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in Mathematics by (15 points)
Show that (x-1)/x < ln(x) < x-1 for x>1 ?

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1 Answer

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by (210 points)
edited by
according to the mean value theorem we have:

 \( f(x)=\ln x \; \; \; \; \; f^{'}(c)=\frac{1}{c} \; \; , \; \; [1,x] \)

\( 1<c<x \Rightarrow \frac{1}{x}<\frac{1}{c}<1 \; \; (*) \)

\( f^{'}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow f^{'}(c)=\frac{f(x)-f(1)}{x-1} \Rightarrow f^{'}(c)=\frac{\ln x}{x-1} \)

\( \overset{(*)} \Longrightarrow \frac{1}{x}<\frac{\ln x}{x-1}<1 \overset{x>1} \Longrightarrow \frac{x-1}{x}<\ln x<x-1 \)

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