Limit lim x → 0 tan 3 ⁡ x − sin 3 ⁡ x x 5 without l'Hôpital's rule.

Question

Limit       lim          x      →      0                          tan        3            ⁡      x      −              sin        3            ⁡      x              x      5       without l&#039;Hôpital&#039;s rule.

EreneDreaceaw · Accepted Answer

HINT:      lim          x      →      0            (                  (                                            tan              ⁡              x                        x                          )            3        ⋅                  1                  x          2                      −                  (                                            sin              ⁡              x                        x                          )            3        ⋅                  1                  x          2                      )   is of the form   ∞  −  ∞See List of indeterminate formsUse            1                        cos          3                ⁡        x              ⋅            (                                    sin            ⁡            x                    x                    )        3    ⋅                    1        −        cos        ⁡        x                    x        2              ⋅  (  1  +  cos  ⁡  x  +      cos    2    ⁡  x  )

Villaretq0 · Accepted Answer

lim          x      →      0                                    tan          3                ⁡        x        −                  sin          3                ⁡        x                    x        5              =      lim          x      →      0                                                              sin              3                        ⁡            x                                              cos              3                        ⁡            x                          −                  sin          3                ⁡        x                    x        5              =      lim          x      →      0                                    sin          3                ⁡        x        (        1        −                  cos          3                ⁡        x        )                              x          5                          cos          3                ⁡        x            with   cos  ⁡  x  ≈  1  −      1    2        x    2   as   x  →  0      lim          x      →      0                                    sin          3                ⁡        x                    x        3              ×                    1        −        (        1        −                  1          2                          x          2                          )          3                                      x          2                ×        1              =      lim          x      →      0            3    2    −      3    4        x    2    +      1    8        x    4    =      3    2

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