Answer to Question #198804 in Real Analysis for Nikhil

Question #198804

Prove that

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



1
Expert's answer
2021-06-16T06:33:48-0400

Solution:-

Let "f(x)=\\log (\\frac1{1-x})" in [0, 1]

Since f(x) satisfies the condition of L.M.V. theorem in [0, 1], there exists "\\theta(0<\\theta<1)" such that

"\\dfrac{f(x)-f(0)}{x-0}=f^{\\prime}(\\theta x)\n\\\\\n\\Rightarrow \\quad \\dfrac{\\log (\\frac1{1-x})}{x}=\\dfrac{}{}\\dfrac{1}{1-\\theta x}"

"\\Rightarrow \\quad \\log (\\frac1{1-x})=\\dfrac{x}{1-\\theta x} \\quad \\ldots (i)"

Now, "\\quad 0<\\theta<1, 0<x<1 \n\\Rightarrow \\theta x<x"

"\\Rightarrow \\quad -\\theta x>-x\n\\\\\\Rightarrow \\quad 1-\\theta x>1-x\n\\\\\n\\Rightarrow \\quad \\frac{1}{1-\\theta x}<\\frac{1}{1-x}\n\\\\\\Rightarrow \\quad \\frac{x}{1-\\theta x}<\\frac{x}{1-x}\n \\quad \\ldots (ii)"

Again "\\quad 0<\\theta<1, 0<x<1"

"\\Rightarrow \\quad \\theta x>0\n\\\\\\Rightarrow \\quad -\\theta x<0\n\\\\ \\Rightarrow \\quad 1-\\theta x<1\n\\\\\n\\Rightarrow \\quad \\frac{1}{1-\\theta x}>1\n\\\\\n\\Rightarrow \\quad \\frac{x}{1-\\theta x}>x \\quad \\ldots (iii)"

From (i), (ii) and (iii), we get, "x< \\log (\\frac1{1-x})<\\dfrac{x}{1-x}, 0<x<1"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS