To compute <munder> <mo movablelimits="true" form="prefix">lim n </munder> ( 1 +

doodverft05 2022-06-14 Answered
To compute lim n ( 1 + n ) 1 ln n without L'Hospital
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Answers (1)

Aaron Everett
Answered 2022-06-15 Author has 18 answers
I always treat limits of this form by computing the limit of the logarithm, so we want to consider
lim n ln ( 1 + n ) ln n
If we consider instead of the sequence the function limit
lim x ln ( 1 + x ) ln x
we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution x = 1 / t and note that
ln ( 1 + 1 / t ) = ln ( t + 1 ) ln t
so the limit becomes
lim t 0 + ln t ln ( 1 + t ) ln t = lim t 0 + ( 1 ln ( 1 + t ) ln t ) = 1
because the fraction is 0 /
Thus we have proved that
lim n ( 1 + n ) 1 / ln n = e 1 = e
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