Well, I've made a double limit using the polar forms. The thing is the limit is wrong, I've made a plot, and then I saw that the limit doesn't exist, and what I wanna know is what I'm reasoning wrong, and some tips to get a deeper comprehension on this limits, and on what I am doing. For the last one I wanna know the limit value, I think it doesn't exists neither. Is it because the sine and cosine oscillates?

$\displaystyle \displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{xy}{xy+(x-y)^2}}$

$\displaystyle \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}$

$\displaystyle \displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2\cos\theta\sin\thet a}{r^2\cos\theta\sin\theta+(r\cos\theta-r\sin\theta)^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2cos\theta\sin\theta }{r^2[\cos\theta\sin\theta(cos\theta-\sin\theta)^2]}}=$

$\displaystyle =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{\ cos\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}=\displaystyle \lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{r ^2-\cos\theta\sin\theta}}=-1$

r is always positive, as we defined it.

$\displaystyle \displaystyle\lim_{(x,y) \to{(-1,3)}}{\displaystyle\frac{\sqrt[ ]{x+y-2}}{(x+1)^2+(y-3)^2}}$

$\displaystyle \begin{Bmatrix}x=-1+r\cos\theta\\y=3+r\sin\theta\end{matrix}$

$\displaystyle \displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\ lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{r}\sqrt[ ]{\cos\theta+\sin\theta}}{r^2}}=\displaystyle\lim_{ r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\ lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{\cos\theta+\sin\theta}}{r^{\frac{3}{2}}}}$

Bye there, thanks for posting.