How to evaluate lim n → ∞ ( ∫ 0 2 π cos ⁡ ( n x ) x 2 + n 2 d x ) w h e r e ∀ n ∈ N

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How to evaluate       lim          n      →      ∞            (          ∫              0                    2        π                                      cos          ⁡          (          n          x          )                                      x                          2                                +                      n                          2                                            d    x    )    w  h  e  r  e     ∀        n  ∈      N

conveneau71 · Accepted Answer

We have:-      sup          x      ∈      [      0      ,      2      π      ]            |              cos      ⁡      (      n      x      )                      x                  2                    +              n                  2                      −  0      |    ≤      sup          x      ∈      [      0      ,      2      π      ]            1                  x                  2                    +              n                  2                      ≤      1          n              2            As       1          n              2              →  0 we have             cos      ⁡      (      n      x      )                      x                  2                    +              n                  2                     is uniformly convergent to 0.As for each   n  ∈      N  .             cos      ⁡      (      n      x      )                      x                  2                    +              n                  2                     is Riemann integrable in   [  0  ,  2  π  ] as each of them is continuous on this compact interval. So applying the theorem We can interchange the order of limit and integral. Thus we have      lim          n      →      ∞            ∫          0              2      π                          cos        ⁡        (        n        x        )                              x                      2                          +                  n                      2                                d  x  =      ∫          0              2      π            lim          n      →      ∞                  cos      ⁡      (      n      x      )                      x                  2                    +              n                  2                      d  x  =      ∫          0              2      π        0    d  x  =  0

Riya Hansen · Accepted Answer

Since   0  ≤  x  ≤  2  π, For large n we have   x  &amp;lt;&amp;lt;  n so we can approximate for bounded x values:      ∫          0              2      π                          cos        ⁡        (        n        x        )                              x                      2                          +                  n                      2                                d  x  ≈      1          n      2            ∫          0              2      π        cos  ⁡  (  n  x  )  (  1  −            x      2              n      2        )  d  x    →      1          n      2            ∫          0              2      π        cos  ⁡  (  n  x  )  d  x  =      1          n      2        ⋅  0  =  0

How to evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXA

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