limit as x approaches 0 of (1-2cos(x)+cos(2x))/(x^2)

Lösung

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

- 1

Schritte zur Lösung

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

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{2 sin ( x ) - 2 sin ( 2 x )}{2 x})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{2 cos ( x ) - 4 cos ( 2 x )}{2})

Setze den Wert

x = 0

ein

= \frac{2 cos ( 0 ) - 4 cos ( 2 \cdot 0 )}{2}

Vereinfache

\frac{2 cos ( 0 ) - 4 cos ( 2 \cdot 0 )}{2} : - 1

= - 1

d er i v a t i v e y = - \frac{2}{5} x - 3

\frac{d}{d x} (e^{c o s (2 x)} sin (2 x))

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

x \to 0 lim (cos (\frac{9}{x}))

\frac{d}{d x} (cos (x - 2))