# Answer in Calculus for Anuj #204749

Prove that

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

**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\nRightarrow 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 nRightarrow theta x<x”

“Rightarrow quad -theta x>-xn\Rightarrow quad 1-theta x>1-xn\nRightarrow 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>0n\Rightarrow quad -theta x<0n\ Rightarrow quad 1-theta x<1n\nRightarrow quad frac{1}{1-theta x}>1n\nRightarrow 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”