{n^2},\frac1n\right)$ ($n\in\mathbb N$).