# Show that if 1>x>0, then x−1>=ln(x)>=1−1/x

Show that if $1>x>0$, then $x-1\ge \mathrm{ln}\left(x\right)\ge 1-\frac{1}{x}$
I know how to prove it using the MVT and I can prove it for $x>1$ but I don't understand how to prove it for $x>0$
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SoroAlcommai9
For $0
$-\mathrm{ln}x={\int }_{x}^{1}\frac{dt}{t}.$
Since
$1⩽\frac{1}{t}⩽\frac{1}{x},$
we have,
${\int }_{x}^{1}dt⩽-\mathrm{ln}x⩽{\int }_{x}^{1}\frac{1}{x}dt,$
and
$1-x⩽-\mathrm{ln}x⩽\frac{1-x}{x}.$
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Ignacio Riggs
Well, all functions are equal at $x=1$. What can you say about the relative size of their derivatives, $1$, $\frac{1}{x}$, and $\frac{1}{{x}^{2}}$ when $0? What can you conclude about the functions?