Ciara Mcdaniel

Ciara Mcdaniel

Answered

2022-07-02

Find the limit:
lim n ( 3 n 2 + 2 n + 1 3 n 2 5 ) n 2 + 2 2 n + 1

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Answer & Explanation

Charlee Gentry

Charlee Gentry

Expert

2022-07-03Added 19 answers

Reduce to the limit for the exponential:
lim n ( 1 + 2 3 n 1 2 n 1 + 5 3 n ) n 2 n 4 4 n + 2 = lim n ( 1 + 2 3 n ) n / 2 exp ( 1 3 ) lim n ( 1 + 2 3 n 1 2 n 1 + 5 3 n ) n 4 4 n + 2 1 lim n ( 1 + 2 3 n 1 + 2 n 1 + 5 3 n 1 2 3 n ) n / 2
Using
1 + 2 3 n 1 + 2 n 1 + 5 3 n 1 2 3 n = 1 22 ( 3 n + 2 ) ( 3 n + 5 )
we conclude that the last limit also equals to 1.

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babyagelesszj

babyagelesszj

Expert

2022-07-04Added 7 answers

The standard trick for dealing with 1 forms is to take logs; it’s very useful if you don’t see anything slicker. Let
L = lim n ( 3 n 2 + 2 n + 1 3 n 2 5 ) n 2 + 2 2 n + 1 ;
then
ln L = ln lim n ( 3 n 2 + 2 n + 1 3 n 2 5 ) n 2 + 2 2 n + 1 = lim n ln ( 3 n 2 + 2 n + 1 3 n 2 5 ) n 2 + 2 2 n + 1 = lim n ( n 2 + 2 2 n + 1 ) ln ( 3 n 2 + 2 n + 1 3 n 2 5 ) ,
where the second step uses the continuity of the log function. This is an 0 form, which you can easily convert to a 0 0 form:
lim n ln ( 3 n 2 + 2 n + 1 3 n 2 5 ) 2 n + 1 n 2 + 2 .
Once you know ln L, recovering L is trivial; just remember to do it!

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