I have a concrete example thou: lim x → 0 + x ( 1 + 4 x 2 ) 1 2

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I have a concrete example thou:      lim          x      →              0        +              x            (      1      +              4                  x          2                    )                      1        2

Kamora Greer · Accepted Answer

Rewrite the problem to      lim          x      →              0        +                                (        1        +                  4                      x            2                          )                              1          2                            (      1              /            x      )      which puts it in the form             ∞      ∞        .   Then, invoke L&#039;Hopital&#039;s Rule.Taking the derivative of the numerator and denominator separately, you have                    (1)                                                        1              2                                                      [                1                +                                  4                                      x                    2                                                  ]                                            −                (                1                                  /                                2                )                                                    [              4              (              −              2              )                              x                                  (                  −                  3                  )                                            ]                                            −                          1                              x                2                                                    .            (1) above can be re-written as :                    (2)                                                        1              2                                                      [                1                +                                  4                                      x                    2                                                  ]                                            −                (                1                                  /                                2                )                                                    [              (              −              8              )                              x                2                            ]                                            (            −            1            )                          x              3                                      .            (2) above can be re-written as :                    (3)                                          4                                          [                1                +                                  4                                      x                    2                                                  ]                                            −                (                1                                  /                                2                )                                              x                .            (3) above can be re-written as :                    (4)                                          4                                          [                                                                            x                      2                                        +                    4                                                        x                    2                                                  ]                                            −                (                1                                  /                                2                )                                              x                .            (4) above can be re-written as :                    (5)                                          4                                          [                                  x                  2                                ]                                            (                1                                  /                                2                )                                                          x            (                          x              2                        +            4                          )                              (                1                                  /                                2                )                                                    .            Noting that the limit is being taken as   x  →      0    +  (5) above can be re-written as :                    (6)                              4                      (                          x              2                        +            4                          )                              (                1                                  /                                2                )                                                    .            Noting that      lim          x      →              0        +                        x      2        +    4    =  2  ,you are left with      lim          x      →              0        +                  4                  x        2            +      4        =      4    2    =  2.

I have a concrete example thou: <munder> <mo movablelimits="true" form="prefix">lim <mro

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