differentiable function proofs Im not sure if this appeared somewhere before but
$f$ is a differentiable function and if $\lim_\limits{x\to \infty}f'(x)=l$ how to show that $\lim_\limits{x\to \infty} \frac {f(x)}{x} = l$ ? 
Basically I used MVT and got stuck.. any clues? I think I need to work on this based on the epsilon definition of limits and combine it with the MVT. 
 A: Let $\varepsilon > 0$. Then there exists $x_0$ such that $|f'(t) - l| < \varepsilon/2$ for $t\ge x_0$. Now, for each $x > x_0$ there exists $t_x\in (x_0,x)$ such that $f(x) = f(x_0) + f'(t_x)(x-x_0)$. Hence,
$$
\frac{f(x)}x - l = \frac 1 x\left(f(x_0) - x_0f'(t_x)\right) + (f'(t_x) - l).
$$
Now, since $f'$ is bounded on $[x_0,\infty)$, there exists $x_1\ge x_0$ such that $\frac 1 x\left|f(x_0) - x_0f'(t_x)\right| < \frac{\varepsilon}2$ for $x > x_1$. So, applying the absolute value to the above line yields that $|\frac{f(x)}x - l| < \varepsilon$ for $x > x_1$. But that means $f(x)/x\to l$ as $x\to\infty$.
A: For any $\epsilon > 0$ there exists $x_1 > 0$ such that for $x > x_1$ we have $l - \epsilon <f'(x) < l + \epsilon.$
By the mean value theorem there exists $\xi$ between $x_1$ and $x$ such that
$$l - \epsilon < \frac{f(x)-f(x_1)}{x - x_1} = f'(\xi) < l + \epsilon.$$
Hence
$$l - \epsilon < \frac{\frac{f(x)}{x}-\frac{f(x_1)}{x}}{1 - \frac{x_1}{x}} < l + \epsilon.$$
Since $f(x_1)/x \to 0$ and $x_1/x \to 0$ as $x \to \infty$ we have
$$l - \epsilon < \liminf_{x \to \infty} \frac{f(x)}{x} \leqslant  \limsup_{x \to \infty} \frac{f(x)}{x} < l + \epsilon.$$
Since $\epsilon > 0$ can be arbitrarily small it follows that
$$\lim_{x \to \infty} \frac{f(x)}{x} = l.$$
