Find the following limit: $\lim \limits_{x,y \to 0,0} \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}$ Recently I came upon a limit which confused me. The reason is that when I try to solve the following limit using polar coordinates I get a constant which I do not know if it gives me information.
Let : $$\lim_{x,y \to 0} \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}$$
Using polar coordnates I get this: 
$$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{\sin\left(r\sin\left(\theta\right)\right)+\log\left(1+r\cos\left(\theta\right)\right)}$$
Which is equal to:
$$\lim_{r \to 0} \frac{r\cos\left(\theta\right)+r\sin\left(\theta\right)-\frac{1}{2}r^2\sin^2\left(\theta\right)}{r\sin\left(\theta\right)+r\cos\left(\theta\right)}=1$$
I already know this limit does not exist. Actually it was quite difficult to find a path for which I get a different limit...
My question is: If I use polar coordinates and the result is not something that depends on $r,\theta$ then what I get is basically useless information? (I know that if that limit goes to infinity the limit of the function does not exist)

Note that


*

*$x=0,\, y=t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}=\frac{t-\frac{1}{2}t^2}{\sin\left(t\right)}\to 1$

*$x=-t+\frac12t^2,\, y=t,\, t\to 0 \implies \frac{x+y-\frac{1}{2}y^2}{\sin\left(y\right)+\log\left(1+x\right)}
=\frac{-t+\frac12t^2+t-\frac12t^2}{\sin\left(t\right)+\log\left(1-t+\frac12t^2\right)}=0$
