How to find the limit of ( cos x ) 1 x 2 as x approaches 0?

Question

How to find the limit of                     (                  cos          x                )                              1                      x            2                               as x approaches 0?

Gaige Villa · Accepted Answer

[      1      ]                                            lim              x        →        0                            (                  cos          x                )                              1                      x            2                              This is an indeterminate form of the type             1      ∞      . You need to first convert it to the form             0      0       or             ∞      ∞       so you can use L&#039;Hopital&#039;s Rule. We can do this by using e and ln.            [      2      ]                                      =          lim              x        →        0                    e              ln                  [                                    (                              cos                x                            )                                                      1                                  x                  2                                                              ]                                [      3      ]                                      =          lim              x        →        0                    e                        (                      1                          x              2                                )                          ln                      (                          cos              x                        )                                =          lim              x        →        0                    e                                    ln                          (                              cos                x                            )                                            x            2                              We can take out e.            [      4      ]                                      =          e                        lim                      x            →            0                                                ln                          (                              cos                x                            )                                            x            2                              This is now an indeterminate form of the type             0      0      . We can use L&#039;Hopital&#039;s Rule now. Get the derivatives of both the numerator and denominator.            [      5      ]                                      =          e                        lim                      x            →            0                                                -                                          sin                x                                            cos                x                                                          2            x                                =          e                        lim                      x            →            0                                    (          -                                    sin              x                                      2              x                              cos                x                                              )                    This is still indeterminate so you must apply L&#039;Hopital&#039;s Rule again.            [      6      ]                                      =          e                        lim                      x            →            0                                    (          -                                    cos              x                                      2                              (                -                x                                  sin                  x                                +                                  cos                  x                                )                                              )                    You can now get the limit by substitution.            [      6      ]                                      =          e              -                              cos            0                                2                          (              -              0                              sin                0                            +                              cos                0                            )                                                      [      7      ]                                      =          e              -                  1          2                                [      8      ]                                      =                            1                                    e

How do you find the limit of (cos x)^(1/(x^2)) as x approaches 0?

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