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

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

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

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

0

Solution steps

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

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

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