Proof: Convexity Implies Existence of Left/Right Derivatives

[blerg]

Homework Statement

Prove that if f(x) is convex on an open interval, then left and right derivatives of f exist at every point.

The Attempt at a Solution

So i have been able to show (from the definition of convex) that for any x1< x < x2 in the interval:
\frac{f(x)-f(x_1)}{x-x_1}\leq \frac{f(x_2)-f(x_1)}{x_2-x_1} \leq\frac{f(x_2)-f(x)}{x_2-x}

in other words, the differential quotient in increasing as x1 increases towards x and decreasing as x2 approaches x.
i'm not sure how to incorporate these facts.

also, should my proof somehow involve that any convex function on an open interval in continuous?

 

[rochfor1]

What do you know about the limits of increasing/decreasing functions bounded above/below?

 

[blerg]

so an increasing/decreasing function that is bounded above/below has a limit.

would i then take lim x1->x- for the left hand limit and lim x2->x+ for right hand limit?

also, for one handed limits, does the fact that it has to be approached from one side gaurentee that the approach is eventually monotone (which would imply the differential quotient is eventually monotone)?