Show that if $1> x>0$, then $x-1 ≥ \ln(x) ≥ 1−\frac{1}{x}$

Question

Show that if $0 < x < 1$, then $$x-1 ≥ \ln(x) ≥ 1−\frac{1}{x}.$$

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 $x > 0$ .

Answer 1

For $0 < x < 1$,

$$-\ln x = \int_x^1 \frac{dt}{t}.$$

Since

$$ 1 \leqslant \frac1{t} \leqslant \frac1{x},$$

we have,

$$\int_x^1dt \leqslant -\ln x \leqslant \int_x^1\frac{1}{x}dt,$$

and

$$1-x \leqslant -\ln x \leqslant \frac{1-x}{x}.$$

Answer 2

For example, define

$$f(x):=\log x+\frac1x-1\implies f'(x)=\frac1x-\frac1{x^2}<0\stackrel{x>0}\iff x-1<0\iff 0<x<1$$

so $\;f\;$ is monotonic decreasing and thus

$$\forall\,0<x<1\;,\;\;f(x)>f(1)=0$$

Try now the second inequality.

Answer 3

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 $0<x<1$? What can you conclude about the functions?

Answer 4

See that, For any $x>0$ We have $$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$

Thus,

$$\frac{x}{x+1} \le \ln(x+1 ) \le x $$