## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to0}\frac{\sin x}{\sin \pi x}=\frac{1}{\pi}$$
*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. $$\lim\limits_{x\to0}\frac{\sin x}{\sin\pi x}$$ $$=\lim\limits_{x\to0}\Bigg(\frac{\sin x}{x}\times\frac{x}{\pi x}\times\frac{\pi x}{\sin\pi x}\Bigg)$$ $$=\lim\limits_{x\to0}\frac{\sin x}{x}\times\lim\limits_{x\to0}\frac{x}{\pi x}\times\lim\limits_{x\to0}\frac{\pi x}{\sin\pi x}$$ $$=1\times\lim\limits_{x\to0}\frac{1}{\pi}\times\frac{1}{\lim\limits_{x\to0}\frac{\sin \pi x}{\pi x}}$$ $$=1\times\frac{1}{\pi}\times\frac{1}{\lim\limits_{x\to0}\frac{\sin\pi x}{\pi x}}$$ $$=\frac{1}{\pi\lim\limits_{x\to0}\frac{\sin \pi x}{\pi x}}$$ Let $\pi x=\theta$. Then as $x\to0$, $\theta$ also $\to0$. So, $$\lim\limits_{x\to0}\frac{\sin x}{\sin \pi x}=\frac{1}{\pi\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta}}$$ $$=\frac{1}{\pi\times1}$$ $$=\frac{1}{\pi}$$