Other approaches to evaluate lim h → 0 4 x + h + 4 x − h − 4 x + 1 2 h 2

Question

Other approaches to evaluate        lim          h      →      0                          4                  x          +          h                    +              4                  x          −          h                    −              4                  x          +                      1            2                                      h      2

Tristian Clayton · Accepted Answer

A technique that requires no L&#039;Hopital, Taylor series or anything more complicated that the definition of limit, knowledge of the natural logarithm (e.g. from       ∫    1    2        d    t      /    t  =  ln  ⁡  2  ) and continuity of   x  ↦      x    2                                4          x                ⋅                  lim                      h            →            0                                                              4              h                        +                          4                              −                h                                      −            2                                h            2                                              =                  4          x                ⋅                  lim                      h            →            0                                                (                          4              h                        −            1                          )              2                                                          h              2                                      4              h                                                                        =                  4          x                ⋅                              (                          lim                              h                →                0                                                                                      4                  h                                −                1                                            h                                  2                  h                                                      )                    2                                                  =                  4          x                ⋅        4        ⋅                              (                          lim                              h                →                0                                                                                      2                  h                                −                                  2                                      −                    h                                                                              2                h                                      )                    2                                                  =                  4                      x            +            1                          ⋅        (        ln        ⁡        2                  )          2

Elisabeth Wiley · Accepted Answer

A possible way is using the Taylor expansion  2  cosh  ⁡  t  =      e    t    +      e          −      t        =  2  +      t    2    +  o  (      t    2    )Hence,                                                        4                              x                +                h                                      +                          4                              x                −                h                                      −                          4                              x                +                                  1                  2                                                                          h            2                                              =                              4          x                                                    e                              h                ln                ⁡                4                                      +                          e                              −                h                ln                ⁡                4                                      −            2                                h            2                                                            =                              4          x                ⋅                              2            +                          h              2                                      ln              2                        ⁡            4            +            o            (                          h              2                        )            −            2                                h            2                                                            =                              4          x                ⋅                  ln          2                ⁡        4        +        o        (        1        )                                                                              ⟶                                      h              →              0                                                            4          x                ⋅                  ln          2                ⁡        4

Other approaches to evaluate lim_(h => 0) (4^(x+h)+4^(x-h)-4^(x+(1)/(2)))/(h^2)

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