I'm trying to use L'Hopital's rule to calculate: lim x → 0 + x − sin ⁡ x ( x sin ⁡ x ) ( 3 / 2 )

Question

I&#039;m trying to use L&#039;Hopital&#039;s rule to calculate:      lim          x      →              0        +                                x        −        sin        ⁡        x                    (        x        sin        ⁡        x                  )                      (            3                          /                        2            )

Kuehne2rcun · Accepted Answer

It might be appropriate to use the limits or infinitesimal properties of   sin  ⁡  x at   x  =  0      x    −    sin    ⁡    x    ∼            x      3              3      !      Then            lim              x      →                        0          +                      ⁡            sin      ⁡      x      −      x                                                          (                              x                sin                ⁡                x                            )                                                          3              2                                            =      1    6              lim              x      →                        0          +                      ⁡                              x          3                                                                        (                              x                sin                ⁡                x                            )                                                          3              2                                          and      1    6              lim              x      →                        0          +                      ⁡                              x          3                                                                        (                              x                sin                ⁡                x                            )                                                          3              2                                            =      1    6                      (                                            lim                                      x              →                                                0                  +                                                              ⁡                      x                                          x                sin                ⁡                x                                                    )            3        =      1    6                      (                                            lim                                      x              →                                                0                  +                                                              ⁡                                    x                              sin                ⁡                x                                                    )            3      Using the known limit      x          sin      ⁡      x        →  1one has            lim              x      →                        0          +                      ⁡            sin      ⁡      x      −      x                                                          (                              x                sin                ⁡                x                            )                                                          3              2                                            =      1    6

I'm trying to use L'Hopital's rule to calculate: <munder> <mo movablelimits="true" form="pre

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