Prove that there exists a sequence $x_n\to\infty$ such that $\lim_{n\to\infty} f(x_n) = b$ and $\lim_{n\to\infty} f'(x_n) = 0$

Question

Let $f(x)$ be a real-valued differentiable function on $[0,\infty)$ and $\limsup_{x\to\infty}f(x)=b>\liminf_{x\to\infty} f(x)=a$.

How to prove that there exists a sequence $x_n\to\infty$ such that $\lim_{n\to\infty} f(x_n) = b$ and $\lim_{n\to\infty} f'(x_n) = 0$?

Analysis:

From $\limsup_{x\to\infty} f(x) > \liminf_{x\to\infty} f(x)$ we know that $f(x)$ must have positive oscillation at infinity.

Since $\limsup_{x\to\infty} f(x)=b$, we can choose a sequence $y_n\to\infty$ such that $\lim_{n\to\infty} f(y_n)=b$. But I am not sure about $\lim_{n\to\infty} f'(y_n)$. For some functions like $f(x)=\sin x$, we can take $x_n = \frac{\pi}{2}+2n\pi$ which is exactly what we want.

Answer 1

I would start with a sequence $(y_n)$ converging to the $\liminf$ instead, and use that to construct a sequence $(x_n)$ where $f$ has local maxima:

Let $(y_n)$ be a strictly increasing sequence with $y_n \to \infty$ and $f(y_n) \to a$. In each interval $[y_n, y_{n+1}]$, let $x_n$ be a point where $f$ attains its maximum on that interval.

Now show that $\limsup_{n \to \infty} f(x_n) = b$, and choose a subsequence $(x_{n_k})$ such that $f(x_{n_k}) \to b$.

Finally show that, for sufficiently large $k$, $x_{n_k}$ is in the interior of the interval $[y_{n_k}, y_{n_k+1}]$, and therefore $f'(x_{n_k}) = 0$. (This is where the condition $a < b$ comes into play.)