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

Lösung

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

0

Schritte zur Lösung

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

Umformung für L'Hopital

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

Wende das L'Hopital Theorem an

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

Vereinfache

\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})

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 0

ein

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

Vereinfache

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

= 0

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

x \to \frac{1}{2} lim (4 x (x - \frac{1}{4}))

x \to 5 lim (\frac{4 x - 10}{4 x - 20})

m \to 0 lim (\frac{5}{m ^{2} + m} - \frac{5}{m})

x \to - \infty lim (e^{6 x})