Limit of a derivative of function at infinity

Question

If we have a differentiable function $f:(0,∞) \rightarrow \mathbb{R}$ such that derivative $f'(t)$ tends to zero as $t \rightarrow ∞$ and for $n \in \mathbb{Z}, f(n) \rightarrow l$ (finite) as $n \rightarrow ∞$, then how do we show that $f(t) \rightarrow l$ as $t (\in \mathbb{R}) \rightarrow ∞$.

What I don't understand is that if for $n \in \mathbb{Z}, f(n) \rightarrow l$ as $n \rightarrow ∞$, then shouldn't this hold for any real number $t$ as well? We can always find an integer greater than any real number. What is the use of limit of derivative in solving this question?

Answer 1

We have that $\forall x\in[n,n+1]$ by MVT

$$\frac{f(x)-f(n)}{x-n}=f'(c) \quad c\in(n,n+1) \implies |f(x)-f(n)|=(x-n)|f'(c)|\le |f'(c)|\to 0$$

Answer 2

Take the function $f(t) = \sin (\pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 \over 2}) = 1$ for all $n$.

You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for large $n$. This is what the $f'(t) \to 0$ condition does.