I'm looking for a solution to the following limit from above: <munder> <mo movablelimits="tr

Daphne Haney

Daphne Haney

Answered question

2022-05-14

I'm looking for a solution to the following limit from above:
lim x 1 4 x 3 2 x x 2 1 ln ( x + x 2 1 ) 2 x 2 x 4 1

Answer & Explanation

Elliana Shelton

Elliana Shelton

Beginner2022-05-15Added 14 answers

To make things a little bit easier, first factor x (whose limit is 1) out of the numerator, then use x 4 1 = ( x 1 ) ( x + 1 ) ( x 2 + 1 ) for the denominator:
lim x 1 4 x 3 2 x x 2 1 ln ( x + x 2 1 ) 2 x 2 x 4 1 = lim x 1 4 x 2 2 x 2 1 ln ( x + x 2 1 ) 2 x x 4 1 = 1 4 lim x 1 4 x 2 2 x 2 1 ln ( x + x 2 1 ) 2 x x 1
Denote t = x + x 2 1
Using ln ( 1 + x ) = x x 2 2 + x 3 3 + o ( x 3 ) we can write
ln ( x + x 2 1 ) = ln ( 1 + ( t 1 ) ) = ( t 1 ) ( t 1 ) 2 2 + ( t 1 ) 3 3 + o ( ( t 1 ) 3 ) .
Notice that o ( ( t 1 ) 3 ) = o ( ( x 1 ) 3 / 2 ) so that substituting this back into the limit we get
lim x 1 4 x 2 2 x 2 1 ( x + x 2 1 1 ( x + x 2 1 1 ) 2 2 + ( x + x 2 1 1 ) 3 3 + o ( ( x 1 ) 3 / 2 ) ) 2 x x 1
where I put the factor 1 4 aside for now. We have
lim x 1 4 x 2 2 x 2 1 o ( ( x 1 ) 3 / 2 ) x 1 = 0
and this is why we had to take 3 terms from the Taylor series. So now we can consider just
lim x 1 4 x 2 2 x 2 1 ( x + x 2 1 1 ( x + x 2 1 1 ) 2 2 + ( x + x 2 1 1 ) 3 3 ) 2 x x 1
which can be solved using standard techniques: simplifying we get
2 3 lim x 1 8 x 4 10 x 3 2 x 2 + 5 x 1 + 2 x 2 1 ( 4 x 3 5 x 2 + x + 4 ) x 2 1
which is the same as
2 3 lim x 1 8 x 4 10 x 3 2 x 2 + 5 x 1 + 2 x 2 1 ( 4 x 3 5 x 2 + x + 4 ) x 1
putting 2 3 aside and using L'Hopital's rule we get
lim x 1 2 x 1 ( 32 x 3 30 x 2 4 x + 5 + 2 x 2 1 ( 12 x 2 10 x + 1 ) + 2 x ( 4 x 3 5 x 2 + x + 4 ) x 2 1 )
Bringing back the constants we took out, the original limit equals 1 4 2 3 16 2 = 4 3

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