Inequality 0<x<1: Is $\frac{x-1}{x}<ln(x)<x-1$ True?

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is

\frac{x-1}{x}<ln(x)<x-1

valid for 0<x<1

I think it is I just want to get a second opinion.
 
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One way to look at these things is to examine the functions:
<br /> f(x)=\frac{x-1}{x}-\ln x\quad g(x)=\ln x-(x-1)<br />
and compute the derivatives f'(x) and g'(x) and examine if f(x) and g(x) are increasing or decreasing functions and examine the values for certain points and the inequality should drop out.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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