Examples of complex-variable functions that fail to have a limit at some point My notes from class have the example $\frac{\overline{z}}{z}$ as z tends to zero.  That the limit does not exist is shown by exhibiting that along the $x$-axis the limit is $1$ and along the $y$-axis it is $-1$.  Also our homework set for the section has only two limit problems.  Can someone please post some examples?  I'm looking for more examples to get comfortable with this idea but when I search online I keep running into this same example.  
Also, is there a way to tell ahead of time that a complex function will not have a limit?  
 A: A couple of worked examples:
$1.$ The function $f: \mathbb{C} \rightarrow \mathbb{C}$, $f(z) = \overline{z}$, is not $\mathbb{C}$-differentiable.
Let us consider the limit
$$\lim_{h \rightarrow 0}\frac{f(z + h) - f(z)}{h} = \lim_{h \rightarrow 0}\frac{(\overline{z} + \overline{h}) + \overline{z}}{h} = \lim_{h \rightarrow 0}\frac{\overline{h}}{h}.$$
But $h = |h|e^{i \theta}$, where $0 \leq \theta \leq 2\pi$ (by using the fact that $z = re^{i\theta}$, where $r = |z|$). So we have that
$$\frac{\overline{h}}{h} = \frac{|h|e^{-i \theta}}{|h|e^{i \theta}} = e^{-2i\theta}.$$
This is bad news because then the limit $\lim_{h \rightarrow 0}\frac{\overline h}{h}$ depends on the angle $\theta$ at which $h$ approaches $0$, which means that $\lim_{h \rightarrow 0}\frac{\overline h}{h}$ does not exist (since if the limit is unique, then it cannot depend on $\theta$).
Indeed, suppose that $h = t \in \mathbb{R}$. Then
$$\lim_{h \rightarrow 0}\frac{\overline h}{h} = \lim_{t \rightarrow 0}\frac{t}{t} = 1.$$
Next, suppse that $h = it$, where $t \in \mathbb{R}$. Then
$$\lim_{h \rightarrow 0}\frac{\overline h}{h} = \lim_{t \rightarrow 0} \frac{-it}{it} = -1.$$
So the limit does not exist.
We can do even worse and instead spiral to $0$. Suppose that $h = te^{it}$, where $t \in \mathbb{R}$. Then
$$\lim_{h \rightarrow 0}\frac{\overline h}{h} = \lim_{t \rightarrow 0}\frac{te^{-\frac{i}{t}}}{te^{\frac{i}{t}}} = \lim_{t \rightarrow 0}e^{-\frac{2i}{t}}.$$
Again, the limit does not exist.
Being $\mathbb{C}$-differentiable is a lot more restrictive than being $\mathbb{R}$-differentible since we can converge to zero in a lot of ways.
It is worth noting that in this particular example, we can establish the lack of $\mathbb{C}$-differentiability of $f$ much more easily through the use of the Cauchy-Riemann equations.
$2.$ The function $f: \mathbb{C} \rightarrow \mathbb{C}$, $f = |z|^2 = z\overline{z}$, is not $\mathbb{C}$-differentiable anywhere except at $z = 0$.
Let us consder the limit
$$\lim_{h \rightarrow 0}\frac{f(z + h) - f(z)}{h} = \lim_{h \rightarrow 0}\frac{(z + h)(\overline{z} - \overline{h}) - z\overline{z}}{h} = \lim_{h \rightarrow 0}(\overline{z} + z\frac{\overline{h}}{h} + \overline{h}).$$
The parts $\overline{z}$ snd $\overline{h}$ are fine and cause no problems, but as we saw in example $1.$, the limit $\lim_{h \rightarrow 0}\frac{\overline{h}}{h}$ does not exist. So, for the whole limit $\lim_{h \rightarrow 0}(\overline{z} + z\frac{\overline{h}}{h} + \overline{h})$ to exist, we require that $z = 0$ to remove the nonexistent limit $\lim_{h \rightarrow 0}\frac{\overline{h}}{h}$.
Similarly to example $1.$, we can also use the Cauchy-Riemann equations here to establish where (if anywhere) $f$ is $\mathbb{C}$-differentiable.
