When a limit does not exist, can its derivative be found?  
I am learning derivatives of complex numbers (functions, actually) and what a learned community member pointed to me was that there is a subtle difference between finding derivatives of real numbers.  
He said that the derivative of a complex function can be calculated iff it satisfies the Cauchy Riemann equations. Which means, limit at the point $z$ exists.  
Back to reals. Have a look at the diagram above.
I can calculate derivatives in both the cases. The 'mechanical' derivative using chain rule, quotient rule, product rule, etc. The derivative will fail if I plug in x = 13 (in both the cases) but for all other values of $x$, I can calculate the derivative (slope).
In other words, $f(x)$ is not differentiable at $x = 13$.  
The same concept applies in case of complex functions, right ?
You use Cauchy Riemann equations to see if $f(z)$ is differentiable at a given $z$. Even if it is not differentiable, I can still calculate the derivative mechanically, right ? And the derivative will fail if I plug in the given $z$ at which it was not differentiable 
Please clarify.
 A: Judging from the captions of the pictures, I think we should still talk about real derivatives for a bit.
Brief answer
Neither of functions depicted in your graphs are going to be differentiable at the discontinuities depicted. After you fill in a removable discontinuity of a function like the one on the left, it could be either differentiable or nondifferentiable at the point. Jump discontinuities of functions on the real line are always nondifferentiable, but they might have one-sided derivatives that are well-defined.
Longer anwer
First of all, remember that the derivative at a point is, intuitively, a "limit of slopes as calculated from the left and from the right." From the left you take a limit of $\frac{f(x)-f(x-h)}{h}$ over very small positive values of $h$, and on the right the same happens with $\frac{f(x+h)-f(x)}{h}$. (It can be the case that both of these can be defined, but they don't match and in that case, the derivative isn't defined at that point.) 
Notice also that it is critical for $f(x)$ to be defined to carry out these computations, and so you won't get anywhere at all without settling on a value for $f(x)$. If you insist that there's no value for $f(x)$, then the slope is formally undefined. If you are willing to fill in removable discontinuities, though, you can proceed. The derivative may or may not exist after the point is filled in (consider $f(x)=|x|$ with the $x=0$ point removed/replaced.)
That leaves the case of the jump discontinuity, which you've depicted in the right hand picture. Jump discontinuities always make one of the slope limits on the right or on the left jump to infinity. Here's what I mean. Suppose $f(x)$ is anywhere exept filling in the lower circle in your right hand picture. Then as you shrink $h$ in $\frac{f(x+h)-f(x)}{h}$, the associated picture is that of a line which always lies on $(x,f(x))$ and $(x+h,f(x+h)$, which lies on the branch on the right. You see as $h$ shrinks, $x+h$ approaches $x$ from the right. Since $f(x)$ is not on that lower empty circle, this line tips ever more steeply as $h$ shrinks. Thus its slope goes to either $+\infty$ or $-\infty$, and the slope there is undefined.
If $f(x)$ happened to land on that empty lower right circle, then you are guaranteed it wouldn't land on the upper left circle, so you would then deduce that the slope estimate from the left would go off to infinity, and the derivative at the point would again not exist.
A: Let's try with a simple example:
$$
f(x)=\begin{cases}
x+1 & \text{if $x>0$}\\
0 & \text{if $x=0$}\\
x-1 & \text{if $x<0$}
\end{cases}
$$
If you compute the derivative “mechanically”, you get
$$
f'(x)=\begin{cases}
1 & \text{if $x>0$}\\
1 & \text{if $x<0$}
\end{cases}
$$
and you might be tempted to say that $f$ is also differentiable at $0$ with derivative $1$, but this would be wrong.
A necessary condition for a function to be differentiable at a point $a$ is that it's continuous at $a$, but the function $f$ is not continuous at $0$.
The proof is simple: if $f$ is differentiable at $a$, then
$$
\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=l
$$
exists (finite). Therefore you can compute
\begin{align}
\lim_{x\to a}f(x)
&=\lim_{x\to a}\left(f(a)+\frac{f(x)-f(a)}{x-a}(x-a)\right)\\
&=f(a)+\left(\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\right)
\left(\lim_{x\to a}(x-a)\right)\\
&=f(a)+l\cdot0=f(a)
\end{align}
because of well known theorems on limits.
Notice that this works in precisely the same way for complex functions; in both cases the function needs to be defined in some neighborhood of $a$ (for instance an open interval containing $a$ in the real case, an open disk containing $a$ in the complex case).
The Cauchy-Riemann equations are a useful tool, but still the definition for the derivative of a complex function $f$ can be given in the same way as in the real case: the complex function $f$, defined in an open neighborhood of $a\in\mathbb{C}$, is differentiable at $a$ if (and only if)
$$
\lim_{z\to a}\frac{f(z)-f(a)}{z-a}
$$
exists (finite).
