siutulysr5

2022-01-21

Prove by Mean Value Theorem

Expert

By the mean value theorem, given $x>0$, there exists $c\in \left(0,x\right)$ such that . Since $0,
$\frac{1}{1+x}<\frac{1}{1+c}<1$. Therefore $\frac{x}{1+x}<\frac{x}{1+c}, i.e.,
$\frac{x}{1+x}<\mathrm{ln}\left(1+x\right)

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