limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$

Question

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*}

Answer 1

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.

Answer 2

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

Answer 3

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 $$\theta=2 \pi$$ $$\theta= \frac{\pi}{4}$$