## Calculus: Early Transcendentals 8th Edition

$$\lim_{\theta\to\pi}\frac{1+\cos\theta}{1-\cos\theta}=0$$
$$A=\lim_{\theta\to\pi}\frac{1+\cos\theta}{1-\cos\theta}$$ No, you do not need L'Hospital's Rule here. Not because there is an elementary method here, but we can replace $\pi$ into $\theta$ right away. In other words, this is no indeterminate form of any kind. $$A=\frac{1+\cos(\pi)}{1-\cos(\pi)}$$ $$A=\frac{1+(-1)}{1-(-1)}$$ $$A=\frac{0}{2}$$ $$A=0$$