limit as x approaches 0 of (sin^2(6x))/x

Lösung

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{x})

0

Schritte zur Lösung

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{x})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{12 sin ( 6 x ) cos ( 6 x )}{1})

Setze den Wert

x = 0

ein

= \frac{12 sin ( 6 \cdot 0 ) cos ( 6 \cdot 0 )}{1}

Vereinfache

\frac{12 sin ( 6 \cdot 0 ) cos ( 6 \cdot 0 )}{1} : 0

= 0

x \to \infty lim (2 x^{x})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 9, y^{^{'}} (0) = - 6

n = 1 \sum \infty n (1 - p)^{n - 1}

a re a 9 x, x^{2} - 10, 0

\int e^{6 x} x^{3} d x