Limit of logarithms without l'Hospital This is my first

tibukooinm

tibukooinm

Answered question

2022-03-23

Limit of logarithms without l'Hospital
This is my first post so I hope you forgive any formatting mistakes. This is a task out of a training exam, I may add that we have not yet introduced l'Hospital or derivatives. We have to determine the following limit:
limnln(2nn)ln(n)ln(3n2n)+ln(n)
I am stuck on this for quite some time now, I tried to apply the ln sum rules, but i can not find a way to solve this one. I'm thankful for advice.

Answer & Explanation

Alexzander Evans

Alexzander Evans

Beginner2022-03-24Added 9 answers

limnln(2nn)ln(n)ln(3n2n)+ln(n)=limnnlnn+ln212lnn2nlnn+ln3+12lnn
Note that, in both the numerator and denominator, the nlnn terms dominate the others in this limit. We can ignore the last two terms in each of the numerator and denominator in this limit, and the result is
limnln(2nn)ln(n)ln(3n2n)+ln(n)=limnnlnn2nlnn=12
To be more specific, we can factor the nlnn terms out:
limnnlnn+ln212lnn2nlnn+ln3+12lnn=limn1+ln2nlnn12n2+ln32nlnn+12n
and see the result
pypberissootcu

pypberissootcu

Beginner2022-03-25Added 14 answers

HINT:
ln(2nn)ln(n)ln(3n2n)+ln(n)=ln2nn12ln3n2n+12
=ln2+(n12)lnnln3+(2n+12)lnn
=2ln2lnn+2n12ln3lnn+4n+1

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