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Peyton Velez

Peyton Velez

Answered question

2022-06-14

Calculate lim x 1 ln ( x + 1 ) ln ( x ) x

Answer & Explanation

Carmelo Payne

Carmelo Payne

Beginner2022-06-15Added 25 answers

If we are allowed to use series expansions, then the problem is fairly trivial. We want to evaluate
lim x 1 x log ( 1 + 1 x ) log ( 1 + 1 x )
The series expansion of log ( 1 + 1 x ) is given by
log ( 1 + 1 x ) = 1 x 1 2 x 2 + 1 3 x 3 1 4 x 4 +
Plugging this in, we find that
1 x log ( 1 + 1 x ) log ( 1 + 1 x ) = 1 x ( 1 x 1 2 x 2 + 1 3 x 3 + ) ( 1 x 1 2 x 2 + 1 3 x 3 + ) = ( 1 2 1 3 x + 1 4 x 2 ) ( 1 1 2 x + 1 3 x 2 + ) x 1 2
Thus,
lim x 1 ln ( x + 1 ) ln ( x ) x = 1 2
Yesenia Sherman

Yesenia Sherman

Beginner2022-06-16Added 5 answers

I shall try to evaluate the limit by using L’Hospital Rule twice.
lim x ( 1 ln ( x + 1 ) ln x x ) = lim x 1 x ln ( x + 1 x ) ln ( x + 1 x ) ( 0 0 ) = lim x x ( x x + 1 ) ( 1 x 2 ) ln ( x + 1 x ) x x + 1 ( 1 x 2 ) = lim x 1 x + 1 ln ( x + 1 x ) 1 x 1 x + 1 ( 0 0 ) = lim x 1 ( x + 1 ) 2 x x + 1 ( 1 x 2 ) 1 x 2 + 1 ( x + 1 ) 2
Simplifying the quotient yields
lim x ( 1 ln ( x + 1 ) ln x x ) = lim x 1 ( x + 1 ) 2 1 x ( x + 1 ) 1 x 2 + 1 ( x + 1 ) 2 = lim x x 2 x ( x + 1 ) ( x + 1 ) 2 + x 2 = lim x x 2 x + 1 = 1 2 + 1 x = 1 2

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