Find lim n → ∞ x n , where: x n = n 2 + 1 + 4 n 2 + 2 n − 27 n 3 + n 2 3 .

Question

Find       lim          n      →      ∞            x    n  , where:      x    n    =            n      2        +    1    +      4          n      2        +    2    n    −            27              n        3            +              n        2              3    .

humbast2 · Accepted Answer

First factor out the highest power of n in each term:      x    n    =  n      1    +          1              n        2              +  2  n      1    +          1              2        n              −  3  n            1      +              1                  27          n                      3  This now puts each expression under each root sign into the form   1  +  t where t will be small as   n  →  ∞, so we can use the binomial theorem, and &quot;...&quot; will represent terms in       1          n      2       and higher that will tend to zero as   n  →  ∞      x    n    =      n    {    (    1    +          1              n        2                    )              1                  /                2              +    2    (    1    +          1              2        n                    )              1                  /                2              −    3    (    1    +          1              27        n                    )              1                  /                3              }        x    n    =      n    {    (    1    +          1      2        ⋅          1              n        2              +    .    .    .    )    +    2    (    1    +          1      2        ⋅          1              2        n              +    .    .    .    )    −    3    (    1    +          1      3        ⋅          1              27        n              +    .    .    .    )    }  If you tidy that up you should obtain       x    n    = (constant) + (terms in 1/n and higher).

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