A non-exponentially bounded analytic function? A function $f:\mathbb R\to\mathbb R$ is said to be exponentially bounded if there is an $n$ such that for sufficiently large $x\in\mathbb R$, $\exp(\exp(\cdots \exp(x)))>f(x)$ (where the $\exp$ is repeated $n$ times).
You know what analytic means.  Is there a "classical" (or easy to describe) non-exponentially bounded function?
I ask because this is related to an open question in model theory, namely whether the real field, expanded with a non-exponentially-bounded function, can be o-minimal.  Most of what we know about o-minimal expansions are related to analytic functions, so I'm interested in what we could be looking for.
 A: Sorry for the delay: the evening was busier than I thought it would.
Note that if $a$ is irrational, then the denominator never vanishes. On the other hand, if $a=\frac pq$ in simplest terms and $p,q$ are odd, then $x=\frac\pi 2q$ is a zero. Now let $F(x)$ be any given continuous function. We shall construct a sequence of nested intervals $I_k$ for $a$ and a sequence of numbers $q_k$ so that the spikes of our function $f_a$ are above $F$ at $\frac\pi 2q_k$ whenever $a\in I_k$. Since we also need to escape a rational value, we'll fix some enumeration $r_k$ of rationals and ensure that $r_k\notin I_k$. 
Put $q_1=1$ and take $I_1$ to be a small interval $[1+\delta_1,1+2\delta_1]$ with very small $\delta>0$ so that $r_1\notin I_1$. Then, if $\delta_1$ is small enough, $f_a(\frac\pi 2q_k)>F(\frac\pi 2q_k)$. 
Now choose any rational fraction $p_2/q_2$ with odd $p_2,q_2$ and $q_2\ge 2$ contained in the interior of $I_1$ and put $I_2=[\frac{p_2}{q_2}+\delta_2,\frac{p_2}{q_2}+2\delta_2]$. Again, we can choose $\delta_2>0$ so that $I_2\subset I_1$, $r_2\notin I_2$, and $f_a(\frac\pi 2q_2)>F(\frac\pi 2q_2)$. Now do $I_3$ using $q_3\ge 3$, $I_4$, etc. in the same way. 
The nested interval lemma then yields $a$ such that $f_a>F$ on a sequence tending to infinity (i.e., beating any prescribed growth control), though this bad sequence itself is hard to discern explicitly. 
A: Well, if you really insist on the domain over all real numbers- just take $f(x)=e^{-x}.$
