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

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

x \to 0 + lim (ln (x) \cdot 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

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 9 x

ln (x - 1)

\frac{d y}{d t} = \frac{e ^{y} + 1}{e ^{y}}

x \to 2 - lim (2 x - 3)

y = 1 - cot (2 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