$f:\mathbb{R} \to \mathbb{R}$ be differentiable and $\lim\limits_{x\to\infty}f'(x)=1$, is $f(x)$ unbounded?

Question

Suppose $f:\mathbb{R} \to \mathbb{R}$ is a differentiable function such that $\lim\limits_{x\to\infty}f'(x)=1$,then is it true necessarily true that $f(x)$ unbounded?

I think that it will always intersect $y=c$ for every $c\in \mathbb{R}$ and thus cannot be bounded!

Answer 1

If the derivative tends to 1 then in particular it will eventually become bigger than 1/2. Therefore the function is dominated from below by $\frac{1}{2}x-c$ for a suitable c, and hence tends to infinity.

More precisely, by definition of limit for every $\epsilon>$ there is an $N$ such that if $x>N$ then $f'(x)>1-\epsilon$. Now choose $\epsilon=1/2$. Then for every $x>N$ we have $f'(x)>1/2$. Now proceed as above.

Answer 2

Choose some $c\in\left(0,1\right)$.

$\lim_{x\rightarrow\infty}f'\left(x\right)=1$ implies the existence of some $x_{0}$ with $x\geq x_{0}\Rightarrow f'\left(x\right)\geq c$.

Then $f\left(x\right)\geq cx+f\left(x_{0}\right)-cx_{0}$ for $x\geq x_{0}$ showing that $f$ is unbounded.