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

Lösung

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

1

Schritte zur Lösung

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

\frac{ln ( 1 + x )}{x} = ln (1 + x) \frac{1}{x}

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

Vereinfache

ln (1 + x) \frac{1}{x} : \frac{ln ( 1 + x )}{x}

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

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 0

ein

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

Vereinfache

= 1

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

x \to 0 lim ((1 + 8 x)^{\frac{1}{x}})

x \to - 1 - lim (- \frac{1}{x + 1})

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

θ \to π lim (θ sec (θ))