limit as t approaches 0 of (ln(2t+1))/t

Lösung

t \to 0 lim (\frac{ln ( 2 t + 1 )}{t})

2

Schritte zur Lösung

t \to 0 lim (\frac{ln ( 2 t + 1 )}{t})

Wende das L'Hopital Theorem an

= t \to 0 lim (\frac{\frac{2}{2 t + 1}}{1})

Setze den Wert

t = 0

ein

= \frac{\frac{2}{2 \cdot 0 + 1}}{1}

Vereinfache

= 2

x \to - 5 lim (sin (\frac{π}{3} x))

x \to - 1 lim (2 x + 7)

x \to 7 + lim (\frac{x - 8}{x - 7})

x \to 6 lim (- x^{2} + 6 x)

x \to 3 lim (3 - 7 x - 5 x^{2})