What is $\lim\limits_{z \to 0} |z \cdot \sin(\frac{1}{z})|$ for $z \in \mathbb C$? What is $\lim\limits_{z \to 0} |z \cdot \sin(\frac{1}{z})|$ for $z \in \mathbb C^*$? I need it to determine the type of the singularity at $z = 0$. 
 A: We have $a_n=\frac{1}{n} \to 0$ as $n\to \infty$ and
$$
\lim_{n \to \infty}\Big|a_n\sin\Big(\frac{1}{a_n}\Big)\Big|=\lim_{n \to \infty}\frac{|\sin n|}{n}=0,
$$
but
$$
\lim_{n \to \infty}\Big|ia_n\sin\Big(\frac{1}{ia_n}\Big)\Big|=\lim_{n \to \infty}\frac{e^n-e^{-n}}{2n}=\infty.
$$
Therefore $\lim_{z \to 0}|z\sin(z^{-1})|$ does not exist.

Notice for every $k \in \mathbb{N}$ the limit $\lim_{z\to 0}|z\sin(z^{-1})|$ does not exist. In fact
$$
\lim_{n \to \infty}\Big|a_n^{k+1}\sin\Big(\frac{1}{a_n}\Big)\Big|=\lim_{n \to \infty}\frac{|\sin n|}{n^{k+1}}=0,
$$
but
$$
\lim_{n \to \infty}\Big|(ia_n)^{k+1}\sin\Big(\frac{1}{ia_n}\Big)\Big|=\lim_{n \to \infty}\frac{e^n-e^{-n}}{2n^{k+1}}=\infty.
$$
It follows that $z=0$ is neither a pole nor a removable singularity. Hence $z=0$ is an essential singularity.
A: It is undefined. Consider the function $g(z)=\frac{\sin{z}}{z}$. This can be made an entire function by setting $g(0)=0$. But then you are looking for $\lim_{z\to\infty} g(z)$. Any non-polynomial entire function must have an essential singularity at $\infty$, so this limit does not exist.
Alternatively:
For $x$ real, $\lim_{x\to\infty} g(x)$ is clearly zero, since $|\sin x|\leq 1$.  
But when $z=ix$ is imaginary, then $g(z)=\frac{e^x-e^{-x}}{2x}$, which has infinite limit as $x\to\infty$.
