## Calculus (3rd Edition)

(a) $L=\lim _{\theta \rightarrow 0} \frac{14 \sin \theta}{\theta}$ (b) $L=14$.
Given $$L= \lim _{x \rightarrow 0} \frac{\sin 14x}{x}.$$ (a) Putting $\theta =14x$, when $x\to 0$ then $\theta \to 0$ . Hence, we have $$L= \lim _{x \rightarrow 0} \frac{\sin 14x}{x}=\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta/14}=\lim _{\theta \rightarrow 0} \frac{14 \sin \theta}{\theta}.$$ (b) $$L= \lim _{\theta \rightarrow 0} \frac{14 \sin \theta}{\theta}=14 \lim _{\theta \rightarrow 0} \frac{ \sin \theta}{\theta}=14.$$ Where we used the fact that $\lim _{\theta \rightarrow 0} \frac{ \sin \theta}{\theta}=1$.