Calculate limit involving $\sin$ function

Question

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!

Answer 1

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 $\Phi(150)=25$.

Answer 2

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, $$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}=\lim_{x \rightarrow 0} \frac{x-x+150\frac{x^3}{6}+O(x^5)}{x^3}=\lim_{x \rightarrow 0}(25+O(x^2))=25$$

Answer 3

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 $\lim_{x\to0}\frac{S_k(x)}{x}=\lim_{x\to0} \frac{C_1(x)\dotsm C_k(x)}{1}=1$.

Answer 4

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 $$ \lim_{x\to 0}\frac{x-\sin^{(150)} x}{x^3}\stackrel{\eqref{4}}=\frac{150}{6}=25. $$