Can someone show me how you'd prove:
using Maclaurin's theorem please?
For any $\displaystyle 0\not = x>-1$ we can find $\displaystyle y$ between $\displaystyle 0$ and $\displaystyle x$ so that:
$\displaystyle \log (1+x) - x = \frac{f''(y)}{2!}x^2 < 0$
Since the second derivative is negative.
Therefore, $\displaystyle \log (1+x) < x$ for $\displaystyle -1<x\not = 0$.