Prove that a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ which has horizontal asymptotes has a point $c$ such that $f(c) = c$.

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function.
The function $f$ has horizontal asymptotes to $+\infty$ and to
$-\infty$. Prove that there $\exists$ $c \in \mathbb{R}$ such that
$f(c) = c$.

I don't see how I could prove this. I really like this quote:

"One should never try to prove something that is not almost obvious." – Alexander Grothendieck

Well, what is asked of me to prove does not seem obvious at all. I mean, all we have is continuity and horizontal asymptotes. So we know that the functions is defined and continuous on the set $\mathbb{R}$ and we also know:

$$\lim_{x \to \infty} f(x) \ne \pm \infty$$

$$\lim_{x \to -\infty} f(x) \ne \pm \infty$$

The above is because if the $2$ limits would be $\pm \infty$, then we wouldn't have a horizontal asymptote in that direction, we'd have a slant asymptote or no asymptote at all.

That's all we have. Given so little information, it doesn't seem obvious to me that we have at least one point $c \in \mathbb{R}$ such that the relation $f(c) = c$ is true.

So, how could I prove this statement, and more importantly, how can I see and understand that the statement has to be true?

Answer 1

Let $g(x)=f(x)-x$. Since

• $g$ is continuous;
• $\lim_{x\to\infty}g(x)=-\infty$;
• $\lim_{x\to-\infty}g(x)=\infty$,

there is, by the intermediate value theorem, some $c\in\mathbb R$ such that $g(c)=0$. But $g(c)=0\iff f(c)=c$.