Prove $\lim_{x\to\infty} f(x)/x = K$ If f: $\mathbb{R} \rightarrow \mathbb{R}$, $f'$ exists (finite or infinite) for all reals, and $\lim_{x\to\infty} f'(x) = K$. 
Prove $\lim_{x\to\infty} \frac{f(x)}{x} = K$
I started trying to prove it for $K = 0$.
By definition of infinity limit: $\lim_{x\to\infty} f'(x) = K \rightarrow$ given $\epsilon > 0$, $\exists x_{0} \in  \mathbb{R}$ such that $\forall x > x_{0}$ we have $|f'(x) - 0| < \epsilon$.
By the Mean Value Theorem: for $a,b > x_{0}$, $\exists c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
Since $c > x_{0}$, we have that $|f'(c)| = |\frac{f(b)-f(a)}{b-a}| < \epsilon$.
not sure where to go from here
...
Conclude: $|f(x)/x| < \epsilon \rightarrow \lim_{x\to\infty} f'(x) = 0 = K$.
 A: Let $\epsilon > 0$. Since $\lim\limits_{x\to \infty} f'(x) = K$, there is a positive number $M$ such that $\lvert f'(x) - K\rvert < \frac{\epsilon}{2}$ whenever $x \ge M$. Let $x > M$. By the mean value theorem, there is a point $c\in (M,x)$ such that $f(x) = f(M) + f'(c)(x - M)$. Write $f'(c) = K + u$ where $\lvert u \rvert < \frac{\epsilon}{2}$. Then
$$\frac{f(x)}{x}  - K = \frac{f(M)}{x} - \frac{KM}{x} + u\frac{x-M}{x}.$$
Since $\frac{f(M)}{x} - \frac{KM}{x} \to 0$ as $x\to \infty$, there is a positive number $N$ such that for all $x$, $x > N$ implies $$\left\lvert \frac{f(M)}{x} - \frac{KM}{x}\right\rvert < \frac{\epsilon}{2}.$$ If $x > \max\{M, N\}$, then 
$$\left\lvert \frac{f(x)}{x} - K\right\rvert \le \left\lvert \frac{f(M)}{x} - \frac{KM}{x}\right\rvert + \lvert u\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Since $\epsilon$ was arbitrary, $\frac{f(x)}{x}\to K$ as $x\to \infty$.
A: This is a typical
good-part/bad-part argument.
The good part
is that where
we choose a $c > 0$
and find a
$x(c)$ such that
$|f'(x)-K| < c$
for $x > x(c)$.
The bad part
is where
$x \le x(c)$.
Then
$f(x)
=\int_0^x f'(t) dt
=\int_0^{x(c)} f'(t) dt
+\int_{x(c)}^x f'(t) dt
$
so that
$f(x)-Kx
=\int_0^{x(c)} (f'(t)-K) dt
+\int_{x(c)}^x (f'(t)-K) dt
=I_1(c)+I_2(c, x)
$.
$I_1(c)$ just depends on $c$.
$|I_2(c, x)|
\le (x-x(c))c
$.
Therefore
$|f(x)-Kx|
=|I_1(c)+I_2(c, x)|
\le|I_1(c)|+|I_2(c, x)|
\le |I_1(c)|+|(x-x(c))c|
$
so,
dividing by $x$,
$|\dfrac{f(x)}{x}-K|
\le \dfrac1{x}|I_1(c)|+|(1-x(c)/x)c|
\lt \dfrac1{x}|I_1(c)|+|c|
$.
For any $c > 0$,
we can choose $x$ large enough
so that
$\dfrac1{x}|I_1(c)|
< c$
(by choosing
$x > \dfrac{I_1(c)}{c}$).
Therefore,
for any $c > 0$,
for $x > \dfrac{I_1(c)}{c}$,
$|\dfrac{f(x)}{x}-K|
< 2c$.
Therefore
$\lim_{x \to \infty} \dfrac{f(x)}{x}
= K$.
