How to calculate lim x → 0 ( 1 + tan ⁡ x 1 + sin ⁡ x ) 1 sin ⁡ x

Question

How to calculate      lim          x      →      0                  (                          1        +        tan        ⁡        x                    1        +        sin        ⁡        x                                )                            1                  sin          ⁡          x

Sawyer Day · Accepted Answer

By taking cases   x  ≤  0 and   x  ≥  0 we obtain the limits      |    tan  ⁡  x      |    ≥      |    sin  ⁡  x      |      ⟹              (                        1          +          tan          ⁡          x                          1          +          sin          ⁡          x                    )                      1                  sin          ⁡          x                      ≥  1      {                            sec          ⁡          x          +          tan          ⁡          x          ≥          1          +          tan          ⁡          x                                      cos          ⁡          x          +          tan          ⁡          x          ≤          1          +          tan          ⁡          x                            ⟹              (                        1          +          tan          ⁡          x                          1          +          sin          ⁡          x                    )                      1                  sin          ⁡          x                      ≤            (      sec      ⁡      x      )                      1                              |                    sin          ⁡          x                      |                              which means with your substitution we have      lim          t      →      0        1  ≤  L  ≤      lim          t      →      0                  (      1      −              t        2            )              −              1                  2                      |                    t                      |                              which means with your substitution we have      lim          t      →      0        1  ≤  L  ≤      lim          t      →      0                  (      1      −              t        2            )              −              1                  2                      |                    t                      |                              The limit on the left is 1 and the limit on the right is      lim          t      →      0                  (      1      −              t        2            )              −                                    |                    t                      |                                    2                      t            2                                =      lim          t      →      0                  (                        (          1          −                      t            2                    )                                      1                          t              2                                          )              −                                    |                    t                      |                          2              =            (              e                  −          1                    )        0    =  1Thus the limit we want is 1 by squeeze theorem.

cazinskup3 · Accepted Answer

We can use Taylor - McLaurin expansion for   sin  ⁡  (  x  ) and   tan  ⁡  (  x  )  tan  ⁡  (  x  )  =  x  +            x      3              3      !        +      o    (      x    3    )And:  sin  ⁡  (  x  )  =  x  −            x      3              3      !        +  o  (      x    3    )So, we have:            1      +      tan      ⁡      (      x      )              1      +      sin      ⁡      (      x      )        −  1  +  1  =            tan      ⁡      (      x      )      −      sin      ⁡      (      x      )              1      +      sin      ⁡      (      x      )        +  1  =                    1        3                    x        3                    1      +      sin      ⁡      (      x      )      We know also that:  ln  ⁡  (  1  +  x  )  ∼  xwith   x  →  0, so:      lim          x      →      0        ln  ⁡      (                            1          3                          x          3                            1        +        sin        ⁡        (        x        )              )    ∼            x      3              3      ⋅      (      1      +      sin      ⁡      (      x      )      )      We have almost done because:      lim          x      →      0                  x      3              3      ⋅      (      1      +      sin      ⁡      (      x      )      )        ⋅      1          sin      ⁡      (      x      )        =      lim          x      →      0                  x      3              3      ⋅      (      1      +      x      −              1                  3          !                            x        3            +      o      (              x        3            )      )        ⋅      1          x      −                        x          3                          3          !                    +      o      (              x        3            )        ∼      lim          x      →      0                  x      3              3      x        =  0Finally, we can say that:      lim          x      →      0                  (                          1        +        tan        ⁡        x                    1        +        sin        ⁡        x                                )                            1                  sin          ⁡          x                      =      lim          x      →      0            e          ln      ⁡              (                              1            +            tan            ⁡            (            x            )                                1            +            sin            ⁡            (            x            )                          )            ⋅              1                  sin          ⁡          (          x          )                      =      e    0    =  1

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

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