Using the mean value theorem of lagrange I need to prove that for all x > 0: $$ \frac{1}{x+1} < ln(x+1) - ln(x) < \frac{1}{x} $$
I was able to reach the following solution: $$f(x) = ln(x+1)$$ The then derive the function: $$f'(x) = \frac{1}{x+1}$$ apply mvt on (0,x): $$f'(c) = \frac{f(x) - f(0)}{x} = \frac{ln(x+1)}{x}$$ $$o<c<x$$ $$f'(x) = \frac{1}{c+1}$$ $$\frac{1}{x+1} < \frac{ln(x+1)}{x} < 1$$ $$\frac{x}{x+1} < \ln(x+1) < x$$ $$\frac{x}{x+1} - ln(x)< \ln(x+1) - ln(x)< x - ln(x)$$
But I can't find way to continue from here.
