limit as x approaches 0 of (ln(\frac{1+x)/(1-x))-2x}{x-sin(x)}

Lösung

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

4

Schritte zur Lösung

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

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{\frac{4 x}{( x + 1 ) ^{2} ( - x + 1 ) ^{2}}}{sin ( x )})

Wende Bruchregel an:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= x \to 0 lim (\frac{4 x}{sin ( x ) ( x + 1 ) ^{2} ( - x + 1 ) ^{2}})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{4}{cos ( x ) ( x + 1 ) ^{2} ( - x + 1 ) ^{2} + ( 4 x ^{3} - 4 x ) sin ( x )})

Setze den Wert

x = 0

ein

= \frac{4}{cos ( 0 ) ( 0 + 1 ) ^{2} ( - 0 + 1 ) ^{2} + ( 4 \cdot 0 ^{3} - 4 \cdot 0 ) sin ( 0 )}

Vereinfache

\frac{4}{cos ( 0 ) ( 0 + 1 ) ^{2} ( - 0 + 1 ) ^{2} + ( 4 \cdot 0 ^{3} - 4 \cdot 0 ) sin ( 0 )} : 4

= 4

d er i v a t i v e f (t) = 80 t + t^{2} - \frac{1}{48} t^{3}

\int_{1}^{16} \frac{8}{x} d x

d er i v a t i v e f (x) = e^{- 4 x}

\frac{d}{d x} (\frac{14 x + 8 x ^{2}}{( 7 + 8 x ) ^{2}})

\int 5 e^{x} + 3 sin (x) d x