Ngô Quốc Anh

September 11, 2008

Một bài giới hạn khó về hàm lượng giác

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 22:02

Show that for any irrational \alpha the limit

\displaystyle\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)

does not exist.

Solution. If the limit existed then we would get

\displaystyle 0 = \mathop {\lim }\limits_{n \to \infty } \left( {\sin \left( {\left( {n + 2} \right)\alpha \pi } \right) - \sin \left( {n\alpha \pi } \right)} \right) = 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)

and consequently, \mathop {\lim }\limits_{n \to \infty } \cos \left( {n\alpha \pi } \right) = 0. Similarly,

\displaystyle0 = \mathop {\lim }\limits_{n \to \infty } \left( {\cos \left( {\left( {n + 2} \right)\alpha \pi } \right) - \cos \left( {n\alpha \pi } \right)} \right) = - 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)

which is impossible because \sin^2x+\cos^2x=1 for all x \in \mathbb R. Therefore the limit does not exist.

4 Comments »

  1. Quoc Anh,
    Sua lai dang thuc thu 2 kia. la sin chu k phai sin dau a.
    Dang Tuan Hiep

    Comment by dangtuanhiep — September 13, 2008 @ 0:35

  2. Ờ đúng rồi, cos – cos phải là -2 sin sin chứ🙂. Cảm ơn Hiệp nhá.

    Comment by Ngô Quốc Anh — September 13, 2008 @ 0:36

  3. IC bi am nen ghi nham ca cong thuc luong giac kia .

    Comment by Hue Tran — October 12, 2008 @ 23:00

  4. Cảm ơn H..

    Comment by Ngô Quốc Anh — October 12, 2008 @ 23:08


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