Proof of non-trivial fact to show singularity is essential In this question the asker wanted to show that the function
$$f:z\mapsto\frac{(e^{iz}-1)e^{1/z}}{\sin z}$$
Has an essential singularity at $0$. User @wisefool came up with two complex sequences $z_n=1/n$ and $w_n=i/n$ that both converge to $0$, and showed that
$$\lim_{n\to\infty}|f(z_n)|\neq \lim_{n\to\infty}|f(w_n)|$$
And used this to conclude the singularity at $0$ is essential. My question is... why? This really doesn't seem obvious to me, and he neither gave nor linked to any proof. Could someone help me construct a proof for this?

Let $f$ an analytic function with an isolated singularity at $w$
By definition of a pole of order $n \ge 1$, we have $f(z)=\frac{g(z)}{(z-w)^n}$ for $z$ in a small neighborhood of $w$ where $g$ is analytic there and $g(w) \ne 0$, This clearly implies that $|f(w)| \to \infty, z \to w$ as $1/|z-w|^n$ does so while $g(w) \ne 0$
By definition of a removable singularity, $f$ extends analytically hence continuously to $w$ so $f(z) \to f(w)$ which implies $|f(z)| \to |f(w)|, z \to w$
So in both cases above we get a unique limit, infinity for poles, and arbitrary finite for removable singularities.
The only case where two distinct limits are possible is when the singularity is essential
