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

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.

• you may not plug in the value in a limit unless the function is continuous. The whole point is that the function is not continuous at 0. You are trying to determine how bad this discontinuity is Dec 29, 2017 at 11:25

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.
• I couldn't understand how is this a non-isolated singularity. Could you please explain again? I am convinced that it is not a removable singularity. We say, for example, $f(z) = cot z$ has isolated singularity at $z_0 = n\pi$. So how is this non-isolated? Thank you. Dec 29, 2017 at 12:22
• @user517123 Because $0$ is a singularity and, as I proved, there are singularities arbitrarily close to $0$. That's what “non-isolated” mean. Dec 29, 2017 at 12:28
• @Unknownx The function $f$ has no Lauret expansion around $0$. Jun 14, 2020 at 6:58
• $\sin \left(\frac{1}{\cos \left(\frac{1}{z}\right)}\right)=\sec \left(\frac{1}{z}\right)-\frac{\sec ^3\left(\frac{1}{z}\right)}{3!}+\frac{\sec ^5\left(\frac{1}{z}\right)}{5!}-...\infty =\left(1+\frac{\left(\frac{1}{z}\right)^2}{2!}+\frac{5\left(\frac{1}{z}\right)^4}{4!}+...\right)-\frac{\left(1+\frac{\left(\frac{1}{z}\right)^2}{2!}+\frac{5\left(\frac{1}{z}\right)^4}{4!}+...\ \infty \right)^3}{3!}+...\infty$ Jun 14, 2020 at 13:48