limit as h approaches 0 of ((sin(2(x+h))-sin(2x)))/h

Lösung

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

2 cos (2 x)

Schritte zur Lösung

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

Wende das L'Hopital Theorem an

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

Setze den Wert

h = 0

ein

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

Vereinfache

= 2 cos (2 x)

x \to 4 - lim (\frac{4 - x}{∣ 4 - x ∣})

x \to 0 lim (e^{\frac{1}{x}} - ln (x))

x \to 0 + lim (x - 2)

x \to 0 + lim (x + 3)

x \to - 4 lim ((2 x^{3} + 5 x)^{2})