Calculate the Limit as x approaches 0: lim x → 0 ln ⁡ ( 1 + sin ⁡ x ) sin ⁡ ( 2 x )

Question

Calculate the Limit as x approaches 0:       lim          x      →      0                  ln      ⁡      (      1      +      sin      ⁡      x      )              sin      ⁡      (      2      x      )

laure6237ma · Accepted Answer

Hint:   sin  ⁡  (  2  x  )  =  2  sin  ⁡  x  cos  ⁡  x and                     log        ⁡        (        1        +        sin        ⁡        x        )                    sin        ⁡        x              →  1

Semaj Christian · Accepted Answer

Notice,      lim          x      →      0                  ln      ⁡      (      1      +      sin      ⁡      x      )              sin      ⁡      (      2      x      )        =      lim          x      →      0                  ln      ⁡      (      1      +      sin      ⁡      x      )              2      sin      ⁡      x      cos      ⁡      x        =      1    2        lim          x      →      0                  ln      ⁡      (      1      +      sin      ⁡      x      )              sin      ⁡      x        ⋅      1          cos      ⁡      x      let   sin  ⁡  x  =  t    ⟹    t  →  0 as   x  →  0,  =      1    2        (          lim              t        →        0                            ln        ⁡        (        1        +        t        )            t        )    ⋅      (          lim              x        →        0                    1              cos        ⁡        x              )    =      1    2    (  1  )  ⋅  (  1  )  =            1      2

Calculate the Limit as x approaches 0: <munder> <mo movablelimits="true" form="prefix">lim

Answered question

Answer & Explanation

New Questions in Differential Calculus