Since $\tan x$ is surjective onto the set of reals, given any $k \in \mathbb{R}$, in each interval $\left(\dfrac{n\pi}{2},\dfrac{(n+1)\pi}{2}\right)$ , there exists $a_n$ such that $\tan(a_n) = ka_n$.

$\displaystyle \lim_{n \to \infty} a_n = \infty$

So, since $\displaystyle \lim_{n \to \infty} \dfrac{\tan(a_n)}{a_n} = k$ you know $\displaystyle \lim_{x \to \infty} \dfrac{\tan x}{x}$ does not exist.

Edit: made the argument more general.