limit as x approaches 0 of (sin(6x^2))/(1-cos(3x))

Lösung

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

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

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

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{cos ( 6 x ^{2} ) \cdot 12 x}{3 sin ( 3 x )})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{12 ( - 12 x ^{2} sin ( 6 x ^{2} ) + cos ( 6 x ^{2} ) )}{9 cos ( 3 x )})

Setze den Wert

x = 0

ein

= \frac{12 ( - 12 \cdot 0 ^{2} sin ( 6 \cdot 0 ^{2} ) + cos ( 6 \cdot 0 ^{2} ) )}{9 cos ( 3 \cdot 0 )}

Vereinfache

\frac{12 ( - 12 \cdot 0 ^{2} sin ( 6 \cdot 0 ^{2} ) + cos ( 6 \cdot 0 ^{2} ) )}{9 cos ( 3 \cdot 0 )} : \frac{4}{3}

= \frac{4}{3}

\int (1 + x^{2})^{- 2} d x

(x^{6})^{^{'}}

\frac{d}{d x} (9 + x^{2})

\frac{d}{d x} (e^{x^{4} y})

\int x^{3} (5 + x^{4})^{11} d x