$f'\rightarrow 0$ , if $f(n)\rightarrow l (n\in\mathbb{Z})$, then $f(x)\rightarrow l$?

Question

Let $f:(0,\infty)\rightarrow \mathbb{R}$ be differentiable and $f'(t)\rightarrow 0$ as $t\rightarrow \infty$.

If $f(n)\rightarrow l$ as $n\rightarrow \infty (n\in \mathbb{Z})$, then $f(t)\rightarrow l$ as $t\rightarrow \infty$ ($t\in\mathbb{R})$.

I am currently stuck at this problem.

How to prove this? I am having difficulties relating the limit for integers and the limit for reals when $f'\rightarrow 0$.

Thanks in advance!

Answer 1

Let $\lfloor t \rfloor$ denote the largest integer smaller than $t$. In general, we have $$|f(t) - l| <|f(t) - f(\lfloor t \rfloor)| + |f(\lfloor t \rfloor) - l|.$$

• Because $f(n) \to l$, the second term can be made small if $t$ is large enough.
• The first term can be made small for large $t$ by the mean value theorem. It is equal to $$|f(t) - f(\lfloor t \rfloor)| = |t-\lfloor t \rfloor||f'(c)| \le |f'(c)|$$ for some $c$ between $t$ and $\lfloor t \rfloor$. When $t$ is large, $|f'(c)|$ is small because of the convergence of the derivative to zero. [The intuition here is that $f$ gets flatter and flatter as $t \to \infty$, so the function cannot vary very much between two integers.]