Showing that if $\lim_{x\to\infty}f'(x)=L$ then $\lim_{x\to\infty}\frac{f(x)}{x} = L$. Let $f:[0,\infty)\to\mathbb{R}$ differentiable and suppose that $$\lim_{x\to\infty}f'(x)=L.$$ How can I prove that $$\lim_{x\to\infty}\frac{f(x)}{x} = L\;?$$
I have solved some similar problems using the Mean Value Theorem, and I am trying to use it again in this one, but nothing works. For example, I tried to apply the MVT in $[x, 2x]$ but it does not work. Some hint?
 A: Hint: If for $x > N > 0$, $ f'(x) < L+\epsilon$, then for such $x$, 
$f(x) < f(N) + (x-N) (L + \epsilon)$ and
$$\dfrac{f(x)}{x} < L + \epsilon + \dfrac{f(N)- N(L+\epsilon)}{x} $$
What's the limit of the right side as $x \to \infty$?
Similarly in the other direction...
A: Since $\lim_{x\to\infty}f'(x)=L$, for any $\epsilon>0$, there is $M>0$ such that for any $x>M$, there is
$$
|f'(x)-L|<\epsilon
$$
For any $x>M$, by Lagrange mean value theorem
$$
f(x)=f(M)+f'(ξ)(x−M)\quad\text{and}\quad f(x)−Lx=(f′(ξ)-L)(x−M)+f(M)−LM
$$
Where $M<\xi<x.\:$ Thus
\begin{align}
\left|\frac{f(x)}{x} - L\right|&=\left|\frac{f(x) - Lx}{x}\right|
\\
&=\left|\frac{(f'(\xi) - L)(x-M)+f(M)-LM}{x}\right|
\\
&<|f'(\xi) - L|+\left|\frac{f(M)-LM}{x}\right|
\\
&<\epsilon+\left|\frac{f(M)-LM}{x}\right|
\end{align}
And we have
$$
\varlimsup_{x\to\infty}\left|\frac{f(x)}{x} - L\right|\leqslant\varlimsup_{x\to\infty}\epsilon+\varlimsup_{x\to\infty}\left|\frac{f(M)-LM}{x}\right|=\epsilon
$$
Since $\epsilon$ is arbitrary, this means
$$
\varlimsup_{x\to\infty}\left|\frac{f(x)}{x} - L\right|=0\quad\text{and}\quad\varlimsup_{x\to\infty}\left|\frac{f(x)}{x} - L\right|=\varliminf_{x\to\infty}\left|\frac{f(x)}{x} - L\right|
$$
Therefore
$$
\lim_{x\to\infty}\left|\frac{f(x)}{x} - L\right|=0\quad\text{or}\quad\lim_{x\to\infty}\frac{f(x)}{x} = L
$$
A: From the basic definition of convergence, with $\varepsilon$'s:
Fix $\varepsilon > 0$. By definition, there exists $a \geq 0$ such that, for all $ x\geq a$, $L-\varepsilon \leq f^\prime(x) \leq L+\varepsilon$.
For $x\geq a$, write
$$
f(x) - f(a) = \int_a^x f^\prime
$$
which gives
$$
(L-\varepsilon)(x-a) \leq f(x) - f(a) \leq (L+\varepsilon)(x-a)
$$
or, equivalently,
$$
(L-\varepsilon)\left(1-\frac{a}{x}\right) + \frac{f(a)}{x} \leq \frac{f(x)}{x} \leq (L+\varepsilon)\left(1-\frac{a}{x}\right)  + \frac{f(a)}{x}.
$$
Since $\frac{a}{x}\xrightarrow[x\to\infty]{} 0$ and $\frac{f(a)}{x}\xrightarrow[x\to\infty]{} 0$, there exists $b\geq 0$ such that, for $x\geq b$ we have $\lvert \frac{f(a)}{x}\rvert, \lvert \frac{a}{x}\rvert \leq \varepsilon$. It follows that for $x\geq \max(a,b)$,
$$
(L-\varepsilon)\left(1-\varepsilon\right) + \varepsilon \leq \frac{f(x)}{x} \leq (L+\varepsilon)\left(1-\varepsilon\right)  + \varepsilon.
$$
which implies
$$
L-L\varepsilon \leq \frac{f(x)}{x} \leq L+2\varepsilon.
$$
This is easily seen to be equivalent to showing (e.g., by replacing $\varepsilon$ with $\varepsilon^\prime = \min(\frac{\varepsilon}{2}, \frac{\varepsilon}{L})$ in the beginning), since $\varepsilon$ was arbitrary, that $\frac{f(x)}{x}\xrightarrow[x\to\infty]{} L$.
A: Maybe you're searching for a direct computation.
Let us check the case $L\neq0$.
\begin{align*}
L=\lim_{x\to\infty}f'(x)
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\\
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(x)-f(y)}{x-y}\\
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}\\
\end{align*}
Provided the limit interchange, this last one equals to 
$$
\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}
$$
Now $L\neq0$ by hypotesis, then $f(x)\stackrel{x\to\infty}{\to}{\infty}$, thus 
$$
\left[1-\frac{f(y)}{f(x)}\right]\to1
$$
as well as $1-\frac yx\to1$ as $x\to \infty$. Hence 
$$
\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}
=\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)}{x}
=\lim_{x\to\infty}\frac{f(x)}{x}
$$
which concludes the case $L\neq 0$.
$L=0$ is even simpler.
