Prove this convex function must be a straight line 
Let $f$ be a convex differentiable function on $[x_1,\infty).$ Assume $x_2 > x_1.$ If
$$\lim_{x\to \infty} f'(x) = \frac{f(x_2)- f(x_1)}{x_2-x_1},$$
then on $[x_1,\infty),$ $f$ is the line through $(x_1,f(x_1))$ and $(x_2,f(x_2)).$
It is visible from the graph but how to write a formal proof. 
 A: Let $l(x)$ be the line through the two given points; let $m$ be the slope of $l.$
Claim 1: $f(x) \le l(x)$ for all $x\ge x_1.$ If this were not true, then $f(x) > l(x)$ for some $x>x_1.$ The line through $(x_1,f(x_1)),(x,f(x))$ then has slope $m'>m.$ By the MVT, there exists $y\in (x_1,x)$ such that $f'(y) = m'.$ But the convexity of $f$ implies $f'$ is increasing. Therefore $f'>m'$ on $[y,\infty).$ This implies $f'$ does not have limit $m$ at $\infty,$ contradiction, and Claim 1 is proved.
Claim 2: $f(x) \ge l(x)$ for all $x\ge x_1.$ Let's prove this first for $x\in [x_1,x_2].$ Suppose there is $x\in (x_1,x_2)$ such that $f(x) < l(x).$ Then the line through $(x,f(x)),(x_2,f(x_2))$ has slope greater than $m.$  Just as in Claim 1, this leads to a contradiction.
If $x > x_2$ and $f(x)<l(x),$ then the point $(x_2,f(x_2))$ lies above the chord between $(x_1,f(x_1)), (x,f(x)).$ This contradicts the convexity of $f,$ hence there can be no such $x.$ Thus Claim 2 holds.
The two claims imply the desired result.
A: I think the claim might not be true. What you can show is the following: By
the mean value theorem $ \exists \xi \in ( x_{1} , x_{2} ) $ such that $
f'(\xi) = \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} $. Additionally, we know
that for any convex function, which is differentiable, the derivative is
increasing. This implies that for $ x \ge \xi $, we have $ f '(x) = f(\xi)
$. Otherwise, we obtain a contradiction to
  \begin{align*}
      f'(x) 
      & \stackrel{x \to \infty}{\to} 
            \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} .
    \end{align*}
    This implies that for $ x \ge \xi $, $ f $ coincides with the line through $
(x_{1} , f(x_{1})) $ and $ ( x_{2} , f(x_{2}) ) $. In particular, $ f''(x) =
0 $ for any such $ x $. However, for values $ x < \xi $ the function does
    not have to coincide with this line. A counterexample might look somewhat
    like this:

