Calculating lim x → 0 sec ⁡ ( x ) − 1 x 2 sec ⁡ ( x ) The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since sec ⁡ ( x ) = 1 cos ⁡ ( x ) we have 1 cos ⁡ ( x ) − 1 x 2 1 cos ⁡ ( x ) And that is 1 − cos ⁡ ( x ) cos ⁡ ( x ) x 2 cos ⁡ ( x ) Then ( 1 − cos ⁡ ( x ) ) ⋅ cos ⁡ ( x ) cos ⁡ ( x ) ⋅ x 2 And ( 1 − cos ⁡ ( x ) ) x 2 What was wrong with my procedure?

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Calculating       lim          x      →      0                  sec      ⁡      (      x      )      −      1                      x        2            sec      ⁡      (      x      )      The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since   sec  ⁡  (  x  )  =      1          cos      ⁡      (      x      )       we have                    1                  cos          ⁡          (          x          )                    −      1                      x        2                    1                  cos          ⁡          (          x          )                    And that is                    1        −        cos        ⁡        (        x        )                    cos        ⁡        (        x        )                            x        2                    cos        ⁡        (        x        )            Then            (      1      −      cos      ⁡      (      x      )      )      ⋅      cos      ⁡      (      x      )              cos      ⁡      (      x      )      ⋅              x        2            And            (      1      −      cos      ⁡      (      x      )      )              x      2      What was wrong with my procedure?

aletantas1x · Accepted Answer

1  −  cos  ⁡      x    =  2      sin    2    ⁡            1      2        x   , so            1      −      cos      ⁡              x                    x      2        =      1    2              (                        sin          ⁡                                    1              2                        x                                    x                      /                    2                    )        2    ,and you probably know that      lim          y      →      0                  sin      ⁡              y              y    =  1  ,so the limit is   1      /    2

Sarai Davenport · Accepted Answer

Note that you get an undefined form :      lim          x      →      0                  1      −      cos      ⁡              (        x        )                    x      2        =      0    0  So you can use l&#039;Hopital rule to get:      lim          x      →      0                  1      −      cos      ⁡              (        x        )                    x      2        =      lim          x      →      0                  sin      ⁡              (        x        )                    2      x        =      1    2  Because as you probably know :      lim          x      →      0                  sin      ⁡              (        x        )              x    =            (              d                  d          x                    )              x      =      0            sin    ⁡          (      x      )        =  c  o  s      (    0    )    =  1You can also use series expansion of cosx near 0:  cos  ⁡      (    x    )    =  1  −            x      2        2    +  o  (      x    4    )To get:      lim          x      →      0                  1      −      cos      ⁡              (        x        )                    x      2        =      lim          x      →      0                  1      −      (      1      −                        x          2                2            +      o      (              x        4            )      )      )              x      2          =      lim          x      →      0                          x        2            2              x      2        =      1    2

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