Show that the function $f(z)=\exp\left(\frac{z}{1-\cos z}\right)$ has essential singularity at $z=0$

Question

At $z=0$ show that the function $f(z)=\exp (\dfrac{z}{1-\cos z})$ has essential singularity.

I want to show that the limit does not exist at $z=0$.

Put $z=x\to 0$

$\lim_{x\to 0}\exp(\dfrac{x}{1-\cos x})$.

Let $y=\exp(\dfrac{x}{1-\cos x})\implies \ln y=\dfrac{x}{1-\cos x}$.

Now $\dfrac{x}{1-\cos x}=(0/0)$.so we have $\dfrac{1}{\sin x}\to \infty as x\to 0.$

Hence $y=e^\infty \to \infty$

Put $z=ix\to 0$; $\exp(\dfrac{ix}{1-\cos ix})=\exp(\dfrac{ix}{1-{\dfrac{e^{ix}+e^{-ix}}{2}}})$.

I think the above limit tends to $\infty $ as $x\to 0$.

I am unable to prove the limit does not exist.

Answer 1

It's $\enspace\displaystyle \frac{z}{1-\cos z}=\frac{2}{z}\left(\frac{z/2}{\sin(z/2)}\right)^2\enspace$ with $\enspace\displaystyle \frac{z/2}{\sin(z/2)}\to 1\enspace$ for $\enspace z\to 0$ .

This means that your question is equivalent to the question, if $\enspace\displaystyle e^{1/z}\enspace$ has an essential singularity.

And this is true as we can see with $\enspace\displaystyle e^{1/z}=\sum\limits_{k=0}^\infty \frac{1}{k!z^k}\enspace$ (Laurent series).