Showing that $\cos(z)$ has an essential singularity at $\infty$ 
Problem: Show that $\cos(z)$ has an essential singularity at $\infty$.

EDIT: I just realized that step (2) is definitely wrong, as both those limits are undefined.  Still, the sum of two undefined limits can still be defined so I'm not sure how to proceed.
Proof:


*

*First we note that
$$
\lim_{z \rightarrow \overline{0}} e^{1 / iz}  = 0
$$
$$
\lim_{z \rightarrow \overline{0}} e^{-1/iz}  = \infty
$$
so that
$$
\lim_{z \to \overline{0}} \left| \underbrace{e^{1 / iz}}_{\rightarrow 0} + \underbrace{e^{-1 / iz}}_{\rightarrow \infty} \right| = \infty
$$

*Similarly, we have that
$$
\lim_{z \rightarrow 0+} e^{1 / iz}  = 0
$$
$$
\lim_{z \rightarrow 0+} e^{-1/iz}  = 0
$$
so that
$$
\lim_{z \to 0+} \left| \underbrace{e^{1 / iz}}_{\rightarrow 0} + \underbrace{e^{-1 / iz}}_{\rightarrow 0} \right| = 0
$$

*Putting this together yields that
$$
0 \ne \lim_{z \rightarrow 0} \left| e^{1 / iz} + e^{-1 / iz} \right| \ne \infty
$$
so that
$$
0 \ne \lim_{z \to 0} |z|^{\alpha} \left| e^{1 / iz} + e^{-1 / iz} \right| \ne \infty \text{ for any } \alpha \in \mathbb{R}
$$

*Then we have showed that the function $e^{1/iz} + e^{-1/iz}$ has an essential singularity at $0$.  This then implies that 
$$
{e^{1/iz} + e^{-1/iz} \over 2}
$$
also has an essential singularity at $0$.

*Yet by definition this means that
$$
\cos(z) = {e^{iz} + e^{-iz} \over 2}
$$
has an essential singularity at $\infty$.
 A: You know that, $f(z)$ has an essential singularity at $z=\infty \Leftrightarrow f(1/z)$ has an essential singularity at $z=0$.
You know that, $\displaystyle \cos z = \sum_{n=0}^\infty (-1)^n \frac{z^{2n}}{2n!}$.
i.e., $\displaystyle \cos (\frac{1}{z}) = \sum_{n=0}^\infty \frac{(-1)^n}{2n! z^{2n}}$.
Now you can see the -ve powers of $z$ in the expansion of $\cos(\frac{1}{z})$ continues indefinitely. 
Thus $\cos(\frac{1}{z})$ has an essential singularity at $z=0$, thus $\cos z$ has an essential singularity at $z=\infty$.
A: The function $f(z)=\cos z$ is entire. If it didn't have an essential singularity at $\infty$, then it would either have a removable singularity (which would imply that is bounded, so by Liouville it is constant, a contradiction), or it would have a pole, so $\lim_{z\to \infty} f(z)= \infty$. 
In the second case, consider the function $g(z)=1/f(1/z)$ which satisfies $\lim_{z\to 0} g(z)=0$, so it has a removable singularity around $0$, and in particular, it is holomorphic there. Therefore, there exists $r>0$ such that for $|z|\leq r$ we have
$$ g(z)=z^n h(z)$$
for some function $h$ which is holomorphic for $|z|\leq r$, and $h(0)\neq 0$. This in particular implies that  $h$ is bounded below for $|z|\leq r$, so $|h(z)|\geq M$ for all $|z| \leq r$. Hence we have
$$ \frac {1}{|f(1/z)|} = |z^n h(z)|\geq M|z|^n $$
for $|z|\leq r$. Replacing $z$ by $1/z$ we obtain
$$\frac{1}{|f(z)|} \geq \frac{M}{|z|^n}$$
or equivalently
$$ |f(z)| \leq \frac{1}{M} |z|^n$$
for all $|z|\geq 1/r$. Now using the Cauchy integral formula for $f^{(n)}$ we conclude that $f$ is a polynomial of degree at most $n$, which is a contradiction, since the $(n+1)$-derivative of $\cos z$ is non-zero. 
Concluding $\cos z$ has an essential singularity at $\infty$.
A: If a function has a pole or a removable singularity at an isolated singularity then the function has a limit (finite or $\infty$). $\cos z$ has no limit at $\infty$ even if you restrict the function to the real line, so $\cos z$ has an essential singularity at $\infty$.
A: If it didn't, then $\cos(z):\hat{\mathbb{C}}\to\hat{\mathbb{C}}$ would be a holomorphic function between compact Riemann surfaces (since $\infty$ would either be a pole or a removable singularity). In particular its zeros should be discrete, which is an obvious contradiction with the fact that $\cos(z)$ has infinite zeros. 
This answer is obviously a little bit less elementary, but the other answers are already great and I think this offers an interesting point of view.
A: The equation
$$
w=\cos(z)\iff 0=e^{2iz}-2e^{iz}w+1=(e^{iz}-w)^2+1-w^2
$$
has for any complex $w$ infinitely many solutions 
$$
z=-i\,Ln(w\pm\sqrt{w^2-1})+k\cdot2\pi 
$$
selecting appropriate main branches for the square root and complex logarithm. Since $k\in\Bbb Z$ ranges over all integers, these solutions are arbitrarily large. Which means that for any ball $\Bbb C\setminus B(0,R)$ around infinity, the range encompasses all complex numbers. Which is the geometrical characterization of an essential singularity. 
A: This looks like an answer now.
The answer to the question "Can we prove that all zeros of entire function cos(x) are real from the Taylor series expansion of cos(x)?" may help to understand the behavior of $\cos(z)$ at $z=\infty$
-mike
EDIT:
I copied several formulas from my answer in the link above:

First we rewrite the expression as
$$ \cos(z) =\lim_{n\to\infty}g_n(z/(2n)) $$
$$ g_n(z/(2n)):=\sum_{n=0}^n (-1)^k \binom{2n}{2k}\frac{z^{2k}}{(2n)^{2k}} $$
The function $g_n(z)$ is called the Jensen polynomial associated with entire function $\cos(z)$. Its closed-form expression is given by
$$g_n(z/(2n))=\frac{1}{2}\left(1+\frac{iz}{2n}\right)^{2n}+\frac{1}{2}\left(1+\frac{-iz}{2n}\right)^{2n}$$
Let $\omega_{k}$ and $-\omega_{k}$ with $n=1,2,...,n $ be the $2n$ roots of $y^{2n}=-1$, they are given by:
$$\omega_{k}=\exp\left({i\pi}\frac{2k+1}{2n}\right)$$
then roots of $g_n(z/(2n))$ are given by:
$$z_k=-(2in)\frac{\omega_{k}-1}{\omega_{k}+1}=2n\tan\left(\frac{\pi(2k+1)}{4n}\right)$$
$$z_{n+k}=-(2in)\frac{-\omega_{k}-1}{-\omega_{k}+1}=-2n\cot\left(\frac{\pi(2k+1)}{4n}\right)$$
When $n\to\infty$ the first $n=\infty$ zeros survived 
$$\lim_{n\to\infty}z_k=\lim_{n\to\infty}2n\tan\left(\frac{\pi(2k+1)}{4n}\right)=\frac{\pi}{2}(2k+1)$$
the last $n=\infty$ zeros are pushed to a single point $-\infty$
$$\lim_{n\to\infty}z_{n+k}=\lim_{n\to\infty}(-2n)\tan\left(\frac{\pi}{2}-\frac{\pi(2k+1)}{4n}\right)=\lim_{n\to\infty}(-2n)\tan\left(\frac{\pi}{2}\right)=-\infty$$
The fact, that there are infinite number of zeros clustered on a single point, might be the reason why infinity is an essential singularity for $\cos(z)$.
-mike
