## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to0}\csc x\sin(\sin x)=1$$
$$A=\lim\limits_{x\to0}\csc x\sin(\sin x)$$ $$A=\lim\limits_{x\to0}\frac{\sin(\sin x)}{\sin x}$$ Let $\sin x=\theta$, then as $x\to0$, $\sin x\to0$, so $\theta\to0$ as well. Therefore, $$A=\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1$$