# 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 $x$.

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 $L = \lim \sin x_n = \sin L$.