Ngô Quốc Anh

January 8, 2011

A funny limit involving sine function

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:32

Today, I have been asked to calculate the following limit

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))

for each fixed x \in [0,2\pi]. From the mathematical point of view, we can assume x \in (-\frac{\pi}{2}, \frac{\pi}{2}) as we just replace x by \sin (\sin x)) if necessary.

There are three possible cases

Case 1. x \in (0, \frac{\pi}{2}). In this case, it is well known that function \frac{\sin x}{x} is monotone decreasing since

\displaystyle {\left( {\frac{{\sin x}}{x}} \right)^\prime } = \frac{{x\cos x - \sin x}}{{{x^2}}} = \frac{{\cos x}}{{{x^2}}}\left( {x - \tan x} \right) \leqslant 0

in its domain. Consequently, it holds

\displaystyle 0 \leqslant \frac{{\sin (\sin \overbrace {(...(}^n\sin x)...))}}{{\sin \underbrace {(...(}_n\sin x)...)}} \leqslant \frac{{\sin (\sin \overbrace {(...(}^{n - 1}\sin x)...))}}{{\sin \underbrace {(...(}_{n - 1}\sin x)...)}} \leqslant \cdots \leqslant \frac{{\sin (\sin (x))}}{{\sin x}} \leqslant \frac{{\sin x}}{x}.

It turns out that

\displaystyle 0 \leqslant \sin (\sin \overbrace {(...(}^n\sin x)...)) = \frac{{\sin (\sin \overbrace {(...(}^n\sin x)...))}}{{\sin \underbrace {(...(}_n\sin x)...)}}\frac{{\sin (\sin \overbrace {(...(}^{n - 1}\sin x)...))}}{{\sin \underbrace {(...(}_{n - 1}\sin x)...)}} \cdots \frac{{\sin (\sin (x))}}{{\sin x}}\sin x \leqslant {\left( {\frac{{\sin x}}{x}} \right)^{n - 1}}\sin x.

Keep in mind that

\displaystyle {\left( {\frac{{\sin x}}{x}} \right)^{n - 1}}\sin x \to 0

as n \to \infty. Thus

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))=0.

Case 2. x \in (-\frac{\pi}{2}, 0). Due to the fact that \sin x is odd with respect to x, we can replace x by -x to reach to the result

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))=0.

Case 3. x=0. This is trivial.

Taking into account above cases we deduce that

\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))=0

for any x.


  1. Hi! Here’s more simple proof.
    Set x_n := \sin (x_{n-1}), x_1 := \sin x. The claim is that that the limit exists and is zero. To prove it, we may assume that 0 \le x_1\le1. Since 0\le x_1 < \pi/2, (x_n) is a decreasing sequence of nonnegative numbers, hence has a limit. The limit, L, must be zero, which follows from L = \lim \sin x_n = \sin L.

    Comment by Volodja — January 11, 2011 @ 3:23

    • Oh yes, that was the traditional way we all know from the high school. Thanks a lot guy.

      Comment by Ngô Quốc Anh — January 11, 2011 @ 11:46

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