Prove that: $\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))'$

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.

-
In effect, you're asking for conditions such that the limit at infinity of a function's forward difference and its derivative are equal. I'll just offer one observation: in evaluating the limit of an indeterminate form, one might choose to either use l'Hôpital (derivatives) or Stolz-Cesàro (forward differences), so at least in that respect, the behavior of the forward difference and the derivative are related. – J. M. Aug 5 '12 at 10:16
@J. M.: you're perfectly right if I take into account the function $\frac{f(x)}{x}$. – user 1618033 Aug 5 '12 at 10:19
I think it is enough $f$ satisfies the hypothesis of the Mean-value theorem: so $f$ must be differentiable over $(0,+\infty)$. – Romeo Aug 5 '12 at 10:20
I think that this rule has a good helping potential when dealing with some limits. – user 1618033 Aug 5 '12 at 10:31

1. $f \colon [0,+\infty) \to \mathbb{R}$ is differentiable;
2. $\lim_{x \to +\infty} f'(x)$ 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.
Yes, the existence of $\lim_{n\rightarrow \infty} f(n+1)-f(n)$ does not imply the existence of $\lim_{x\rightarrow \infty} f'(x)$, as can be seen with $f(x) = \sin(\pi x)$. – James Fennell Aug 5 '12 at 11:05