Prove for $x>0$ $$ \frac{x}{1+x}<\ln(1+x)<x $$ I tried writing $\ln(1+x)=\ln(1+x)-\ln(1)$ and using the MVT for the $(1,1+x)$ interval. I eventually could prove the inequality but how do I have to prove even for $(0,1)$
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can it be done without Taylor? – user224677 Mar 18 '15 at 18:08
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I know but we haven't quite studied Taylor series so if you could help with some hint, I just don't know how to prove for (0;1) interval because the rest I have done – user224677 Mar 18 '15 at 18:11
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I will think on it; I don't currently have it solved either :) – graydad Mar 18 '15 at 18:13
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See also: Showing $\frac{x}{1+x}<\log(1+x)<x$ for all $x>0$ using the mean value theorem – Martin Sleziak Aug 31 '19 at 23:16
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By the mean value theorem, given $x > 0$, there exists $c \in (0,x)$ such that $f(x) - f(0) = f'(c)x$, i.e., $\ln(1 + x) = \frac{x}{1 + c}$. Since $0 < c < x$, $\frac{1}{1 + x} < \frac{1}{1 + c} < 1$. Therefore $\frac{x}{1 + x} < \frac{x}{1 + c} < x$, i.e.,
$$\frac{x}{1 + x} < \ln(1 + x) < x.$$
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