Limits and differentiability Suppose that $f$ is differentiable on $(0, \infty )$ and $\lim\limits_{x \to \infty} f(x) = 1$.
If $\lim\limits_{x \to \infty} f'(x) = c$, show that $c = 0$.
I know that since $f$ is differentiable, then it is also continuous on $(0,\infty)$. I also know that, by definition of limits, given any $\epsilon > 0$, there is some $M$ such that whenever $x>M$, $|f(x)-1|<\epsilon$.
Any hints on how to start/which theorems to use would be very helpful!
 A: $\lim\limits_{x \to \infty}f'(x)=\lim\limits_{x \to \infty} \lim\limits_{x \to h} \frac{f(x+h)-f(x)}{h}  $
f and f' are continuous so we can switch the limits order :
$$\lim\limits_{x \to \infty}f'(x)=\lim\limits_{x \to h} \lim\limits_{x \to \infty} \frac{f(x+h)-f(x)}{h}=\lim\limits_{x \to h}\frac{1-1}{h} =0 $$
A: By Mean Value Theorem, there exist a sequence $ \{a_n\}_{n \in \mathbb{N} }$, $n<a_n<n+1$, such that $f^\prime (a_n) = \frac{f(n+1)-f(n)}{(n+1)-n}=f(n+1)-f(n)$. Take limits on both sides and use continuity
A: Outline: (i) Given $\epsilon \gt 0$, $|f(x)-1|\lt \epsilon/2$ for large enough $x$. (ii) By the Mean Value Theorem, for any $x$, we have $f(x+1)-f(x)=f'(c_x)$ for some $c_x$ between $x$ and $x+1$. (iii) So for large enough $x$, there is a $c_x\gt x$ such that $|f'(c_x)|\lt \epsilon$. (iv) Since the limit of $f'(x)$ by assumption exists, it must be $0$.
Remark:
Note that the existence of the limit of $f'(x)$ is an essential assumption. It is not hard to find differentiable $f(x)$ such that $\lim_{x\to\infty} f(x)=1$ but $\lim_{x\to\infty}f'(x)$ does not exist.
