If $ \lim_{x\to\infty} \frac{f(x)}{x} = 1$, then $\exists x_n \rightarrow \infty$ such that $\lim_{n\to\infty} f'(x_n) = 1$.

Question

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function. If
$$
\lim_{x\to\infty} \frac{f(x)}{x} = 1,
$$
then there exists a sequence $(x_n)$ such that $x_n \rightarrow \infty$ as $n \rightarrow \infty$ and
$$
\lim_{n\to\infty} f'(x_n) = 1.
$$

We tried to use the Mean Value Theorem. For each $n$, there exists $x_n \in [0,n]$ such that
$$
f'(x_n) = \frac{f(n) – f(0)}{n}.
$$
Hence, $\lim_{n\rightarrow \infty} f'(x_n)=1$. But, we are not sure that $x_n \rightarrow \infty$.

Thank you for any hint.

Answer 1

You can apply the mean-value theorem to the intervals $[n, 2n]$: $$ f'(x_n) = \frac{f(2n)-f(n)}{2n-n} = 2 \frac{f(2n)}{2n} - \frac{f(n)}{n} \to 2-1 = 1 $$ and $x_n \ge n \to \infty$.