Prove that $\displaystyle \lim_{x \to 0} \frac{\sin(\sin(x))}{x} = 1$

How can I prove that $\displaystyle \lim_{x \to 0} \frac{\sin(\sin(x))}{x} = 1$

Can I say that

$$\lim_{x \to 0} \sin(\sin (x)) = \lim_{x \to 0} \sin (\lim_{x \to 0} \sin (x))$$?

If that's the case, I can see the solution, because:

$$\lim_{x \to c} \sin(x) = \sin(c)$$

but I can't find a property of limits that says that:

$$\lim_{x \to 0} \sin(x) = \lim_{x \to 0} \sin ( \lim_{x \to 0} x)$$

I'm probably getting this thing all wrong.... I'm sorry if this question is too stupid. I'm learning Calculus by myself....

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You can say that $\lim \sin (\sin x) = \sin (\lim \sin x)$ because $\sin$ is continuous function. – Cortizol May 10 '14 at 21:53
Is it true then that $\lim_{x\to 0}(\sin ( \sin ( \sin \ldots \sin x )))/x= 0$? – Dmoreno May 10 '14 at 22:31
@Dmoreno If there are finitely many sines, yes, of course, since $\sin x \sim x$ for $x\rightarrow 0$. – Jean-Claude Arbaut May 11 '14 at 0:01
Thanks @Jean-ClaudeArbaut! That approach was what I was thinking about. – Dmoreno May 11 '14 at 9:29

$$\lim_{x\rightarrow0}\frac{\sin{\sin x} }{x}=\lim_{x\rightarrow0}\frac{\sin{\sin x} }{\sin x} \frac {\sin x}{x}=\lim_{\ sin x\rightarrow0}\frac{\sin{\sin x} }{\sin x} \lim_{x \rightarrow 0}\frac {\sin x}{x}=1 \cdot1$$

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Sorry for the stupidity: I understand that lim x-> 0 sin(x)/x = 1 I also see that you basically multiplied the lim by sin(X)/(x), but where the (sin (sin(x))/sin(x) comes from? – user3347814 May 10 '14 at 22:04
@user3347814 He's just multiplying the argument of the limit by ${\sin(x) \over \sin(x)}$ and rearranging factors in the same step. – DanielV May 10 '14 at 22:10
I like how this answer generalizes easily to any number of applications of $\sin(\sin(\sin(...)))$ – DanielV May 10 '14 at 22:12
OK... so I have another question.. and sorry gain for bothering you.. but why (sin(sinx))/sin(x) is equal 1? – user3347814 May 10 '14 at 22:13
@user3347814 Consider $y = \sin(x)$, then you are looking at $\sin(y) \over y$, which you know the value of if $y \rightarrow 0$ – DanielV May 10 '14 at 22:14

Since plugging in $x=0$ yields an indeterminant form $\dfrac{0}{0}$, apply L'Hôpital's rule to see

$$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(\sin(x))}{x} \stackrel{\mathrm{L.H.}}{=} \displaystyle\lim_{x \rightarrow 0} \dfrac{\cos(x)\cos(\sin(x))}{1}=\cos(0)\cos(0)=1.$$

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I believe you can also look at the DC term of the taylor expansion:

$$\frac{y - {y^3 \over 3!} + {y^5 \over 5!} ...}{x}\mid_{y = \sin(x)}$$

$$\frac{\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right) - {\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right)^3 \over 3!} + {\left(x - {x^3 \over 3!} + {x^3 \over 5!}+...\right)^5 \over 5!} ...}{x}$$

$$1 + O(x^2)$$

...which is $1$ at $x = 0$.

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By definition, your limit is the derivative of $\sin(\sin(x))$ at $0$. Applying the chain rule, $\frac{d}{dx} \sin(\sin(x)) = \cos(\sin(x))\cos(x)$, so the value of the limit is $\cos(\sin(0))\cos(0) = 1$.

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