How do you calculate this limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}$? How do you calculate this limit
$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}?$$
without derivatives please.
Thanks.
 A: Note that :


*

*$$\sin(\sin{x}) = \sin{x} - \frac{(\sin{x})^{3}}{3!} + \frac{(\sin{x})^{5}}{5!} + \cdots $$

*$\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1$.
A: Here is a page with a geometric proof that
$$
\lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0}\frac{\tan(x)}{x}=1
$$
You can skip the Corollaries.
Then you can use the fact that $\lim_{x\to 0}\sin(x)=0$ and the fact mentioned by J.J. and Zarrax that
$$
\lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1
$$
A: Write the limit as
$$\lim_{x \to 0} \frac{\sin(\sin x)}{\sin x} \cdot \frac{\sin x}{x}.$$
It is well-known that
$$\lim_{x \to 0} \frac{\sin x}{x} = 1,$$
and since $\sin x \to 0$ as $x \to 0$, we get that also
$$\lim_{x \to 0} \frac{\sin(\sin x)}{\sin x} = 1.$$
Therefore the limit is $1 \cdot 1 = 1$.
A: 
Edit: The solution below should not does not follow the OPs guidelines that derivatives not be used. However, I will leave it since it's correct and shows how L'Hôpital's rule makes the problem much easier. If you think this answer should be deleted, please let me know why and I'll consider it.

Since this limit is of $\frac{0}{0}$ form, we can apply L'Hôpital's rule, which yields $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$
