TITLE: Is it always possible to calculate the limit of an elementary function?
QUESTION [33 upvotes]: I already asked this question on https://math.stackexchange.com/questions/2691331/is-it-always-possible-to-calculate-the-limit-of-an-elementary-function but as I received no answer; maybe it is not as obvious as I originally thought.
The precise formulation of my question is: define an "strong elementary function" by only admitting rational for the "constant function" in the usual definition of "elementary" function (see for example: https://en.wikipedia.org/wiki/Elementary_function). Let $a$ be an "elementary real" if the constant function $f(x)=a$ is a strong elementary function. With this definition some non rational reals are elementary (for example $\pi=4⋅arctan(1)$); but there are reals that are not elementary. Now let $f(x)$ be a strong elementary function defined in an open interval of an elementary real $a$ with the possible exception of $a$. Suppose that
$lim_{x\rightarrow a} f(x)$
exists. Is this limit necessarily an elementary real?
The idea behind this question is this: it seems that "limits" can always be calculated with some simple tricks (Hospital rule, etc...) in elementary calculus; but is there a general argument that shows that it is always possible? (the precise formulation of the question does not ask for an algorithm, I expect a positive answer to be constructive, but that is not entirely clear).
Update: clarification of the notion of elementary functions.
Definition
By function $\mathbb{R}\rightarrow\mathbb{R}$; I mean a partial function.
The class of (strong) elementary functions is the smallest class of functions such that:

the constant function $f(x)=1$ is elementary.
if $f$ and $g$ are elementary; so is $f+g$; $f-g$; $f\cdot g$; $f/g$. (the domain of $f/g$ is ${\rm dom}(f)\cap {\rm dom}(g)\cap \{x \ | \ g(x)\neq 0\}$.
if $n$ is a natural number; $f(x)=x^n$ and $f(x)=\sqrt[n]{x}$ are elementary; the domain of the later is $\mathbb{R}^+$.
$\sin$, $\cos$, $\tan$ are elementary
$\arcsin$, $\arccos$ and $\arctan$ are elementary. ($\arcsin:[-1\ 1]\rightarrow[-\pi\ \pi]$; $\arccos(x)={\pi\over 2}-\arcsin(x)$; $\arctan: ]-\infty\ +\infty[\rightarrow]-\pi\ \pi[$)
$\exp$ is elementary.
$\ln$ is elementary. ($\ln: \mathbb{R}^+\rightarrow \mathbb{R}$).
if $f$ and $g$ are elementary; so is $f\circ g$; the domain of the latter is $\{x\ | \ x\in{\rm dom}(g) \wedge g(x)\in{\rm dom}(f)\}$.

I have not tried to avoid redundancy but 
note that the point 3 is not redundant because of the domain of the functions considered; for example $f(x)=x^2$ is defined on $\mathbb{R}$ but $f(x)=\exp(2\cdot\ln(x))$ is defined on $\mathbb{R}_0^+$. Also $\sqrt[n]{x}$ is defined for $x=0$ but not $\exp(\frac{\ln(x)}{n})$.
I think this is the class of functions we consider in the Risch algorithm (https://en.wikipedia.org/wiki/Risch_algorithm) except that I do not take all constant functions as elementary; that would obviously make no sense for my question.
I hope I have not missed something obvious. I do not think a small modification of my definition will make any difference; if it is it would be interesting to discus.

REPLY [6 votes]: In fact, zero-equivalence for combinations of polynomial and sine functions is undecidable, which means that it is undecidable whether the limit is zero in that setting (This goes back to at least Paul Wang's 1974 paper The undecidability of the existence of zeros of real elementary functions).