limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$ Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?!
\begin{equation*}
\lim_{(x,y)\to(2,
0)}{xy-2y\over x^2+y^2-4x+4}.
\end{equation*}
 A: Set $z=x-2$ then $z\to 0$ as $x\to 2$ and
$\lim\limits_{(x,y)\to (2,0)}\frac{(x-2)y}{(x-2)^2+y^2}=\lim\limits_{(z,y)\to (0,0)}\frac{zy}{z^2+y^2}=\lim\limits_{(my,y)\to(0,0)}\frac{my^2}{(1+m^2)y^2}=\frac{m}{(1+m^2)}$
Thus there is no limit.
A: Take polar coorinates: $$x=2+r \cos{\theta}$$ $$y=r \sin{\theta}$$
$r \rightarrow 0 \Rightarrow x \rightarrow 2$
Thus $$\lim_{r \rightarrow 0}f(r, \theta)=\lim_{r \rightarrow 0} \frac{(2+r\cos{\theta}-2)(r \sin{\theta})}{(2+r\cos{\theta})^2+r^2 \sin^2{\theta}-4(2+r\cos{\theta})+4}= \lim_{r \rightarrow 0} \frac{r^2 \cos{\theta} \sin{\theta}}{4+2r\cos{\theta}+r^2\cos^2{\theta}+r^2\sin{\theta}-8-4r\cos{\theta}+4}= \lim_{r \rightarrow 0} \frac{r^2\cos{\theta}\sin{\theta}}{r^2}=\cos{\theta}\sin{\theta}$$
This limit does not exist because for different values of $\theta$ we have different results.
Take for instance $$\theta=2 \pi$$ $$\theta= \frac{\pi}{4}$$ 
A: You can use Polar coordinates .And whenever you see  squares .Polar coordinates make it easier because of the  $cos^2x+sin^2x=1$.
