$$\frac{x-1}{x} \leq \ln(x) \leq x-1 \forall\ x>0$$

Consider \(f(x)=\ln(x)-\frac{x-1}{x}\)

\(f'(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{x-1}{x^2}\)

Now consider the following two cases:

Case A: \(0 < x \leq 1\) and Case B: \(1 < x < \infty\)

Then for Case A: \(x-1 \leq 0\) and hence \(f'(x)\leq 0\) \(\implies\) \(f(x)\) is a non-increasing function in \((0,1]\) and hence \(f(x)\geq f(1) \forall x \in (0,1]\)

So for \(x \in (0,1]\), \(f(x) \geq f(1)\) \(\implies\) \(\ln(x)-\frac{x-1}{x} \geq 0\) \(\implies\) \(\frac{x-1}{x} \leq \ln(x)\)

Now for Case B:

\(1 < x < \infty\) \(\implies\) \(x-1>0\)

Thus, in this region \(f'(x)>0\) and hence \(f(x)\) is a strictly increasing function in \((1,\infty)\)

Thus, for \(1 < x < \infty\), \(f(x)>f(1)\) and hence \(\ln(x)-\frac{x-1}{x} > 0\)

Combine results with that in case A for \(f(x)\) to conclude: \(\frac{x-1}{x} \leq \ln(x)\)

Now define \(f(x)= \ln(x)-x+1\) \(\implies\) \(f'(x) = \frac{1}{x}-1 =\frac{1-x}{x}\) Whole story of Case A and Case B again. But it should be simple to see that for \(0 < x \leq 1\), \(g'(x)\geq 0\) and hence \(g(x) \leq g(1)\) \(\implies\) \(\ln(x)-x+1 \leq 0\) And for \(1 < x < \infty\), \(g'(x)<0\) and hence \(g(x)<g(1)\) \(\implies\) \(\ln(x)-x+1 < 0\)

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Saket Choudhary is a PhD candidate at the University of Southern California

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