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

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

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limit as x approaches 1-of ln(x)*ln(1-x)

x \to 1 - lim (ln (x) \cdot ln (1 - x))

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 1

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

Simplify

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

= 0

x \to 3 lim (x^{2} - x)

z \to 2 lim (\frac{z ^{3} - 8}{z - 2})

x \to \frac{π}{2} + lim (\frac{tan ( x )}{x})

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

x \to 0 lim (\frac{3}{sin ( x )})

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