$\begingroup$I was trying to prove the following proposition:
If $f$ is differentiable and $\displaystyle\lim_{x\to\infty}f'(x)=m\in\mathbb{R}$, then $\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{x}=m$.
I tried with the mean value theorem but I gave up. Can someone help me?
$\endgroup$3 Answers
$\begingroup$Since $f'(x)\to m$, then for all $\epsilon>0$, there exists a number $x_0$ such that $m-\epsilon< f'(x)<m +\epsilon$ whenever $x>x_0$.
From the mean value theorem, we have
$$f(x)=f(x_0)+f'(\xi)(x-x_0)$$
for some $\xi \in (x_0,x)$. But then we can write for $x>x_0$
$$\frac{f(x_0)}{x}+\left(1-\frac{x_0}{x}\right)(m -\epsilon)<\frac{f(x)}{x}<\frac{f(x_0)}{x}+\left(1-\frac{x_0}{x}\right)(m +\epsilon)$$
Letting $x\to \infty$ we see that for all $\epsilon>0$ we have
$$m -\epsilon<\lim_{x\to \infty}\frac{f(x)}{x}<m +\epsilon$$
and hence $\lim_{x\to \infty}\frac{f(x)}{x}=m$ as was to be shown!
$\endgroup$$\begingroup$You could show it using L'Hôpital's rule.
$\endgroup$4$\begingroup$When we have $f'(x)=m+o(1)$ we can integrate it : $\quad f(x)=mx+C+o(x)$
Which is $\displaystyle{\frac{f(x)}{x}=m+\underbrace{\frac{C}{x}}_{\to\; 0}+o(1)}\quad$ or simply $\lim\limits_{x\to\infty}\frac{f(x)}{x}=m$.
In fact Taylor expansions can be integrated because they are nothing more than inequalities
As Dr. MV wrote this is $m-\varepsilon < f'(x) < m+\varepsilon$
Integrating it gives $\displaystyle{\int_{x_0}^{x}(m-\varepsilon)\;dt < \int_{x_0}^{x}f'(t)\;dt < \int_{x_0}^{x}(m+\varepsilon})\;dt$
$\implies(m-\varepsilon)(x-x_0)< f(x)-f(x_0) < (m+\varepsilon)(x-x_0)$
Regrouping $mx+(f(x_0)-mx_0)-(x-x_0)\varepsilon < f(x) < mx+(f(x_0)-mx_0)+(x-x_0)\varepsilon$
And you recognize $C=f(x_0)-mx_0$ and $(x-x_0)\varepsilon=o(x)$
Finally (abusing a little the notations) this is $mx+C-o(x) < f(x) < mx+C+o(x)$ and we arrive at the same conclusion.
