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

Lösung

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

0

Schritte zur Lösung

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

Umformung für L'Hopital

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

Wende das L'Hopital Theorem an

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

Vereinfache

\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})

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 1

ein

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

Vereinfache

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

= 0

x \to 4 lim (3 (x - 4)^{x - 4})

x \to 0 lim (2 x csc (x))

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

x \to 0 lim (ln (sin (∣ x ∣)))

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