Problem: lim n → ∞ n ⋅ sin ⁡ ( n n + 1 π ) = π

Question

Problem:      lim          n      →      ∞        n  ⋅  sin  ⁡  (      n          n      +      1        π  )  =  π

Mekjulleymg · Accepted Answer

Note that  n  ⋅  sin  ⁡  (      n          n      +      1        π  )  =  n  ⋅  sin  ⁡  (  π  −      1          n      +      1        π  )  =  n  ⋅  sin  ⁡  (      1          n      +      1        π  )  .use replacement   u  =      1          n      +      1      , then   u  →  0 when   n  →  ∞, also   n  =            1      −      u        u  . therefore      lim          n      →      ∞        n  ⋅  sin  ⁡  (      n          n      +      1        π  )  =      lim          u      →      0                  1      −      u        u    ⋅  sin  ⁡  (  π  u  )  =  π      lim          u      →      0        (  1  −  u  )                    sin        ⁡        (        π        u        )                    π        u              =  π      lim          u      →      0        (  1  −  u  )  ⋅      lim          u      →      0                          sin        ⁡        (        π        u        )                    π        u              =  π  .      lim          n      →      ∞        n  ⋅  sin  ⁡  (      n          n      +      1        π  )  =      lim          u      →      0                  1      −      u        u    ⋅  sin  ⁡  (  π  u  )  =  π      lim          u      →      0        (  1  −  u  )                    sin        ⁡        (        π        u        )                    π        u              =  π      lim          u      →      0        (  1  −  u  )  ⋅      lim          u      →      0                          sin        ⁡        (        π        u        )                    π        u              =  π  .

Problem: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-OR

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New Questions in Calculus 1