How do i solve this limit: $\lim_{x\to 0}{x-\sin(\sin(...(\sin x)))\over x^{3}}$

Question

I have the next limit : $$\large \lim_{x\to 0}{x-\sin(\sin(\overbrace {\cdot \ \cdot \ \cdot }^n(\sin(x))\overbrace {\cdot \ \cdot \ \cdot }^n))\over x^{3}}$$ $\sin(\sin(...(\sin(x))...))$-is n times.
I have no idea. Someone can help me? Thank you!

Answer 1

In a neighbourhood of the origin we have $$ \sin(x)= x-\frac{x^3}{6}+O(x^5) \tag{1}$$ hence by applying $\sin(\cdot)$ to both terms and exploiting the sine addition formulas and $(1)$ we get $$ \sin\sin(x) = x-\frac{x^3}{3}+O(x^5) \tag{2}$$ as well as $$ \sin\sin\sin(x) = x-\frac{x^3}{2}+O(x^5)\tag{3} $$ and $$ \sin^{[n]}(x) = x-\frac{nx^3}{6}+O(x^5)\tag{4} $$ by induction. It follows that: $$ \lim_{x\to 0}\frac{x-\sin^{[n]}(x)}{x^3} = \color{red}{\frac{n}{6}}\tag{5}$$

Answer 2

$$\lim_{x \rightarrow 0}\frac{1-\cos(x)\cos(\sin(x))......\cos(\sin(\sin..(\sin(x)....))}{3x^2}$$

This tends to $\frac {0}{0}$ Use L'Hospital again

$$\lim_{x \rightarrow 0}\frac{\sin(x)(\cos(\sin(x))......\cos(\sin(\sin..(\sin(x)....)))+\sin(\sin(x)(.......)}{6x}$$

Basically, will give $\sin(x)$ oriented terms to provide you a $\frac{0}{0}$ again. Note: Numerator has $n$ terms

Now, again. when you use L'hospital, it will give you $n$ terms which will be $\cos(x)$ oriented. Thus when $x\rightarrow 0$, these terms will $\rightarrow1$

giving you :

$$\frac{n}{6}$$ as the final answer

Answer 3

Let us rewrite \begin{align} \frac{x-\sin^{[n]}x}{x^3}&=\frac{x-\sin x+\sin x-\sin\sin x+\ldots+\sin^{[n-1]} x-\sin^{[n]} x}{x^3}=\\ &=\frac{x-\sin x}{x^3}+\frac{\sin x-\sin\sin x}{x^3}+\ldots+\frac{\sin^{[n-1]}x-\sin^{[n]}x}{x^3} \end{align} and calculte the limit of each fraction separately.

1. Since $$ \sin t=t-\frac{t^3}{6}+o(t^3) $$ we have the first fraction just as $$ \frac{x-\sin x}{x^3}=\frac{\frac{x^3}{6}-o(x^3)}{x^3}=\frac16+o(x)\to\frac16. $$
2. Similarly with notation $y=\sin^{[k-1]}x$ the $k$-th fraction is $$ \frac{\sin^{[k-1]}x-\sin^{[k]}x}{x^3}=\frac{y-\sin y}{x^3}=\frac{y-\sin y}{y^3}\cdot\frac{y^3}{x^3}=\left(\frac16+o(y)\right)\cdot\left(\frac{y}{x}\right)^3\to\frac16\cdot 1^3=\frac16 $$ because $$ \frac{y}{x}=\frac{\sin^{[k-1]}x}{x}=\frac{\sin^{[k-1]}x}{\sin^{[k-2]}x}\cdot\frac{\sin^{[k-2]}x}{\sin^{[k-3]}x}\cdot\ldots\cdot\frac{\sin x}{x}\to1\cdot 1\cdot\ldots\cdot 1=1. $$
3. Finally the limit is $$ \underbrace{\frac16+\frac16+\ldots+\frac16}_{n\text{ times}}=\frac{n}{6}. $$