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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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&#039;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.

Jayce Bates · Accepted Answer

Using  (  1  +  z      )    n    ≥  1  +                    (                    n      2                      )                  z    2  for   z  &amp;gt;  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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