I am prompted to solve the following limit lim x ↦ 0 ( cos ⁡ ( x ) 1 x 2 )I try to approach this problem by doing lim x ↦ 0 ( − 1 + ( 1 + cos ⁡ ( x ) ) cos ⁡ ( x ) cos ⁡ ( x ) x 2 )where the limit of 1 + cos ⁡ ( x ) 1 cos ⁡ ( x ) is e. So I would have to calcualte the limit of cos ⁡ ( x ) x 2 . Is this approach correct?

Question

I am prompted to solve the following limit      lim          x      ↦      0        (  cos  ⁡  (  x      )          1              x        2              )I try to approach this problem by doing      lim          x      ↦      0        (  −  1  +  (  1  +  cos  ⁡  (  x  )      )                  cos        ⁡        (        x        )                    cos        ⁡        (        x        )                  x          2                      )where the limit of   1  +  cos  ⁡  (  x      )          1              cos        ⁡        (        x        )             is e. So I would have to calcualte the limit of             cos      ⁡      (      x      )              x      2      . Is this approach correct?

Amanda Tate · Accepted Answer

lim              x        →        0              cos    ⁡    (    x          )              1                  x          2                      =          lim              x        →        0                    e                        1                      x            2                          ln        ⁡        cos        ⁡        x                        lim              x        →        0                    1              x        2              ln    ⁡    cos    ⁡    x    =          lim              x        →        0                    1              x        2              ln    ⁡          1      −              sin        2            ⁡      x        =          1      2              lim              x        →        0                                      sin                      2                          ⁡        x                    x                  2                                    ln        ⁡        (        1        −                  sin                      2                          ⁡        x        )                              sin                      2                          ⁡        x              =              1      2              lim              x        →        0                            (                              sin            ⁡            x                    x                )            2              lim              y        →        0                            ln        ⁡        (        1        −        y        )            y        =          1      2        ⋅    1    ⋅    (    −    1    )    =    −          1      2            ⇒          lim              x        →        0              cos    ⁡    (    x          )              1                  x          2                      =          e              −                  1          2                      =          1              e

Davis Osborn · Accepted Answer

(  cos  ⁡  x      )          1              /                    x        2              =  (  1  −      sin    2    ⁡  x      )          1              /            2              x        2              =            [              lim                  sin          ⁡          x          →          0                                      (          1          −                      sin            2                    ⁡          x          )                          (          −          1                      /                                sin            2                    ⁡          x          )                    ]              −                        1          2                    (              lim                  x          →          0                    sin      ⁡      x              /            x              )        2            Now use      lim          h      →      0                  (      1      +      h      )              1              /            h        =      lim          n      →      ∞                  (      1      +                        1          n                    )        n    =  eand      lim          u      →      0                          sin        ⁡        u            u        =  1

I am prompted to solve the following limit <munder> <mo movablelimits="true" form="prefix

Answered question

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