# A challenging (for me) limit calculation

How to calculate the following limit $$\lim_{n\to\infty}\sqrt{n}\underbrace{{}\sin(\sin(\sin(\sin(\cdots\sin(\frac{1}{\sqrt{n}})\cdots))))}_{n \text{ sin's}} \text{?}$$

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• The square root sign is irrelevant and should be removed (in both places). Commented Oct 23, 2023 at 7:07
• Why is the square root sign irrelevant? Commented Oct 23, 2023 at 8:27
• @SamNead: Why? Numerically it seems to go to 0.8660... . But I don't think this is suitable for mathoverflow Commented Oct 23, 2023 at 8:35
• Oh, I was totally wrong about the square root being irrelevant. Perhaps I would have had an easier time with this problem if we had defined $S^{(n)}$ to be sine composed with itself $n$ times, and then asked about the limit of $n \cdot S^{(n^2)}(1/n)$. Sigh. Commented Oct 23, 2023 at 13:08
• see Asymptotic Methods in Analysis by de Bruijn. He does iterated sine in detail, pages 153-166 in the Dover edition. Commented Oct 23, 2023 at 19:31

This limit converges to $$\frac{\sqrt3}2$$. The idea is that $$\sin(x) = x - \frac{x^3}6 + O(x^5)$$, so we start with $$\frac1{\sqrt n}$$ and repeatedly subtract $$\frac{x^3}6$$. We can approximate this discrete system with the differential equation $$f'(x) = -\frac{f(x)^3}6$$, which has a solution $$f(x)=\frac{\sqrt3}{\sqrt{c + x}}$$, and we want $$f(0) = \frac1{\sqrt n}$$, so $$c = 3n$$, and then $$f(n) = \frac{\sqrt3}{2\sqrt n}$$.
For a formal proof, we can define two sequences $$a_0 = \frac1{\sqrt n}, a_{i+1} = \sin(a_i)$$ and $$b_i = \frac{\sqrt3}{\sqrt{3n + i}}$$. We have $$b_{i+1} = b_i \frac{\sqrt{3}}{\sqrt{3 + b_i^2}}$$. Let's bound $$c_i = |a_i - b_i|$$. By looking at the Taylor series, we have $$|\sin(b_{i-1}) - b_i| = O(b_i^5)$$. Additionally, $$|a_i - \sin(b_{i-1})| = |\sin(a_{i-1}) - \sin(b_{i-1})| \leq |a_{i-1} - b_{i-1}|$$, so $$c_i \leq c_{i-1} + O(b_i^5) \leq c_{i-1} + O(n^{-\frac52})$$, or $$c_i \leq O(n^{-\frac32})$$, so the approximation given by $$b_n$$ for $$a_n$$ is good enough.
• Fun fact: replace $\sin$ with $\tan^{-1}$ and the corresponding limit converges to $\sqrt{3/5}$. Since $\tan^{-1} x = x - x^3/3 + O(x^5)$ you have the approximation $f^\prime(x) = -f(x)^3/3$ and the rest of the proof carries through. Commented Oct 23, 2023 at 17:28
• @WillJagy In your formula, what is $g(x)$? Commented Oct 23, 2023 at 21:49
• @TimothyChow turns out it was a constant plus the solution of Abel's equation, $\psi (\sin x) - \psi(x) = 1.$ I had thought, incorrectly, that something bounded was there, but $\psi(x) = \frac{3}{x^2} + \frac{6}{5} \log x + bounded.$ Anyway, what I once knew is in mathoverflow.net/questions/45608/… Commented Oct 23, 2023 at 22:12
• $$\alpha(z) = \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + O(z^6)$$ Commented Oct 23, 2023 at 22:21