What is the limit as x approaches 0+of ln(x)ln(1-x) ?

The limit as x approaches 0+of ln(x)ln(1-x) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0+of ln(x)ln(1-x)

x \to 0 + lim (ln (x) ln (1 - x))

0

Solution steps

x \to 0 + lim (ln (x) ln (1 - x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{l n ( 1 - x )}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{\frac{1}{x}}{\frac{1}{l n ^{2} ( 1 - x ) ( 1 - x )}})

= x \to 0 + lim (\frac{\frac{1}{x}}{\frac{1}{l n ^{2} ( 1 - x ) ( 1 - x )}})

Simplify

\frac{\frac{1}{x}}{\frac{1}{l n ^{2} ( 1 - x ) ( 1 - x )}} : \frac{ln ^{2} ( 1 - x ) ( 1 - x )}{x}

= x \to 0 + lim (\frac{ln ^{2} ( 1 - x ) ( 1 - x )}{x})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{- 2 ln ( 1 - x ) - ln ^{2} ( 1 - x )}{1})

= x \to 0 + lim (\frac{- 2 ln ( 1 - x ) - ln ^{2} ( 1 - x )}{1})

Plug in the value

x = 0

= \frac{- 2 ln ( 1 - 0 ) - ln ^{2} ( 1 - 0 )}{1}

Simplify

\frac{- 2 ln ( 1 - 0 ) - ln ^{2} ( 1 - 0 )}{1} : 0

= 0

x \to 2 lim (\frac{5}{4 x - 8})

x \to 7 lim (13)

x \to - 1 lim (- x + 8)

x \to 0 lim ((1 + x^{2})^{3})

x \to 1 lim (\frac{1}{x} + ln (x))

What is the limit as x approaches 0+of ln(x)ln(1-x) ?
The limit as x approaches 0+of ln(x)ln(1-x) is 0