limit as x approaches 0 of (ax-aln(x+1))/(xln(x+1))

Lösung

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

\frac{1}{2} a

Schritte zur Lösung

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

Klammere

a x

aus

a x - a ln (x + 1)

aus

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

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= a \cdot x \to 0 lim \frac{x ( 1 - \frac{a l n ( x + 1 )}{a x} )}{x ln ( x + 1 )}

Vereinfache

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

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

Wende das L'Hopital Theorem an

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

Wende das L'Hopital Theorem an

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

Setze den Wert

x = 0

ein

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

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

= \frac{1}{2} a

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