## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x \to 0}\frac{\sin 5x}{3x}=\frac{5}{3}$$
*STRATEGY: To deal with these types of exercises that involve finding the limits of a trigonometrical function, try to change the function into one of these 2 forms $\lim\limits_{\theta\to0}\frac{\sin\theta}{\sin}$ or $\lim\limits_{\theta\to0}\frac{\cos\theta-1}{\cos}$, since you already know that the former equals 1 and the latter equals 0. $$\lim\limits_{x \to 0}\frac{\sin 5x}{3x}=\lim\limits_{x \to 0}(\frac{5}{3}\frac{\sin 5x}{5x})=\frac{5}{3}\lim\limits_{x \to 0}\frac{\sin 5x}{5x}$$ Let $\theta$ to be $5x$, then as $x\to0$, $\theta$ also $\to0$ Therefore, $$\lim\limits_{x \to 0}\frac{\sin 5x}{3x}=\frac{5}{3}\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}$$ We already know from equation 2 that $\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}=1$. So, $$\lim\limits_{x \to 0}\frac{\sin 5x}{3x}=\frac{5}{3}\times1=\frac{5}{3}$$