I think this is a nice problem. It is a known fact, just nobody tries to fully prove it.
Are you asking us to justify the definition of the derivative?
If so:
Let f(x) be a function that is continuous on some interval I containing x0. If we wish to find the gradient (or slope) of the line connecting f(x0) to some other point f(x1) (which is the slope of the secant line between these points) we may use the formula:
M = [f(x0) - f(x1)]/[x0 - x1]
Let's redefine x1 as some point h units from x0. Therefore, we have x1 = x0 + h, and the slope of the of the secant line becomes:
M = [f(x0) - f(x0 + h)]/[x0 - (x0 + h)] = [f(x0) - f(x0 + h)]/h
Now, we may find the slope of a line tangent to f(x0) if we allow the distance between x0 and x0 + h to go to zero (in other words, we are letting h (the distance between x0 and x1) go to zero). Doing so we obtain a new limit:
L = lim{h->0} [f(x0) - f(x0 + h)]/h
We define this limit to be the slope of the tangent line to f(x) at x0, which we further define to be the derivative of f(x): f'(x) if we let x0 be any arbitrary value of x on the domain of f(x).
f'(x) = L = lim{h->0} (f(x) - f(x + h))/h
Because I like to do these things very very carefully. Some might say too carefully. (In fact, I think we need to know the functions are continous for you to make such a statement . Thus, in general what you said is probably not true . Yes, here it is).
Maybe I will post my solution.