TITLE: Prove that: $\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $
QUESTION [5 upvotes]: I conjecture that in some specific conditions a differentiating function gives the following equality:
$$\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $$
However, I'm not sure yet what exactly those conditions are in order to precisely know where
I may apply this rule or not. If you wanna take a look over my posted problem here you'll immediately notice that this rule applies for that case. I really appreciate if you help me clarify this.

REPLY [5 votes]: I believe that the minimal assumptions are:

$f \colon [0,+\infty) \to \mathbb{R}$ is differentiable;
$\lim_{x \to +\infty} f'(x)$ exists.

Then you easily check that $$\lim_{n \to +\infty} f(n+1)-f(n)=\lim_{x \to +\infty} f'(x),$$
since you can apply Lagrange's theorem: $$f(n+1)-f(n)=f'(\xi_n)$$ for a suitable $\xi_n \in (n,n+1)$.
I think that the two limits are not equivalent to each other, since it may be impossible to control $f'$ by using the values of $f$ at discrete points.