Bounds of $\frac{\ln(x+1)}{x}\ \forall x>0$ $f:(0,\infty)$,
$f(x)=\frac{\ln(x+1)}{x}$
Prove that for $\forall\ x&gt;0$ that $f(x) \in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}$ which I think simplifies to $\frac{x^3}{x+1}-x^2(\ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1. 

I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.
For $x \in (0,\infty)$, you have
$$\ln (x+1) = \int_0^x \frac{dt}{1+t}$$

Hence

$$0 \le f(x) = \frac{1}{x}\int_0^x \frac{dt}{1+t} &lt; \frac{1}{x}\int_0^x \ dt =1$$

As all considered maps are continuous.
It is much simpler to show that the derivative of $\ln(x+1)$ is always in $(0,1)$ (and it is continuous). 
Then $\ln(x+1)$ is strictly between $0$ and $x$, think about it.