## Calculus 8th Edition

$-\infty$
Given: $\lim_{x\to(\pi/2)^+}\frac{\cos x}{1-\sin x}$ Since $\lim\limits_{x\to(\pi/2)^+}(\cos x)=\cos(\dfrac{\pi}{2})=0$ and $\lim\limits_{x\to(\pi/2)^+}(1-\sin x)=1-\sin(\dfrac{\pi}{2})=1-1=0$, This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{x\to(\pi/2)^+}\dfrac{\frac{d}{dx}(\cos x)}{\dfrac{d}{dx}(1-\sin x)}=\lim\limits_{x\to(\pi/2)^+}\dfrac{(-\sin x)}{0-\cos x}\\=\lim\limits_{x\to(\pi/2)^+}\dfrac{\sin x}{\cos x}$ Here, $x \to \dfrac{\pi}{2}$ from the right side, $\sin x \to 1$, but $\cos x \to 0$. This tells us that the limit must be $\infty$. But $\sin x\gt0$ and $\cos x\lt0$ Therefore, this limit must be $-\infty$ for the positive numerator but the denominator is negative.