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!
 A: 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,
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}=\lim_{x \rightarrow 0} \frac{x-x+150\frac{x^3}{6}+O(x^5)}{x^3}=\lim_{x \rightarrow 0}(25+O(x^2))=25$$
A: The derivative is not at all messy.
Let's denote $N=150$, $S_k(x)$ the $k$-fold sine and $C_k(x)=\cos S_{k-1}(x)$. Note that we have $S_k(0)=0$ and $C_k(0)=1$.
In this notation, the numerator is $x-S_N(x)$. Its derivative is $(x-S_N(x))'=1-C_1(x)\dotsm C_N(x)$. The derivative of this (i.e., the 2nd derivative) is
$$(x-S_N(x))''=\sum_{k=1}^{N} C_1^2(x)\dotsm C_{k-1}^2(x) S_k(x) C_{k+1}(x)\dotsm C_N(x),$$
because $C_k(x)' = -C_1(x)\dotsm C_{k-1}(x) S_k(x)$ and we simply use the product rule. Therefore, the 2nd derivative of the numerator is a sum of $N$ products, each containing one sine and then some cosines.
The second derivative of the denominator is $(x^3)''=6x$. If we prove that $\lim_{x\to0}\frac{S_k(x)}{x}=1$ for all $k$, we receive (applying l'Hospital twice) the result $N/6=25$, which is correct. But this is true as we have $S_k(x)'=C_1(x)\dotsm C_k(x)$ whence $\lim_{x\to0}\frac{S_k(x)}{x}=\lim_{x\to0} \frac{C_1(x)\dotsm C_k(x)}{1}=1$.
A: Here is a more or less elementary calculation of the limit.
A quite commonly known limit is
$$
\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0 }{x-0}=\sin'(0)=\cos(0)=1.
\tag{1}\label{1}
$$
Also, you may prove with L'Hopital that
$$
\lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{1-\cos x}{3x^2}=\lim_{x\to 0}\frac{\sin x}{6x}=\frac{1}{6}\lim_{x\to 0}\frac{\sin x}{x}\stackrel{\eqref{1}}=\frac{1}{6}\cdot 1=\frac{1}{6}.
\tag{2}\label{2}
$$
Using this, we see
$$
\lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{x^3}\\
=\lim_{x\to 0}\left[\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\cdot\left(\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\right]\\
=\left(\lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\\
=\left(\lim_{x\to 0}\frac{x-\sin x}{x^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\stackrel{\eqref{1}\&\eqref{2}}=\frac{1}{6}.
\tag{3}\label{3}
$$
Where $\sin^{(n)}x:=\overbrace{\sin(\sin(...\sin(x)...))}^{n\text{ times}}$ and $\sin^{(0)}x:=x$. We used the fact that if for two functions $f,g$ we have
$$
\lim_{x\substack{\to\\ \neq}a}g(x)=b\qquad\text{and}\qquad \lim_{x\substack{\to\\ \neq}b}f(x)=l
$$
and $g(x)\neq b$ in a neighborhood of $a$ then
$$
\lim_{x\substack{\to\\ \neq}a}f(g(x))=l.
$$
Finally we conclude
$$
\lim_{x\to 0}\frac{x-\sin^{(n)} x}{x^3}=\lim_{x\to 0}\left[\frac{x-\sin^{(1)}x+\sin^{(1)}x-\sin^{(2)}x+...+\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right]\\
=\left(\lim_{x\to 0}\frac{x-\sin^{(1)}x}{x^3}\right)+\left(\lim_{x\to 0}\frac{\sin^{(1)}x-\sin^{(2)}x}{x^3}\right)+...+\left(\lim_{x\to 0}\frac{\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right)\\
=\overbrace{\frac{1}{6}+...+\frac{1}{6}}^{n\text{ times}}\stackrel{\eqref{3}}=\frac{n}{6}
\tag{4}\label{4}
$$
and therefore
$$
\lim_{x\to 0}\frac{x-\sin^{(150)} x}{x^3}\stackrel{\eqref{4}}=\frac{150}{6}=25.
$$
A: Let us denote
$$\phi(x;n):=\sin^n(x)=\sin(\sin(\dots\sin(x)\dots)$$
and 
$$\Phi(n):=\lim_{x\to0}\frac{x-\phi(x;n)}{x^3}.$$
Then you want to compute $\Phi(150)$. First of all, check by L'Hôpital rule that
$$\lim_{x\to 0}\frac{\sin(x)}{x}=1,\quad\text{and}\quad
\lim_{x\to 0}\frac{x-\sin(x)}{x^3}=\frac16.$$
Note that the first limit implies that
$$\lim_{x\to0}\frac{\sin(\sin(x))}{x}=
\lim_{x\to0}\frac{\sin(\sin(x))}{\sin(x)}\frac{\sin(x)}{x}=
\lim_{x\to0}\frac{\sin(\sin(x))}{\sin(x)}
\lim_{x\to0}\frac{\sin(x)}{x}=1$$
and in general (by an inductive argument), 
$$\lim_{x\to0}\frac{\phi(x;n)}x=1.$$
Now, write
$$\begin{align*}
\Phi(n)&=\lim_{x\to0}\frac{x-\phi(x;n-1)+\phi(x;n-1)-\phi(x;n)}{x^3}\\
&=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\phi(x;n)}{x^3}\\
&=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\left(\frac{\phi(x;n-1)}{x}\right)^3\\
&=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\lim_{x\to0}\left(\frac{\phi(x;n-1)}{x}\right)^3\\
&=\Phi(n-1)+\lim_{x\to0}\frac{\phi(x;n-1)-\sin(\phi(x;n-1))}{\phi(x;n-1)^3}\\
&=\Phi(n-1)+\frac16.
\end{align*}$$
This is, $\Phi$ satisfies the recursive relation
$$\Phi(n)-\Phi(n-1)=\frac16.$$
Telescoping we see that
$$\Phi(N)-\Phi(1)=\sum_{n=2}^N(\Phi(n)-\Phi(n-1))=\sum_{n=2}^N\frac16,$$
or
$$\Phi(N)=\Phi(1)+(N-1)\frac16=\frac16+(N-1)\frac16=\frac{N}6.$$
Now set $N=150$ to obtain $\Phi(150)=25$.
