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

Lösung

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

- \frac{1}{2}

Dezimale

- 0.5

Schritte zur Lösung

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

Vereinfache

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

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

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 0

ein

= \frac{- \frac{1}{( 0 + 1 ) ^{2}}}{\frac{0 + 2}{( 0 + 1 ) ^{2}}}

Vereinfache

\frac{- \frac{1}{( 0 + 1 ) ^{2}}}{\frac{0 + 2}{( 0 + 1 ) ^{2}}} : - \frac{1}{2}

= - \frac{1}{2}

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

x \to 0 lim (- \frac{1}{∣ x ∣})

x \to 1 - lim (ln (ln (x)))

x \to 0 lim ((12 + x)^{x})

x \to - 1 lim (3 x + 2)