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

Lösung

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

2

Schritte zur Lösung

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

\frac{ln ( x + 1 )}{x} = ln (x + 1) \frac{1}{x}

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

Vereinfache

ln (x + 1) \frac{1}{x} : \frac{ln ( x + 1 )}{x}

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

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 0

ein

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

Vereinfache

= 2

d er i v a t i v e 3 e^{- x^{2}}

d er i v a t i v e y = x^{3} sin^{2} (x)

d er i v a t i v e f (x) = e^{x^{2} - x}

t an g e n t 3 x + 1, at a = 8

\int sin (\frac{1 - x}{3}) d x