Is it always possible to calculate the limit of an elementary function? 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. 
 A: 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).
A: EDITED. I use the definition of "elementary function" of Liouville and Ritt (also repeated in Wikipedia). See Ritt's papers in TAMS 25 (1923) 211-222, and
TAMS 27 (1925) 68-90. This definition includes the analytic continuation through removable singularities. These elementary functions are analytic and can be multi-valued. Of course, other definitions are possible, and the answer will depend on the definition. Your new definition of "elementary functions" is close to the classes
of "elementary functions" and 
"naive elementary functions", defined by Laczkovich and Ruzsa except that they allow all constants in the definition, and you allow only $1$.
The difference between their classes is in where exactly is
$\arcsin$ defined. So it is important to state a more precise definition.
Ref. M. Laczkovich and I. Ruzsa,  Elementary and integral elementary functions,
Illinois J. Math., 44, 1 (2000) 161-182.
For elementary functions in the sense of Liouville and Ritt, the answer is probably different, depending on what you mean by the limit.
(Existing of a limit as $x\to a$ when $x$ is complex is a much stronger condition then existence of a limit when $x$ is real). 


*

*Complex limits. The answer seems to be yes. What follows is a heuristic argument. Elementary functions are analytic, with at most countably many singularities.
So when the finite limit exists, the singularity is removable or a ramification point. Expanding everything in power series (perhaps with fractional powers), we can compute the limit. The coefficients of these power series are elementary constants, because they are expressed in terms of derivatives. So the answer is probably yes, if we mean complex limits. (This is heuristic because singularities can accumulate so one needs more careful argument). 

*Real limits. The answer seems to be no. Consider a trigonometric sum 
$$f(t)=\sum_{j=1}^n a_j\exp(\lambda_jit),\quad i=\sqrt{-1},$$
and assume that it has no real zeros. Suppose that the $\lambda_j$ are real "elementary constants" but incommensurable. Then $f(t)=r(t)\exp(i\phi(t))$, where $r(t)>0$ and $\phi$ is a well-defined elementary real function (in the sense of Liouville and Ritt). The limit
$$m:=\lim_{t\to+\infty}\phi(t)/t$$ always exists: this is the celebrated Mean Motion Theorem. There is a formula for this limit due to A. Wintner:
$$m=\sum_{j=1}^n\lambda_jW_j,\quad\mbox{where}\quad W_j=\int_{T^n}\Re\frac{a_j\exp(i\theta_j)}{\sum_ka_k\exp(i\theta_k)}d\theta_1\ldots d\theta_n,$$
where $T^n=[0,2\pi]^n$.
These integrals are probably not "elementary constants", Weyl computed them:
$$W_j=a_j\int_0^\infty J_1(a_jx)\prod_{k\neq j}J_0(a_kx)\, dx.$$ 
where $J$ are Bessel functions. Again, this argument is incomplete, because one has to
prove that the $W_j$ can indeed be "non-elementary" constants. These 
"elementary constants" are discussed in this paper
and the author writes that no single explicit example of non-elementary number is known! We know that they exist only because the set of elementary numbers is countable.
But at least the argument shows that to compute a limit of an elementary function
it is not enough to use such things as l'Hopital rule, one may need to compute definite integrals.
References on the mean motion theorem: S. Sternberg, Celestial Mechanics, Part I, Benjamin, 1969, H. Weyl, Mean motion I, Amer. J. Math., 60, 4(1938) 889-896.
UPDATE. For any counterexample to your conjecture, there is a formidable difficulty that not a single explicit number in known to be non-elementary.
Therefore I propose to modify the question: 

Let $f(x,a)$ be an elementary
  function of two variables. Is it true that $g(a)=\lim_{x\to x_0} f(x,a)$
  is an elementary function of $a$, provided that the limit exists for 
  all $a$ on some interval? 

This seems to be in the same spirit, but more accessible than what you asked.
Because we know quite a lot about elementary functions but really nothing
about elementary numbers.
For this question, perhaps it is possible to prove rigorously that the answer is no, with
Liouville-Ritt elementary functions. But so far I was unable to
prove that $W_j$ is non-elementary as a function of, say $a_1$,
when $n\geq 3$. (When $n\leq 3$ it is elementary, and it was known that
the mean motion has an elementary expression when $n\leq 3$ (Bohl).
With "purely real elementary functions", in the spirit of your definition,
the answer is still no, if we allow the $\arcsin{}$ in your definition
to have the domain $(-1,1)$. Then all elementary functions are analytic (as compositions of analytic functions),
but we have a non-analytic limit: ($x\to+\infty$ along the positive ray)
$$\lim_{x\to+\infty}\arctan(ax)=\frac{\pi}{2}{\mathrm{sgn}}(a),\quad \lim_{x\to+\infty}(1+a^x)^{1/x}.$$
If you allow $\arcsin$ on $[-1,1]$ then elementary functions can be discontinuous like $\arcsin(\sin x)$, and the above counterexample does
not work.
If you remove trigonometric functions from your class, the answer is yes. And this was essentially proved by Hardy in
his book Orders of Infinity.
