# 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:

1. the constant function $f(x)=1$ is elementary.
2. if $f$ and $g$ are elementary; so is $f+g$; $f-g$; $f\cdot g$; $f/g$. (the domain of $f/g$ is ${\rm dom}(f)\cap {\rm dom}(g)\cap \{x \ | \ g(x)\neq 0\}$.
3. if $n$ is a natural number; $f(x)=x^n$ and $f(x)=\sqrt[n]{x}$ are elementary; the domain of the later is $\mathbb{R}^+$.
4. $\sin$, $\cos$, $\tan$ are elementary
5. $\arcsin$, $\arccos$ and $\arctan$ are elementary. ($\arcsin:[-1\ 1]\rightarrow[-\pi\ \pi]$; $\arccos(x)={\pi\over 2}-\arcsin(x)$; $\arctan: ]-\infty\ +\infty[\rightarrow]-\pi\ \pi[$)
6. $\exp$ is elementary.
7. $\ln$ is elementary. ($\ln: \mathbb{R}^+\rightarrow \mathbb{R}$).
8. 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.

• If I understand correctly, the "elementary" class is not supposed to be closed under local inversion. I don't know if it matters, though. – YCor Mar 17 '18 at 10:49
• In your limit, $x\to a$ you admit all complex $x$ or only real? If complex, then $a$ can be only an algebraic singularity (a branch point pole, no worse), and then the limit can be always found by l'Hospital rule. – Alexandre Eremenko Mar 17 '18 at 15:09
• @ Alexandre Eremenko. My question concerns the reals. A comment on the original question asked on "math stack exchange" already did notice that the answer is true in the complex plane. But this result does not appear to transfer for real defined functions. Many "counter-example in analysis" do not exists anyway in complex analysis (smooth non-analytical function, etc.) – Olivier Esser Mar 17 '18 at 15:29
• @oliver: If you are interested in real limits, see my answer part 2. However I am not 100% sure about the answer with complex limits. A singularity of an elementary function may not be isolated, so classification "removable, essential, ramification point" is not applicable. One needs some more sophisticated argument. – Alexandre Eremenko Mar 17 '18 at 16:38
• @Gro-Tsen but then I'm confused by the relation to your problem just to clarify: olivier asked this question, while I'm Olivier (not at all the same MO user). – Olivier Mar 20 '18 at 15:08

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).

1. 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).

2. 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.

• @olivier : Really? You don't recognize representing $f$ in polar coordinates? – Eric Towers Mar 17 '18 at 19:27
• But how do you ensure that the functions are elementary? OK for $r(t)$ but how do you get $\phi(t)$? You can't make piecewise definition. You use what is commonly known as $atan2(x,y) = \pm arctan(y/x)$ plus the special treatment for x=0. That's not elementary. – Olivier Esser Mar 17 '18 at 19:44
• @oliver: You do not give a definition of elementary function, referring to Wikipedia. Wikipedia does not have a precise definition, but gives complex logarithm as an example. It refers to Ritt's papers, and I follow the Ritt definition. – Alexandre Eremenko Mar 17 '18 at 23:56
• @oliver: "multivalued function" is not a function, therefore you cannot integrate it. What you can integrate is a branch. There are infinitely many branches of arctan on the real line, they are different functions and each of them you can integrate. "Elementary functions" (as defined by Liouville and Ritt) are multivalued. If you propose some other approach, please give a precise definition of an "elementary function". Explain exactly what you mean by $x^{1/3}$ to begin with, and then think of $\arctan(1/x)$. – Alexandre Eremenko Mar 18 '18 at 16:47
• Regarding the definition of "elementary real," I gave one possible definition in this paper: timothychow.net/closedform.pdf I think that this definition is close to, if not exactly the same as, what olivier wants. The definition has the advantage of sidestepping some of the issues about domains of definition of elementary functions. On the other hand, difficulties remain; e.g., it remains unclear whether $\gamma$ is an elementary real. – Timothy Chow Mar 18 '18 at 19:28

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).

• Yes, it is undecidable whether the limit is 0, and I suppose undecidable whether the limit exists. But this is a somewhat different problem: we want to know: IF the limit exists is it elementary. – Alexandre Eremenko Mar 21 '18 at 12:23
• If it is undecidable whether the limit is 0; then it is undecidable whether the limit exists since obviously $\lim_{x\rightarrow 0} f(x)\neq 0$ iff $\lim_{x\rightarrow 0} {1\over f(x)}$ exists. But I agree this is a different (although interesting) problem. – Olivier Esser Mar 21 '18 at 20:18