So i have this limit: lim x → ∞ x ( x 2 + 4 − x 2 + 2 )

Question

So i have this limit:      lim          x      →      ∞        x  (            x      2        +    4    −            x      2        +    2    )

Samuel Vang · Accepted Answer

We have                              x          2                +        4            +                        x          2                +        2                                      x          2                +        4            +                        x          2                +        2              =  1so  x  (            x      2        +    4    −            x      2        +    2    )    =  x  (            x      2        +    4    −            x      2        +    2    )  ⋅                              x          2                +        4            +                        x          2                +        2                                      x          2                +        4            +                        x          2                +        2                =            x      (      (                        x          2                +        4                    )        2            −      (                        x          2                +        2                    )        2            )                                x          2                +        4            +                        x          2                +        2                =            2      x                                x          2                +        4            +                        x          2                +        2            and therefore  lim  x  (            x      2        +    4    −            x      2        +    2    )  =  lim            2      x                                x          2                +        4            +                        x          2                +        2             if x tends to an arbitrary value. We have  x  (            x      2        +    4    −            x      2        +    2    )    =      x    3    (      1    +          4              x        2              −      1    +          2              x        2              )and therefore  lim  x  (            x      2        +    4    −            x      2        +    2    )    =  lim      x    3    (      1    +          4              x        2              −      1    +          2              x        2              )but we have  x  (            x      2        +    4    −            x      2        +    2    )  ≠  1  (      1    +          4              x        2              −      1    +          2              x        2              )except for x=1, so there is no reason to assume that the limits of the RHS and LHS of this inequality are equal except if   x  →  1A different example is      lim          x      →      ∞                                    x          2                +        4            −                        x          2                +        2              x  here you have  =      lim          x      →      ∞                  x              1        +                  4                      x            2                              −              1        +                  2                      x            2                                x      =      lim          x      →      ∞        (      1    +          4              x        2              −      1    +          2              x        2              )    =      1    +          lim              x        →        ∞                    4              x        2              −      1    +          lim              x        →        ∞                    2              x        2              =  0But you also can use the conjugate trick from the first example:                              x          2                +        4            −                        x          2                +        2              x      =                              x          2                +        4            −                        x          2                +        2              x                                x          2                +        4            +                        x          2                +        2                                      x          2                +        4            +                        x          2                +        2                =            (                        x          2                +        4                    )        2            −      (                        x          2                +        2                    )        2                    x      (                        x          2                +        4            +                        x          2                +        2            )          =      2          x      (                        x          2                +        4            +                        x          2                +        2            )        ∈  [      2          x      ⋅      2      (      x      +      1      )        ,      2          x      ⋅      2      x        ]      →    [  0  ,  0  ]    (      x    →    ∞    )

So i have this limit: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="M

Answered question

Answer & Explanation

New Questions in Calculus 1