Analyzing the singularity of $f(z)= \sin\big(\frac{1}{\cos(\frac1z)}\big)$ at $z=0$

Question

For the function :

$$f(z)= \sin\bigg(\frac{1}{\cos(\frac1z)}\bigg)$$

the point $z=0$ is:

1)a removable singularity

2)a pole

3)an essential singularity

4)a non-isolated singularity

The answer seems to be 3)an essential singularity.

But I arrived at 1)removable singularity because when $f(z)$ has removable singularity,

$\lim\limits_{z\to0}$ $(z-z_0)$$f(z)=0$. (Since $\lim\limits_{x\to a}$$f(x)=f(a)$.)

Can someone help me, pointing out where I had gone wrong? Thanks in advance.

Answer 1

It has a non-isolated singularity at $0$, since it has a singularity at every point of the form $\frac1{\pi/2+n\pi}$ ($n\in\mathbb N$).

It is not true that $\lim_{z\to0}zf(z)=0$. Actually, this limit does not exist.