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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.

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6 Answers 6

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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.

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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.

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  • $\begingroup$ Is $sin(x)\sim x$ means $\lim_{x\to 0}(\sin(x)-x)=0$? $\endgroup$
    – MathNerd
    Sep 25, 2014 at 8:55
  • $\begingroup$ It can be, but it espacially mean that $$\lim_{x\to 0}\frac{\sin x}{x}=1$$ ;-) $\endgroup$
    – idm
    Sep 25, 2014 at 9:01
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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 $$\lim_{x\to0}\frac{\sin_{n}x}{\sin_{n-1}x}.$$

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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

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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.

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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 $\sin_{n}'\left(0\right)=1$.

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  • $\begingroup$ Why apply L'Hôpital's rule? This is directly the definition of the derivative... $\endgroup$ Sep 25, 2014 at 9:54
  • $\begingroup$ @NajibIdrissi Ah yes, you are correct. So it is even more simple. In cases like this my mind is kind of 'preoccupied with l'Hôpital' (no valid excuse, though). I will do some repair. Thank you! $\endgroup$
    – drhab
    Sep 25, 2014 at 9:57

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