I have to evaluate lim n → ∞ ( − 1 ) n − 1 sin ⁡ ( π n 2 + 0.5 n + 1 ) w h e r e n ϵ N

Question

I have to evaluate      lim          n      →      ∞        (  −  1      )          n      −      1        sin  ⁡  (  π            n      2        +    0.5    n    +    1    )      w  h  e  r  e      n    ϵ    N

Maggie Bowman · Accepted Answer

The problem is that as   n  →  ∞,             n      2        +    0.5    n    +    1   diverges, and             n      2        +    0.5    n    +    1    −  n  ≠  0. We have                              lim                      n            →            ∞                                                n            2                    +          0.5          n          +          1                −        n                            =                  lim                      n            →            ∞                                    (          n          +          0.25                      )            2                    +          1          −                      0.25            2                          −        n                                          =                  lim                      n            →            ∞                                                (                                          n                2                            +              0.5              n              +              1                        −            n            )            (                                          n                2                            +              0.5              n              +              1                        +            n            )                                                              n                2                            +              0.5              n              +              1                        +            n                                                            =                  lim                      n            →            ∞                                                0.5            n            +            1                                                              n                2                            +              0.5              n              +              1                        +            n                          =                  1          4

Esmeralda Lane · Accepted Answer

Before   1  +      0.5    n    +      1          n      2       can &quot;become&quot; 1, it is greater than 1. And so when it multiplies by       n    2  , you get something greater than       n    2  . And it turns out that it is still greater enough than       n    2   to make a difference.Use       n    2    +  0.5  n  +  1  =  (  n  +  0.25      )    2    +      15    16  . Then you have   (  n  +  0.25  )      1    +          15              16        (        n        +        0.25                  )          2                    . At this point your intuition works out to be correct, and   1  +      15          16      (      n      +      0.25              )        2             goes to 1 quickly enough. But more should be done to prove this.

I have to evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-

Answered question

Answer & Explanation

New Questions in Calculus 1