"Simple" condition that would prove a function transcendental I conjectured that for every algebraic function $f(x)$ that is differentiable on $\mathbb{R}$, its $\lim_{x\to\infty}$ is either $\infty$, $-\infty$, or a finite value, so:

If $f(x)$ is differentiable everywhere on $\mathbb{R}$ and its $\lim_{x\to\infty}$ is not $\infty$, $-\infty$, nor a finite value, then $f(x)$ is transcendental.

If this is true, how could it be proved?
This question was asked on MathOverflow.
 A: Note: An accepted answer to this question is found at: "Simple" condition that would prove a function transcendental
Iosif Pinelis's answer:

$\newcommand\R{\mathbb R}$Your conjecture is true. Indeed, suppose
that a function $f\colon\R\to\R$ is continuous and algebraic, so that
$$\sum_{j\in[n]_0}p_j(x)f(x)^j=0\tag{1}$$ for all $x\in\R$, where
$[n]_0:=\{0,\dots,n\}$, $n$ is a natural number, and, for each
$j\in[n]_0$, $p_j$ is a polynomial function of some degree $m_j$, so
that for some real $a_j\ne0$ $$p_j(x)=(a_j+o(1))x^{m_j}\tag{2}$$ as
$x\to\infty$.
Suppose now that for some real $c$ and some sequence $(x_m)$ in $\R$
converging to $\infty$ we have $f(x_m)\to c$. Then, by (1) and (2),
$$p(c):=\sum_{j\in J}a_j c^j=0,\tag{3}$$ where $J$ is the (nonempty)
set of all $j\in[n]_0$ such that for all $i\in[n]_0$ we have $m_j\ge
> m_i$. So, $c$ must be in the finite set of the roots of the polynomial
$p$.
On the other hand, if your conjecture were false, then, by the
intermediate value theorem, there would be infinitely (even
uncountably) many real $c$ such that for some sequence $(x_m)$ in $\R$
converging to $\infty$ we have $f(x_m)\to c$.
So, your conjecture is true.

