Let $f:[0,\infty)\rightarrow R $ be a continuously differentiable function. Show that if
$ \lim_{x \rightarrow \infty} f'(x)<1 $
then
$ f(x_0)<x_0 $ for some $x_0$ large enough. (An example of a function that satisfies these assumption is $f(x) = \sqrt x $). I am struggling with the proof. I've tried with the mean value theorem:
Choose $0<x<y$. Then by the MVT there esists a number $c \in (x,y)$ such that
$ \dfrac{f(y)-f(x)}{y-x} = f'(c) $
But at this point I got stuck… The other info I have is that there exists a number $M>0$ such that if $x>M$ then $f'(x)<1$ (the limit condition stated in the hp). Is there a way to combine these two facts in order to prove what I want?
Best Answer
Take $r\in\left(\lim_{x\to\infty}f'(x),1\right)$. Since $\lim_{x\to\infty}f'(x)<r$, there is a $M>0$ such that $x\geqslant M\implies f'(x)<r$. Then, when $x>M$, by the mean value theorem$$\frac{f(x)-f(M)}{x-M}=f'(c)<r,$$for some $c$ between $M$ and $x$. So,$$x>M\implies f(x)<f(M)+r(x-M).$$But, if $x$ is large enough, we have $f(M)+r(x-M)<x$, since $0<r<1$, and therefore, if $x$ is large enough, we have $f(x)<x$.