exp(1/z) has an essential singularity at 0

Example

Let $f(z) = e^{1/z}$ for $z \in \mathbb{C} \setminus \{0\}$, where $e^w$ is the complex exponential. The function $f$ is holomorphic on $\mathbb{C} \setminus \{0\}$, so $0$ is an isolated singularity of $f$. Its Laurent series at $0$ is $$e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!} z^{-n} = 1 + \frac{1}{z} + \frac{1}{2! \, z^2} + \cdots \qquad (z \neq 0).$$ It has infinitely many nonzero terms with negative powers, so $0$ is an essential singularity of $f$.

Remarks

A direct check: for real $x \to 0^+$, $e^{1/x} \to \infty$, and for real $x \to 0^-$, $e^{1/x} \to 0$. So $f$ has no limit at $0$ and $|f|$ does not tend to $\infty$. In each punctured disc $0 < |z| < \delta$, $f$ takes each value $w \neq 0$ infinitely often: the solutions of $e^{1/z} = w$ are $z = 1 / (\operatorname{Log} w + 2\pi i k)$ with $k \in \mathbb{Z}$ and $\operatorname{Log} w + 2\pi i k \neq 0$, and these points tend to $0$ as $|k| \to \infty$. The value $0$ is never taken. This agrees with Picard's great theorem, which allows one exception. The residue of $f$ at $0$ is the coefficient of $z^{-1}$, which is $1$.

Explanations

L1 human

Near $0$, the number $1/z$ is huge and can point in any direction, so $e^{1/z}$ can be huge, tiny, or anything in between. That is why no single value or infinity describes $e^{1/z}$ near $0$.