Re: MATH1131 help thread
Let f(t) := ln(1+t), for t ≥ 0. Let x > 0 be a fixed real number.
Using MVT, as f is continuous on [0,x] and differentiable on (0,x), we know there is a c in (0,x) such that
f'(c) = [f(x) - f(0)]/(x-0).
But f'(c) = 1/(1+c), and f(0) = ln 1 = 0. So we get
1/(1+c) =...