Calculate limit involving $\sin$ function Calculate the following limit:
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$
I tried applying L'Hospital's rule, but it got too messy.
Thank you in advance!

Use short form of series expansion:
$$
\sin(x)=x-\frac{x^3}{6}+O(x^5)\\
\sin(\sin(x))=\sin(x)-\frac{\sin^3(x)}{6}+O(\sin^5(x))\to\\
\sin(\sin(x))=x-\frac{x^3}{6}+O(x^5)-\frac{(x-\frac{x^3}{6}+O(x^5))^3}{6}+O(x^5)\to\\
\sin(\sin(x))=x-\frac{x^3}{3}+O(x^5)
$$
Important part here is to notice that each additional $\sin(...)$ leads to the expression of the same form, while adding $-\frac{x^3}{6}$.
Then,
$$\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}=x-150\frac{x^3}{6}+O(x^5)$$
Hence,
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}=\lim_{x \rightarrow 0} \frac{x-x+150\frac{x^3}{6}+O(x^5)}{x^3}=\lim_{x \rightarrow 0}(25+O(x^2))=25$$
