Get the limit from lim x → ∞ ( x ∫ π 2 arctan ⁡ ( x ) sin ⁡ ( t 2 ) d t )

Question

Get the limit from      lim          x      →      ∞            (    x          ∫                        π          2                            arctan        ⁡                  (          x          )                      sin    ⁡          (              t        2            )        d    t        )

hofyonlines5 · Accepted Answer

You have                              lim                      x            →            ∞                          x                  ∫                      π                          /                        2                                arctan            ⁡            x                          sin        ⁡        (                  t          2                )                          d                t                            =                  lim                      x            →            ∞                                                              ∫                              π                                  /                                2                                            arctan                ⁡                x                                      sin            ⁡            (                          t              2                        )                                      d                        t                                1            x                                                            =                  lim                      x            →            ∞                                                              1                              1                +                                  x                  2                                                      sin            ⁡                          (                              arctan                2                            ⁡              x              )                                            −                          1                              x                2                                                                                      =        −                  lim                      x            →            ∞                                                x            2                                1            +                          x              2                                      sin        ⁡                  (                      arctan            2                    ⁡          x          )                                                  =        −        sin        ⁡                  (                                    π              2                        4                    )                .

Get the limit from <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-

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