Find limit $\lim_{z\to 0}\frac{\bar z}{z}$ 
Evaluate $\lim_{z\to 0}\frac{\bar z}{z}$ where $z=x+iy$

My try:
$$\lim_{z\to 0}\frac{\bar z}{z}=\lim_{z\to 0}\frac{x-iy}{x+iy}$$
$$=\lim_{z\to 0}\frac{(x-iy)(x-iy)}{(x+iy)(x-iy)}$$
$$=\lim_{z\to 0}\frac{x^2-y^2-2ixy}{x^2+y^2}$$
$$=\lim_{z\to 0}\frac{x^2-y^2}{x^2+y^2}-\frac{2ixy}{x^2+y^2}$$
I can't solve it from here. My teacher says that the limit doesn't exist but I don't know how & why. Please help me solve this question
 A: You don't need simplify
$$\lim_{z\to 0}\frac{\bar z}{z}=\lim_{z\to 0}\frac{x-iy}{x+iy}$$
consider for $x=0$ and $y=0$. Then you have different values.
A: Hint: Try to approach the origin along the real axis, and then along the imaginary axis (i.e. take first $z = t,\; t\in \mathbb{R}$ and let $t \to 0$, and then take $z = it, \; t\in\mathbb{R}$ and let $t \to 0)$.  Do you get the same limit?
A: Yes, starting form your step, the limit
$$\lim_{z\to 0}\frac{\bar z}{z}=\lim_{(x,y)\to (0,0)}\frac{x^2-y^2}{x^2+y^2}-\frac{2ixy}{x^2+y^2}$$
doesn't exist indeed


*

*for $y=0$ $$\implies \frac{x^2-y^2}{x^2+y^2}-\frac{2ixy}{x^2+y^2}=\frac{x^2}{x^2}-\frac{0}{x^2}=1\implies \lim_{z\to 0}\frac{\bar z}{z}=1$$

*for $x=y$ $$\implies \frac{x^2-y^2}{x^2+y^2}-\frac{2ixy}{x^2+y^2}=0-\frac{2ix^2}{2x^2}=-i\implies \lim_{z\to 0}\frac{\bar z}{z}=-i$$
As noticed by Nosrati, we don't need to simplify the original expression to obtain the same result.
A: Just a correction to a comment. Limit does exists, but limit along real component is different than limit along imaginary component. Hence function is not continuous. Right? I wonder if epsilon-delta form of limit might help prove no limit exists.
A: Polar coordinates;
$z=re^{i\phi}$, $r\not =0.$
$\overline{z}=re^{-i\phi}$.
$\dfrac{\overline{z}}{z}= e^{i(-2\phi)}.$
Approaching $0$ , i.e. 
$r \rightarrow 0$, along any $\phi =$const,  
yields a different point $P(e^{i(-2\phi)})$ on the unit circle in the complex plane.
