Let x be a negative real number. I am trying to find the following limit of a sequence: <munder>

Kendrick Hampton

Kendrick Hampton

Answered question

2022-06-24

Let x be a negative real number. I am trying to find the following limit of a sequence:
lim n ( 1 1 e x ( e 1 ) n 1 e ) n .

Answer & Explanation

hofyonlines5

hofyonlines5

Beginner2022-06-25Added 12 answers

I'll call the expression
f ( n ) = ( 1 1 e x ( e 1 ) / n 1 e ) n = ( e e x ( e 1 ) / n e 1 ) n
For a given negative real x, we can consider this f a function of real values n with n > x ( e 1 )
Since the behavior as n looks like a " 1 " case, we can try taking the logarithm and then applying L'Hopital's rule.
ln f ( n ) = ln ( e e x ( e 1 ) / n e 1 ) n ln f ( n ) = n [ ln ( e e x ( e 1 ) / n ) ln ( e 1 ) ] ln f ( n ) = ln ( e e x ( e 1 ) / n ) ln ( e 1 ) 1 / n lim n ln f ( n ) = lim n 1 e e x ( e 1 ) / n ( e x ( e 1 ) / n ) x ( e 1 ) n 2 1 / n 2 lim n ln f ( n ) = lim n x ( e 1 ) e x ( e 1 ) / n e e x ( e 1 ) / n lim n ln f ( n ) = x ( e 1 ) e 0 e e 0 lim n ln f ( n ) = x ln ( lim n f ( n ) ) = x lim n f ( n ) = e x
Quintin Stafford

Quintin Stafford

Beginner2022-06-26Added 4 answers

Notice that as n + we have
e x ( e 1 ) n 1 x ( e 1 ) n + O ( 1 / n 2 )
hence
lim n + ( 1 x ( e 1 ) ( 1 e ) n ) n = lim n + ( 1 + x n ) n = e x

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