Does having multiple limit values at a point imply essential discontinuity? In Complex Analysis, do "jump discontinuities" exist?
If I find that a function of $z$ approaches two different values as z is approached from two different directions, can I immediately conclude that the point is an essential singularity? Or do other possibilities exist?
 A: I'll give a proof of what I've written in the comment to the question (when I wrote that, I was assuming in my mind than the singularity was isolated one).
Suppose that we have an holomorphic function $f\colon U\setminus \{p\} \longrightarrow \mathbb{C}$, where $U\subseteq \mathbb{C}$ is an open subset and $p\in U$ a point. We want to prove that


*

*$p$ is a removable singularity if and only if $\lim_{z\to p}f(z)=c\in \mathbb{C}$

*$p$ is a pole if and only if $\lim_{z\to p}f(z)= \infty$

*$p$ is an essential singularity if and only if $\lim_{z\to p}f(z)$ does not exists
We can suppose up to translation that $p=0$.
First, if $0$ is a removable singularity, then by definition there is an holomorphic function $g\colon U \longrightarrow \mathbb{C}$ that extends $f$, so that $\lim_{z\to 0 }f(z)=g(0)\in \mathbb{C}$. Conversely, if the limit $\lim_{z\to 0}f(z)=c$ exists and belongs to $\mathbb{C}$, then we can define a function $g\colon U \longrightarrow \mathbb{C}$ by
$$ g(z)=\begin{cases} f(z) & \text{ if } z\ne 0 \\ c & \text{ if } z=0 \end{cases} $$
It is clear that this function is continuous and holomorphic on $U\setminus \{ p\}$ so that it is holomorphic on the whole of $U$, by Morera's theorem (this can be found for example in Cartan's "Elementary Theory of Analytical Functions of One or Several Complex Variables").
Now, if $f$ has a pole in $0$: this means that the Laurent series of $f(z)$ in $0$ is of the form
$$ f(z)=\sum_{n\geq N}a_n z^n $$
for a certain $N\geq 1$. This implies that the function $g(z)=z^Nf(z)$ is holomorphic, so that $\lim_{z\to 0}z^Nf(z)=a\in \mathbb{C}$, but since $N\geq 1$, this implies that $\lim_{z\to 0}f(z)=\infty$.
Instead, if $f$ has an essential singularity, then Casorati-Weierstrass' Theorem or Picard's Theorem tell us that the image through $f$ of an arbitrarily small neighborhood of $0$ is dense in $\mathbb{C}$, so that the limit $\lim_{z\to 0}f(z)$ does not exists.
Conversely, if $\lim_{z\to 0}f(z)=\infty$,then $0$ cannot be neither a removable singularity nor an essential singularity, for what we have proven before, so that it must be a pole. Now a similar reasoning proves that, if the limit does not exists, then the point must be an essential singularity,
A: Your question did not specify the domain of definition of your function, so it's hard to give a precise answer. (You have also used $z$ to mean two different things in the same sentence.) However, your function definitely does not have a limit at $z$, which does imply that $z$ is an essential singularity of your function.
(I suspect you mean "essential singularity" rather than "essential discontinuity"; the latter means, by definition, that the limit does not exist, and so your function certainly has an essential discontinuity as well.)
A: Assuming the singularity is isolated, yes. However, take $f(z)=\log z$ (principal branch) and consider what happens when $z \to -1$ along paths in the upper and lower halfplanes respectively.
