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*}
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.
You can use Polar coordinates .And whenever you see  squares .Polar coordinates make it easier because of the  $cos^2x+sin^2x=1$.