lim x → 0 1 − cos ⁡ ( x ) x cos ⁡ ( x ) without L'hopital

Question

lim          x      →      0                  1      −      cos      ⁡      (      x      )              x      cos      ⁡      (      x      )       without L&#039;hopital

Karley Atkinson · Accepted Answer

HINT:                    (1)                                          1            −            cos            ⁡            (            x            )                                x            cos            ⁡            (            x            )                          =                              2                          sin              2                        ⁡            (            x                          /                        2            )                                x            cos            ⁡            (            x            )                              SPOILER ALERT: Scroll over the highlighted area to reveal the solutionUsing (1) we have                              lim                      x            →            0                                                1            −            cos            ⁡            (            x            )                                x            cos            ⁡            (            x            )                                              =                  lim                      x            →            0                                                2                          sin              2                        ⁡            (            x                          /                        2            )                                x            cos            ⁡            (            x            )                                                                          =                  (                      lim                          x              →              0                                                          sin              ⁡              (              x                              /                            2              )                                      x                              /                            2                                )                          (                      lim                          x              →              0                                                          sin              ⁡              (              x                              /                            2              )                                      cos              ⁡              (              x              )                                )                                                                =        (        1        )        (        0        )                                                        =        0

stud4kuj5bwn · Accepted Answer

lim          x      →      0                  1      −      cos      ⁡      x              x      cos      ⁡      x        =      lim          x      →      0                  1      −      cos      ⁡      x              x      cos      ⁡      x        ⋅            1      +      cos      ⁡      x              1      +      cos      ⁡      x        =      lim          x      →      0                  1      −              cos        2            ⁡      x              x      cos      ⁡      x      (      1      +      cos      ⁡      x      )        =      lim          x      →      0                          sin        2            ⁡      x              x      cos      ⁡      x      (      1      +      cos      ⁡      x      )      Note that       lim          x      →      0                  sin      ⁡      x        x    =  1, that       lim          x      →      0        sin  ⁡  x  =  0, that       lim          x      →      0        cos  ⁡  x  =  1, and that       lim          x      →      0        (  1  +  cos  ⁡  x  )  =  2 giving the final result      lim          x      →      0                  1      −      cos      ⁡      x              x      cos      ⁡      x        =  0

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

Answered question

Answer & Explanation

New Questions in Differential Calculus