Answer to Question #177916 in Real Analysis for Nikhil Singh

Question #177916

Prove that

x< log(1/1-x)< x/1-x ; 0<x<1


1
Expert's answer
2021-05-02T07:45:54-0400

"\\bigstar" To show

x< log(1/1-x)< x/1-x ; 0<x<1


We can prove this by using


"\\bigstar" Mean Value Theorem


for all x>0

"\\bull" using the mean value theorem


"\\boxed{x< log({1\\over1-x})}"

Here x can't equal to 1


Let

"\\boxed{f(x)=x - log({1\\over1-x})}" ..............1


Since

f(0) = 0


"\\boxed{f'(x) = 1+{1\\over(1-x})}" .............2


for all x>0

 f(x)<0




Now



"\\boxed{log({1\\over1-x})<{ x\\over1-x }}" ............3


let

"\\boxed{f(x) = log({1\\over1-x}) - {x\\over1-x }}" ............4


"\\bigstar" according to the mean value theorem, an x0

between a and x

with

"\\boxed{f'(x_o ) = { f(x)-f(a)\\over x-a}}" ...........5

"\\bull" Using the property

, we can prove the inequality


By using (1) ,(2) ,(3), (4) ,(5)


We got

"\\bigstar"

"\\boxed{\nx< log({1\\over1-x})< {x\\over1-x }; 0<x<1}"





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