Show that $\lim_{x \rightarrow \infty} f'(x)<1$ implies $f(x_0)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? 
 A: 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 $f(x)<x$.
A: Let $g(x) =f(x) - x$ so that $g'(x) \to L<0$ as $x\to\infty$ and therefore $g'(x) $ is negative as $x\to\infty $. It thus follows that $g(x)$ eventually decreases and hence either tends to a limit $M$ or to $-\infty$. If it tends to $-\infty$ then $g$ is eventually negative and we are done. But if $g(x) \to M$ then by mean value theorem $g'(\xi) = g(x+1)-g(x)\to 0$ and this contradicts that $g'(x) \to L<0$.
Note: There is no need to assume that $f$ is continuously differentiable and we just need the hypotheses that $\lim_{x\to\infty} f'(x) $ exists and this limit is less than $1$.
A: So I think it is safe to say that the hypothesis
$\displaystyle \lim_{x \to \infty} f'(x) < 1 \tag 1$
gives us two important facts concerning the function $f(x)$:
first, that $\lim_{x \to \infty} f'(x)$ actually exists, that is, there is some $L \in \Bbb R$ such that, given any $\epsilon > 0$, there also exists $M_\epsilon \in \Bbb R$ such that
$x \ge M_\epsilon \Longrightarrow \vert f'(x) - L \vert < \epsilon; \tag 2$ 
second, that in fact
$L < 1; \tag 3$
then
$1 - L > 0, \tag 4$
so we can choose, say 
$\epsilon \le \dfrac{1 - L}{2} = \dfrac{1}{2}(1 - L); \tag 5$
then we may write (2) in the form
$x \ge M_\epsilon \Longrightarrow L - \epsilon < f'(x) < L + \epsilon \le L + \dfrac{1}{2}(1 - L) = \dfrac{1}{2}(L + 1) < 1; \tag 6$
the key here is that by taking $x$ sufficiently large, we may ensure $f'(x)$ is bounded by a number, in this case $(1/2)(L + 1)$, which is strictly less than $1$ which, intuitively, shows us that $f(x)$ grows more slowly than $x$ for $x$ large enough.  
Now consider the function $x - f(x)$; for $\xi \ge M_\epsilon$ we have
$x - f(x)  - (\xi - f(\xi)) = \displaystyle \int_\xi^x (s - f(s))' \; ds$
$=\displaystyle \int_\xi^x (1 - f'(s)) \; ds \ge \int_\xi^x \dfrac{1}{2}(1 - L) \; ds = \dfrac{1}{2}(1 - L)(x - \xi), \tag 7$
or
$x - f(x) \ge \dfrac{1}{2}(1 - L)(x - \xi) +(\xi - f(\xi)) \to \infty \; \text{as} \ x \to \infty \tag 8$
since
$\dfrac{1}{2}(1 - L) > 0, \tag 9$
that is, the right-hand side of (8) describes a line of positive slope; we further see from (8) that this line intersects the $x$-axis at that unique $x$ such that
$\dfrac{1}{2}(1 - L)(x - \xi) +(\xi - f(\xi)) = 0, \tag 9$
i.e., where
$x = -\dfrac{2(\xi - f(\xi))}{1 - L} + \xi; \tag{10}$
therefore
$x > f(x) \tag{11}$
for 
$x > -\dfrac{2(\xi - f(\xi))}{1 - L} + \xi. \tag{12}$
Nota Bene:  It is not necessary to assume $f(x)$ actually has a limit as $x \to \infty$; scrutiny of the above argument reveals that the desired result, $f(x) < x$ for sufficiently large $x$, binds under the weaker assumption that
$f'(x) \le L < 1, \; \forall x \; \text{"big enough"}. \tag{13}$
End of Note.
