Consider lim x → 0 sin ⁡ ( tan ⁡ x ) sin ⁡ x . The answer is 1. This is clear intuitively since tan ⁡ x ≈ x for small x. How do you show this rigorously?

Question

Consider       lim          x      →      0                  sin      ⁡      (      tan      ⁡      x      )              sin      ⁡      x      . The answer is 1. This is clear intuitively since   tan  ⁡  x  ≈  x for small x. How do you show this rigorously?

Lavizzariym · Accepted Answer

sin      ⁡      (      tan      ⁡      x      )              sin      ⁡      x        =            sin      ⁡      (      tan      ⁡      x      )              tan      ⁡      x                  tan      ⁡      x              sin      ⁡      x        =            sin      ⁡      (      tan      ⁡      x      )              tan      ⁡      x            1          cos      ⁡      x        .As   x  →  0  ,  tan  ⁡  x  →  0  ,, hence the first fraction on the right   →  1. We also know   cos  ⁡  x  →  1  , so the second fraction on the right   →  1. The limit is therefore   1  ⋅  1  =  1

Monserrat Sawyer · Accepted Answer

lim          x      →      0                          sin        ⁡        (        tan        ⁡        (        x        )        )                    sin        ⁡        (        x        )              =      lim          x      →      0                          sin        ⁡        (        tan        ⁡        (        x        )        )                    sin        ⁡        (        x        )                  /                x              ⋅            1      x        =      lim          x      →      0                          sin        ⁡        (        tan        ⁡        (        x        )        )                  /                tan        ⁡        (        x        )                    sin        ⁡        (        x        )                  /                x              ⋅                    tan        ⁡        (        x        )            x        .Now      lim          x      →      0        sin  ⁡  (  tan  ⁡  (  x  )  )      /    tan  ⁡  (  x  )  =      lim          tan      ⁡      (      x      )      →      0        sin  ⁡  (  tan  ⁡  (  x  )  )      /    tan  ⁡  (  x  )  =  1  ,      lim          x      →      0        sin  ⁡  (  x  )      /    x  =  1and you can use this hint:            lim              x        →        0                    (                        tan          ⁡          (          x          )                x            )        =          lim              x        →        0                    (                        tan          ⁡          (          x          )          −          0                          x          −          0                    )        =          lim              x        →        0                    (                        tan          ⁡          (          x          )          −          tan          ⁡          (          0          )                          x          −          0                    )        =    ⋯        lim          x      →      0        tan  ⁡  (  x  )      /    x  =  sec  ⁡  (  0  )  =  1  .

Consider <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-

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