# How to ensure when the derivative approaches zero, the integral approaches a constant?

I asked a very similar question here. But now this is different. Suppose $$f(t)$$ is differentiable and $$c$$ is a finite constant, then the following statement looks correct, but in fact it is not: $$\begin{equation} \lim\limits_{t \to \infty} f'(t) = 0 \implies \lim\limits_{t \to \infty} f(t)=c \end{equation}$$ A counter-example is $$f(t)=\ln(t)$$. Now the question is, what condition should be used for the above statement to be true?

• The inverse implication is always true: $$\begin{equation} \lim\limits_{t \to \infty} f(t)=c\implies \lim\limits_{t \to \infty} f'(t) = 0 \end{equation}$$ Thus, we find that $\lim\limits_{t \to \infty} f'(t) = 0$ is a necessary condition for the function being convergent at $x\to\infty$. I think that what you're actually looking for is a statement $A$ so that $$\left( \lim\limits_{t \to \infty} f'(t) = 0\right) \land A \implies \lim\limits_{t \to \infty} f'(t) = 0$$ is true, but so that the formula $$A \implies \lim\limits_{t \to \infty} f'(t) = 0$$ doesn't hold in general. – Sudix Jul 24 '19 at 15:32
• The inverse implication is NOT true. See the link in my question. – winston Jul 27 '19 at 21:01

An equivalent condition is that $$\int_a^\infty f'(t)\,dt$$ converges as an improper Riemann integral for some $$a\in\Bbb R$$. This happens for instance if $$|f'(x)|\le C\,x^{-p}$$ for some $$C\ge0$$ and $$p>1$$.

• Is there any proof of your result? – winston Dec 18 '18 at 16:39
• $$f(x)=f(a)+\int_a^xf'(t)\,dt,\quad x>a.$$ – Julián Aguirre Dec 18 '18 at 16:40
• True if $f'$ is Riemann integrable on bounded intervals. – zhw. Dec 18 '18 at 17:14
• Does it mean that $\lim\limits_{t \to \infty} f'(t)=0$ does not make any difference to the condition? Doesn't it somehow relax your condition in some way? – winston Dec 18 '18 at 20:21
• If $\lim_{x\to\infty}f'(x)$ exists, it must be $0$. But it may happen that the limit does not exist. – Julián Aguirre Dec 19 '18 at 8:28

One way you could potentially change your statement so that it is true is to assert that the function is bounded asymptotically (as $$t \rightarrow \infty$$).

If $$\lim\limits_{t \rightarrow \infty} f'(t) = 0$$ and $$\exists n,M$$ such that $$\forall t > n, |f(t)| \le M$$, then $$\exists c$$ such that $$\lim\limits_{t \rightarrow \infty} f(t) = c$$

I don't have a proof of this currently, but I believe it should be true, and in fact I believe that this also works in reverse. I will provide a proof if I can come up with one.

Two ways for a differentiable (and hence, continuous) function $$f(t)$$ to satisfy $$\lim_{t \to \infty} f(t) = c$$:

(1) $$f(t) = c$$; i.e., is a constant function

(2) $$f(t)$$ has a horizontal asymptote $$y=c$$ in the positive x-direction.

Hence, restrict $$f(t)$$ to be either of these two function types and your original assertion $$\lim_{t \to \infty} f'(t) \Rightarrow \lim_{t \to \infty} f(t)$$ will be true.