limit as x approaches 0 of (sin(x)-arctan(x))/(x^2ln(1+x))

Lösung

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

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

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

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

= x \to 0 lim \frac{- cos ( x ) + \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}}{\frac{2 x ^{2} + 6 x + 6}{( x + 1 ) ^{3}}}

Setze den Wert

x = 0

ein

= \frac{- cos ( 0 ) + \frac{2 ( - 3 \cdot 0 ^{2} + 1 )}{( 0 ^{2} + 1 ) ^{3}}}{\frac{2 \cdot 0 ^{2} + 6 \cdot 0 + 6}{( 0 + 1 ) ^{3}}}

Vereinfache

\frac{- cos ( 0 ) + \frac{2 ( - 3 \cdot 0 ^{2} + 1 )}{( 0 ^{2} + 1 ) ^{3}}}{\frac{2 \cdot 0 ^{2} + 6 \cdot 0 + 6}{( 0 + 1 ) ^{3}}} : \frac{1}{6}

= \frac{1}{6}

x \to 5 lim (x + 5)

x \to 1 lim (\frac{3 x + 2}{1 - x})

x \to 0 lim (sec (x) - tan (x))

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

x \to 3 + lim (\frac{x ^{2} + 9}{x - 3})