Let $f: \mathbb R \to \mathbb R$ a differentiable function. Prove that if $\lim_{x \to +\infty}f(x)= 1$ and $\lim_{x \to +\infty}f'(x) = c$ then $c=0$.
It seems pretty intuitive, but I am having difficulty proving this rigorously. Let $\epsilon >0$. Then by what's given there is $M$ such that for all $x, y > M, |f(x)-f(y)| < \epsilon$. I am not sure how to proceed from now. The difficulty is that, when we write $|f'(x)|$ in terms of limits, there is $x-y$ term at the denominator.
On
More generally, it suffices to assume that $f$ is differentiable and BOUNDED in $[a,+\infty)$ for some real number $a$. Then if $$\lim_{x \to +\infty}f'(x) = c\in\mathbb{R}$$ by L'Hopital's rule (the denominator diverges and no additional assumption is needed about the limit of the numerator): $$0=\lim_{x \to +\infty}\frac{f(x)}{x}=\lim_{x \to +\infty}\frac{f'(x)}{1}=c.$$
Given $\epsilon>0$, find an $M>0$ such that $|f(x)-1|<\epsilon$ and $|f'(x)-c|<\epsilon$ for all $x\geq M$, then $|f(x+1)-f(x)|\leq|f(x+1)-1|+|f(x)-1|<\epsilon$. But $|f(x+1)-f(x)|=|f'(\eta_{x})|$ for some $\eta_{x}\in(x,x+1)$ by Mean Value Theorem so $|f'(\eta_{x})|<\epsilon$. Hence $|f'(\eta_{x})-c|<\epsilon$ and hence $|c|\leq|f'(\eta_{x})-c|+|f'(\eta_{x})|<3\epsilon$, so $c=0$.