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

Lösung

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

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{cos ( h + 2 x )}{1})

Setze den Wert

h = 0

ein

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

Vereinfache

= cos (2 x)

x \to 0 + lim (x^{\frac{6}{1 + 2 l n ( x )}})

x \to 0 lim (1 x^{2})

x \to 8 lim (7 - \frac{1}{2} x)

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

x \to - \frac{1}{2} lim (\frac{2 x ^{2} + 7 x + 3}{2 x ^{2} + 3 x + 1})