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polivijuye 2022-06-27 Answered
Evaluate:
lim n 0 ( 1 + x n ) n sin x n d x .
I have tried to use the (Lebesgue) dominated convergence theorem to evaluate the same. At first I noticed that:
| ( 1 + x n ) n sin x n | | ( 1 + x n ) n | 1  for all positive  x  and for all natural  n .
Since g ( x ) = 1 is Lebesgue integrable for each x [ 0 , ) . So by DCT, the given limit equals:
0 lim n ( 1 + x n ) n sin x n d x = 0 e x 0 d x = 0.
This question was asked in our end term exam this semester and so I don't know the correct answer to this question. Therefore, just to check whether I have evaluated the limit correctly I am posting the same here on MSE. If I have gone wrong somewhere, please point out and give some insights. Thanks in advance.
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Answers (1)

Jayce Bates
Answered 2022-06-28 Author has 18 answers
Using
( 1 + z ) n 1 + ( n 2 ) z 2
for z > 0, one has
( 1 + x n ) n n ( n 1 ) 2 x 2 n 2 = 1 + 1 2 ( 1 1 n ) x 2 1 + 1 4 x 2
for n 2. Then define
f n ( x ) = ( 1 + x n ) n sin ( x n )
to estimate f n as
| f n ( x ) | 4 4 + x 2
for x 1 and then one can use LDC.
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