Evaluate lim x → 0 ( 1 ln ⁡ ( 1 + x ) +...

Boilanubjaini8f

Boilanubjaini8f

Answered

2022-06-25

Evaluate
lim x 0 ( 1 ln ( 1 + x ) + 1 ln ( 1 x ) )

Answer & Explanation

Josie123

Josie123

Expert

2022-06-26Added 16 answers

What's wrong with L'Hopital's?
lim x 0 1 ln ( 1 + x ) + 1 ln ( 1 x ) = lim x 0 ln ( 1 x 2 ) ln ( 1 + x ) ln ( 1 x ) = lim x 0 2 x / ( 1 x 2 ) ln ( 1 x ) / ( 1 + x ) ln ( 1 + x ) / ( 1 x ) = lim x 0 2 x ( 1 x ) ln ( 1 x ) ( 1 + x ) ln ( 1 + x ) = lim x 0 2 1 ln ( 1 x ) 1 ln ( 1 + x ) = 2 2 = 1.
skylsn

skylsn

Expert

2022-06-27Added 4 answers

Since log ( 1 + x ) = x x 2 2 + O ( x 3 ), we have log ( 1 x ) = x x 2 2 + O ( x 3 ). Therefore,
1 log ( 1 + x ) + 1 log ( 1 x ) = 1 x ( 1 x 2 + O ( x 2 ) ) + 1 x ( 1 + x 2 + O ( x 2 ) ) ( ) = 1 + x 2 + O ( x 2 ) x 1 x 2 + O ( x 2 ) x = x + O ( x 2 ) x
where ( ) is because 1 1 + x = 1 x + O ( x 2 )
Therefore,
lim x 0 1 log ( 1 + x ) + 1 log ( 1 x ) = 1

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