A Funny Limit Involving Sine Function

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.
Nhãn:

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