Why $\overline{z}$ is not differentiable? 
*

*By definition the derivative of $\overline{z}$ is:


\begin{align}
\lim_{\Delta h\to 0}\frac{\overline{z+\Delta h}-\overline{z}}{\Delta h}&=\lim_{\Delta h\to 0}\frac{\overline{z}+\overline{\Delta h}-\overline{z}}{\Delta h}
\\&=\lim_{\Delta h\to 0}\frac{\overline{\Delta h}}{\Delta h}
\\&=\lim_{\Delta h\to 0}\frac{{\Delta x-i\Delta y}}{\Delta x+i\Delta y}
\\&=\lim_{\Delta h\to 0}\frac{r(\cos\alpha-i\sin\alpha)}{r(\cos\alpha+i\sin\alpha)}
\\&=\lim_{\Delta h\to 0}\frac{\cos\alpha-i\sin\alpha}{\cos\alpha+i\sin\alpha}
\end{align}
How should it be continue with turning over to $\frac{e^{-i\alpha}}{e^{i\alpha}}$?


*Is there an intuitive explanation why the conjugate function is not differentiable? As it is just "reflection" on $x$- axis 

 A: Intuitively, a function $f$ of one variable is differentiable at a point $p$ if for some tiny region $U$ around $p$, applying $f$ looks like multiplying by some number / scalar (in other words, scaling and rotating), relative to $p$ and $f(p)$. No matter how you scale and rotate, you can never flip. Thus complex conjugation doesn't look like scalar multiplication locally, and therefore is not differentiable.
A: This will not expand on your limit proof in your question. But there is this nice theorem in complex analysis, called the identity theorem:

Theorem. Given two complex differentiable functions $f$ and $g$ and a set $D$ with some limit point (e.g. an open set, a line, ...). If $f(z)=g(z)$ for all $z\in D$ then we already have $f=g$.

Your function $f(z)=z$ and its conjugate $\bar f$ agree on the real line, i.e. $\Bbb R\subset\Bbb C$. The real line is a set with limit point, so when both functions are complex differentiable, then $z=\bar z$ for all $z\in\Bbb C$. Obviously wrong. But because we know that $z$ is complex differentiable, $\bar z$ can not.
This works for all non-constant functions $f$ for which the restriction to $\Bbb R $ is real, i.e. $f|_\Bbb R:\Bbb R\to\Bbb R$. For example, this shows that $\overline\sin (z)$ is not complex differentiable.
A: You can see that by approaching to the origin first on the real axis and second on the complex axis, that means: Consider for the first limit $\Delta x=0$ and for the second limit consider $\Delta y=0$. You get two different limits, and therefore the limit does not exist. Hence $f(z)=\overline{z}$ is not differentiable.
A: Conjugation is real-differentiable, in the sense that all partial derivatives of it exist everywhere. However, this does not mean it's complex differentiable. A real differentiable function $x+yi\mapsto u(x,y)+iv(x,y)$ is complex differentiable provided that
$$\begin{align}\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}\\
\frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x}\end{align}$$
These are the Cauhcy-Riemann eComplex conjugation does not satisfy this, since $u(x,y)=x, v(x,y)=-y$. 
Intuitively, $z\mapsto \overline z$ is not differentiable since its directional derivatives are different depending on their direction, as opposed to a function like $z^2$. 
