TITLE: Prove complex function is not holomorphic using definition (without Cauchy-Riemann equations)
QUESTION [5 upvotes]: I'm just starting to learn complex analysis and currently reading the book "Complex analysis" by Stein and Shakarchi.
Since the fact of the function being holomorphic appears to be pretty important, I'm starting with it.
Here's the definition:
The function $f$ is holomorphic at the point $z_0 \in \Omega$ if the quotient $$\frac{f(z_0+h)-f(z_0)}{h}$$ converges to a limit when
$h \to 0$.
Then we are given two examples. First of the function $f(z)=z$ being holomorphic on any open set in $\Bbb C$, and $f'(z) = 1$. And second of the function $f(z)=\bar z$ being not holomorphic.
While I think I understand first example, I cannot see how the second will be different, how the fact of the complex conjugate changes the set up? Basically, why $f(z)=\bar z$ is not holomorphic by definition? 
I am aware of the Cauchy-Riemann equations, but I want to see the proof based on the definition (especially that the fact is claimed at this point in the book).
I tried to search for an answer but didn't find anything particularly related. I also think I might miss some very basic concept. With all that said I will really appreciate your input.

REPLY [2 votes]: Long comment about problems like that ("prove that ... isn't complex differentiable without using C-R").
Let be $c = a + bi$. It's easy to see that the condition
$$\exists \lim_{z\to c}\frac{f(z) - f(c)}{z - c}$$
is equivalent to "$f$ considered as function of two real variables is $\Bbb R$-differentiable at $(a,b)$ and the differential is $\Bbb C$-linear". And differential being $\Bbb C$-linear $\iff$ C-R equations. Then, there are two possibilities for $f$ not being $\Bbb C$-differentiable at some point $c$:

$f$ considered as function of two real variables isn't $\Bbb R$-differentiable at $(a,b)$: the $\Bbb R$-differential does not exists.
$f$ considered as function of two real variables is $\Bbb R$-differentiable at $(a,b)$ but fails to verify the C-R equations: the $\Bbb R$-differential does exists but isn't $\Bbb C$-linear.

The examples usually seen (conjugation, modulus...) are $\Bbb R$-differentiable at (almost) every point but they do not satisfy C-R.
But what are the partial derivatives? Limits in a subspace. When we check that  $\lim_{z\to c}\frac{f(z) - f(c)}{z - c}$ does not exists finding that the limits in different subspaces are different (like in the other answer), we are actually checking C-R.
Bottom line: posing the problems of this type without clarification can mislead the students.