For any integer $n\geq 1$, define $\sin_n=\sin\circ ... \circ \sin$ ($n$ times). Prove that $\lim_{x\to 0}\frac{\sin_nx}{x}=1$ for all $n\geq 1$ I got this problem:
For any integer $n\geq 1$, define $\sin_n=\sin\circ ... \circ \sin$ ($n$ times). Prove that $\lim_{x\to 0}\frac{\sin_nx}{x}=1$ for all $n\geq 1$.
Some hints will be appreciated.
 A: We'll prove it by induction:
We'll show that $\forall n\in\Bbb{Z}^+, \lim_{x\to 0} \frac{\sin_nx}{x}=1$:
If $n=1$ then we get that $\lim_{x\to 0} \frac{\sin_nx}{x}=\lim_{x\to 0} \frac{\sin x}{x}=1$.
Now suppose that for $k\in\Bbb{Z}^+, \lim_{x\to 0} \frac{\sin_kx}{x}=1$. (Induction hypothesis).
And we'll show that $\lim_{x\to 0} \frac{\sin_{k+1}x}{x}=1$:
Since we know that $\lim_{x \to 0}  \sin x =0$ and since by induction hypothesis we know that $\lim_{y\to 0} \frac{\sin_{k}y}{y}=1$, we'll define a function $g:\Bbb{R }\to\Bbb{R}$ by the rule $g(x )=1$ if $x=0$ and $g( x) =\frac {\sin_kx}{x }$ if $ x\neq 0$. Now since $ g$ is continuous in $\Bbb{R}$ we get that $g $ is continuous at $0$ and so $\lim_{x\to 0}g(\sin x)= 1$ But since $\forall x\in\Bbb{R}-\{0\}, g( x) =\frac {\sin_kx}{x }$ we get that $\lim_{x\to 0}\frac {\sin_k(\sin x)}{\sin x }=1$ and so  $\lim_{x\to 0}\frac {\sin_{k+1 } x}{\sin x }=1$. 
Now $$\lim_{x\to 0}\frac {\sin_{k+1 } x}{x }=\lim_{x\to 0}\frac {\sin_{k+1 } x}{\sin x }\frac {\sin x}{ x }=\lim_{x\to 0}\frac {\sin_{k+1 } x}{\sin x } \lim_{x\to 0}\frac {\sin x}{ x }=\\=1 \cdot 1 =1$$ 
as was to be shown.
A: Hint: $\sin(x)\sim x$ then, $\sin(\sin(x))\sim\sin x\sim x$ then $\sin_n(x)\sim\sin_{n-1}(x)\sim...\sim x$. I let you conclude.
A: Hint : multiply and divide by $\sin_{n-1}x$ and then try to simplify then again multiply and divide by $\sin_{n-2} x$ and now I think you can carry out from here
A: Hint: this is true for $n=1$. Note that $\sin_0x=x$ and therefore it gives you $$\lim_{x\to0}\frac{\sin_1 x}{\sin_0 x}=1.$$
This is probably enough to figure out that you should consider
$$\lim_{x\to0}\frac{\sin_{n}x}{\sin_{n-1}x}.$$
A: Okay, here's how I see it (I can never remember which O to use, here I mean the one that goes to zero quicker than x i.e. $lim_{x => 0} \frac{O(x^2)}{x} = 0$
Now, we  have by our friend taylor;
$sin(x) = x + O(x^3)$
So therefore we have;
$sin(x)^n = (x+O(x^2))^n = x^n + O(x^4)$
So $lim_{x=>0} \frac{sin(x)^n}{x} = lim_{x=>0} \frac{x^n + O(x^3)}{x} = lim_{x=>0} x^{n-1} + \frac{O(x^4)}{x} = 0+0$
So you're never going to be able to prove it because the statement is only true for n=1.
Are you sure you copied this question down correctly?  
==edit==
Nevermind, I copied the question down incorrectly;
Try this though;  
$\sin(x) = x+O(x^3)$
$\sin(\sin(x)) = sin(x) + O(\sin(x)^3) = x + O(x^3)$
$\sin(\sin...(\sin(x)) = \sin(\sin..(x)) = x + O(x^3)$
$\lim_{x\to0} \frac{\sin_n(x)}{x} = \lim_{x\to0} \frac{x+O(x^3)}{x} = 1 + 0$
Taylor series are pretty useful when doing this kind of thing, it's a shame that you hardly ever see them being used.
A: Note that $\sin_{n}\left(0\right)=0$ so that $\lim_{x\rightarrow0}\frac{\sin_{n}\left(x\right)}{x}=\lim_{x\rightarrow0}\frac{\sin_{n}\left(x\right)-\sin_{n}\left(0\right)}{x-0}=\sin_{n}'\left(0\right)$.
On base of $\sin_{n}\left(x\right)=\sin_{n-1}\left(\sin x\right)$
we find  $\sin_{n}'\left(x\right)=\sin_{n-1}'\left(\sin x\right)\cos x$ and
consequently $\sin_{n}'\left(0\right)=\sin_{n-1}'\left(0\right)$.
Then with induction it can be shown that $\sin_{n}'\left(0\right)=1$.
