## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to(\pi/2)^+}\frac{\cos x}{1-\sin x}=-\infty$$
$$A=\lim_{x\to(\pi/2)^+}\frac{\cos x}{1-\sin x}$$ In this exercise, the elementary method seems to lead nowhere. Therefore, we will only try applying L'Hospital's Rule. Since $\lim_{x\to(\pi/2)^+}(\cos x)=\cos(\frac{\pi}{2})=0$ and $\lim_{x\to(\pi/2)^+}(1-\sin x)=1-\sin(\frac{\pi}{2})=1-1=0$, this limit is an indeterminate form of $\frac{0}{0}$, so we can apply L'Hospital's Rule. $$A=\lim_{x\to(\pi/2)^+}\frac{\frac{d}{dx}(\cos x)}{\frac{d}{dx}(1-\sin x)}$$ $$A=\lim_{x\to(\pi/2)^+}\frac{(-\sin x)}{0-\cos x}$$ $$A=\lim_{x\to(\pi/2)^+}\frac{\sin x}{\cos x}$$ As $x$ approaches $\frac{\pi}{2}$ from the right side, $\sin x$ approaches $1$, but $\cos x$ approaches $0$. A glance look tells us the result of this limit should be $\infty$. However, as $x$ approaches $\frac{\pi}{2}$ from the right side, which means $x$ is close to the point $\frac{\pi}{2}$, but still $x\gt\frac{\pi}{2}$, we have $$\sin x\gt0\hspace{1cm}and\hspace{1cm}\cos x\lt0$$ Therefore, this limit is $-\infty$ for the numerator is positive but the denominator is negative.