limits with double variable 
Evaluation of $$\lim_{(u,v)\rightarrow (0,0)}\frac{v^2\sin(u)}{u^2+v^2}$$

We will calculate the limit along different path.
*Along $u$ axis, put $v=0$, we get limit $=0$
*Along $v$ axis, put $u=0$, we get limit$=0$
*Along $v=mu$ lime, we get $\displaystyle \lim_{u\rightarrow 0}\frac{m^2\sin(u)}{(1+m^2)}=0$
So the limit $$\lim_{(u,v)\rightarrow (0,0)}\frac{v^2\sin(u)}{u^2+v^2}=0$$
But walframalpha shows limit does not exists.
please help me where i am wrong . Thanks
 A: The limit is $0$ indeed. If $(u,v)\ne(0,0)$, then$$0\leqslant\left|\frac{v^2\sin(u)}{u^2+v^2}\right|\leqslant|\sin(u)|\leqslant|u|$$and so, since $\lim_{(u,v)\to(0,0)}|u|=0$, it follows from the squeeze theorem that your limit is $0$ indeed.
However, your justification is wrong. It is not enough to show that the limit is $0$ if $(u,v)$ approaches $(0,0)$ along some paths to deduce that.
A: In order for a limit to exist, it has to be path independent, meaning no matter how the point (u,v) approaches (0,0), the limit exists and is always the same. You have just checked the $u$- and $v$-axis and every straight line into the origin. However, you could for example approach the origin in a spiral and might get a different result.
A: As mentioned before, the limit should be zero for every path. But you can rewrite your function as following:
$$
\frac{v^2\sin(u)}{u^2+v^2}=\frac{\sin(u)}{v^2(u^2+v^2)}=\frac{\sin(u)}{\frac{1}{v^2}(u^2+v^2)}=\frac{\sin(u)}{\frac{u^2}{v^2}+1}
$$
Since we are calculating the limit $+1$ in the denominator really doesn't bother us. But we can see, that the function $(u/v)^2$ depends only on path, no mather how close to zero $u$ and $v$ are. So we can choose path, so that expression $\frac{\sin(u)}{\frac{u^2}{v^2}+1}$ has some non-zero constant value:
$$
\frac{\sin(u)}{\frac{u^2}{v^2}+1}=A
$$
$A$ can't be arbitrary big (at least I can say,that $A<1$), but than we can have path parameterised as:
$$
v^2=\frac{u^2}{\sin(u)-A}
$$
This path would than lead to non-zero limit.
