How can I find lim n → ∞ n + 2 3 − n + 1 3 n + 2 − n + 1 n − 3 6 ?

Question

How can I find      lim          n      →      ∞                                    n          +          2                3            −                        n          +          1                3                            n        +        2            −              n        +        1                        n      −      3        6    ?

popman14ee · Accepted Answer

Your idea of multiplying and dividing by       n    +    2    +      n    +    1   (or by the conjugate) is a good one, and we can do the same thing for the conjugate of the numerator. Since the denominator had square roots, its conjugate came from the difference of squares formula  a  −  b  =  (      a    −      b    )  (      a    +      b    )Similarly, since the numerator has cube roots, its conjugate must come from the difference of cubes formula  a  −  b  =  (      a    3    −      b    3    )  (            a      3        2    +      a    3        b    3    +            b      3        2    )Applying these factors the expression, we obtain      lim          n      →      ∞                                    n          +          2                3            −                        n          +          1                3                            n        +        2            −              n        +        1              ⋅                    n        +        2            +              n        +        1                            n        +        2            +              n        +        1              ⋅                                          n            +            2                    3                2            +                        n          +          2                3                              n          +          1                3            +                                    n            +            1                    3                2                                                  n            +            2                    3                2            +                        n          +          2                3                              n          +          1                3            +                                    n            +            1                    3                2              ⋅            n      −      3        6    =      lim          n      →      ∞                          n        +        2            +              n        +        1                                                  n            +            2                    3                2            +                        n          +          2                3                              n          +          1                3            +                                    n            +            1                    3                2              ⋅            n      −      3        6    =      lim          n      →      ∞                          n        +        2            +              n        +        1                                                  n            +            2                    3                2            +                        n          +          2                3                              n          +          1                3            +                                    n            +            1                    3                2              ⋅            n      −      3        6  Then pull factors of n from their respective roots      lim          n      →      ∞                          1        +                  2          n                    +              1        +                  1          n                                                          1            +                          2              n                                3                2            +                        1          +                      2            n                          3                              1          +                      1            n                          3            +                                    1            +                          1              n                                3                2              ⋅            1      −              3        n              6    ⋅                    n                              1            2                                      n                              1            6                                      n                        2          3                      =            1      +      1              1      +      1      +      1        =      2    3

How can I find <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXA

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New Questions in Calculus 1