Answer
$$\lim_{x\to0}\frac{\sin(\sin x)}{x}=1$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\sin(\sin x)}{x}$$
$$A=\lim_{x\to0}\frac{\sin(\sin x)}{\sin x}\times\lim_{x\to0}\frac{\sin x}{x}$$
$$A=\lim_{x\to0}\frac{\sin(\sin x)}{\sin x}\times1$$
As $x\in(-\pi,\pi)$ and $x\ne0$, we have $\sin x\ne0$. So $\lim_{x\to0}\frac{\sin(\sin x)}{\sin x}=1$
$$A=1\times1=1$$
