Find the limit: lim n → ∞ ( 3 n 2 + 2 n + 1 3 n 2 − 5 ) n 2 + 2 2 n + 1

Question

Find the limit:      lim          n      →      ∞                          (                              3                          n              2                        +            2            n            +            1                                3                          n              2                        −            5                          )                                                        n              2                        +            2                                2            n            +            1

Charlee Gentry · Accepted Answer

Reduce to the limit for the exponential:      lim          n      →      ∞                  (      1      +              2                  3          n                                      1          −                      2            n                                    1          +                      5                          3              n                                          )                      n        2            −                        n          −          4                          4          n          +          2                      =                                          lim                          n              →              ∞                                                          (              1              +                              2                                  3                  n                                            )                                      n                              /                            2                                      ⏟                    exp      ⁡              (                  1          3                )              ⋅                                          lim                          n              →              ∞                                                          (              1              +                              2                                  3                  n                                                                              1                  −                                      2                    n                                                                    1                  +                                      5                                          3                      n                                                                                  )                                      −                                                n                  −                  4                                                  4                  n                  +                  2                                                                    ⏟                    1        ⋅      lim          n      →      ∞                  (                        1          +                      2                          3              n                                                          1              +                              2                n                                                    1              +                              5                                  3                  n                                                                              1          −                      2                          3              n                                          )              n              /            2      Using            1      +              2                  3          n                                      1          +                      2            n                                    1          +                      5                          3              n                                                  1      −              2                  3          n                      =  1  −      22          (      3      n      +      2      )      (      3      n      +      5      )      we conclude that the last limit also equals to 1.

babyagelesszj · Accepted Answer

The standard trick for dealing with       1    ∞   forms is to take logs; it’s very useful if you don’t see anything slicker. Let  L  =      lim          n      →      ∞                          (                              3                          n              2                        +            2            n            +            1                                3                          n              2                        −            5                          )                                                        n              2                        +            2                                2            n            +            1                                  ;then                    ln        ⁡        L                            =        ln        ⁡                  lim                      n            →            ∞                                                              (                                                3                                      n                    2                                    +                  2                  n                  +                  1                                                  3                                      n                    2                                    −                  5                                            )                                                                                            n                    2                                    +                  2                                                  2                  n                  +                  1                                                                                                      =                  lim                      n            →            ∞                          ln        ⁡                                            (                                                3                                      n                    2                                    +                  2                  n                  +                  1                                                  3                                      n                    2                                    −                  5                                            )                                                                                            n                    2                                    +                  2                                                  2                  n                  +                  1                                                                                                      =                  lim                      n            →            ∞                                    (                                                    n                2                            +              2                                      2              n              +              1                                )                ln        ⁡                  (                                    3                              n                2                            +              2              n              +              1                                      3                              n                2                            −              5                                )                        ,            where the second step uses the continuity of the log function. This is an   ∞  ⋅  0 form, which you can easily convert to a       0    0   form:      lim          n      →      ∞                  ln      ⁡              (                              3                          n              2                        +            2            n            +            1                                3                          n              2                        −            5                          )                            2        n        +        1                              n          2                +        2                .Once you know   ln  ⁡  L, recovering L is trivial; just remember to do it!

Find the limit: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeX

Answered question

Answer & Explanation

New Questions in Calculus 2