What is the limit as x approaches 0 of (ax-aln(x+1))/(xln(x+1)) ?

The limit as x approaches 0 of (ax-aln(x+1))/(xln(x+1)) is 1/2 a

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

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

\frac{1}{2} a

Solution steps

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

Factor out

a x

from

a x - a ln (x + 1) : 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 )}

Simplify

\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 )})

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= 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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What is the limit as x approaches 0 of (ax-aln(x+1))/(xln(x+1)) ?
The limit as x approaches 0 of (ax-aln(x+1))/(xln(x+1)) is 1/2 a