I have been struggled with the following problem.
Suppose that $f$ is differentiable on $(0, \infty)$. If we have $\lim_{x \rightarrow \infty} f(x) = c_1$ and $\lim_{x \rightarrow \infty} f'(x) = c_2$, where $c_1 c_2$ are two constants.
Can we get $c_2 = 0$？ If not could you please show me a counterexample?
It is true, and here's a hint: if $n \in \mathbb{N}$ and $\lim_{x \to \infty} f(x)$ exists, show that there exists some $M_n$ such that $$x, y > M_n \implies |f(x) - f(y)| < \frac{1}{n}.$$ As such, if $x > M_n$, we get $|f(x + 1) - f(x)| < \frac{1}{n}$. Use the Mean Value Theorem to find a sequence $x_n$ such that $x_n \to \infty$, but $f'(x_n) \to 0$.
Full Answer:
Fix $n \in \mathbb{N}$. Then, since $\lim_{x \to \infty} f(x) = c_1$, there exists some $M_n$ such that $$x > M_n \implies |f(x) - c_1| < \frac{1}{2n}.$$ Then, \begin{align*} x > M_n &\implies x + 1 > M_n \\ &\implies |f(x) - c_1| + |c_1 - f(x + 1)| < \frac{1}{2n} + \frac{1}{2n} \\ &\implies |f(x) - f(x + 1)| < \frac{1}{n}. \end{align*} Let $a_n > \max\{M_n, n\}$. Then $|f(a_n + 1) - f(a_n)| < \frac{1}{n}$. Applying the mean value theorem, we get some $x_n \in (a_n, a_n + 1)$ such that $$|f'(x_n)| = \left|\frac{f(a_n + 1) - f(a_n)}{a_n + 1 - a_n}\right| = |f(a_n + 1) - f(a_n)| < \frac{1}{n}.$$ Hence, by squeeze theorem, $f'(x_n) \to 0$. Note also that $x_n > a_n > n$, so $x_n \to \infty$.
Now, if $f'(x)$ converges as $x \to \infty$, then $f'(x_n)$ must converge to this limit. As we showed, this limit is $0$, so $f'(x) \to 0$.
Use mean value theorem to get $f(x+1)-f(x)=f'(c)$ for some $c$ with $x<c<x+1$ and take limit as $x\to\infty $. You get $c_1-c_1=c_2$.
