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.

but then I'm confused by the relation to your problemjust to clarify: olivier asked this question, while I'm Olivier (not at all the same MO user). $\endgroup$ – Olivier Mar 20 '18 at 15:084more comments