limit as x approaches 0 of (ln(1+mx))/x

Lösung

x \to 0 lim (\frac{ln ( 1 + m x )}{x})

m

Schritte zur Lösung

x \to 0 lim (\frac{ln ( 1 + m x )}{x})

Wende das L'Hopital Theorem an

= x \to 0 lim (\frac{\frac{m}{1 + x m}}{1})

Setze den Wert

x = 0

ein

= \frac{\frac{m}{1 + 0 \cdot m}}{1}

Vereinfache

= m

m \to 0 lim (\frac{sin ( m )}{m})

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

x \to - 6 + lim (\frac{x}{x + 6})

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

x \to 1 lim (3 x - 2)