I've already asked the question on [MSE](https://math.stackexchange.com/questions/3703824/simple-condition-that-would-prove-a-function-transcendental) but there are still no answers, so I'm going to ask it here.


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?



"Simple" condition that would prove a function transcendental
$\newcommand\R{\mathbb R}$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.


