How do I evaluate this without using taylor expansion : lim_(x -> oo)x^2 log((x+1)/(x))-x?

Hanna Webster

Hanna Webster

Answered question

2022-11-18

How do I evaluate this without using taylor expansion : lim x x 2 log ( x + 1 x ) x  ?
How do I evaluate this without using Taylor expansion?
lim x x 2 log ( x + 1 x ) x
Note: I used Taylor expansion at z = 0 and I have got 1 2
Thank you for any help

Answer & Explanation

Izabella Henson

Izabella Henson

Beginner2022-11-19Added 20 answers

An approach is to set u := 1 x , then you are looking for
lim u 0 ( log ( 1 + u ) u u 2 )
then you may conclude with L'Hospital's rule
lim u 0 ( log ( 1 + u ) u u 2 ) = lim u 0 ( 1 1 + u 1 2 u ) = lim u 0 ( 1 ( 1 + u ) 2 2 ) = 1 2 .

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