## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to(\pi/2)^-}\cos x\sec 5x=\frac{1}{5}$$
$$A=\lim_{x\to(\pi/2)^-}\cos x\sec 5x$$ From trigonometric identities, $\sec 5x=\frac{1}{\cos 5x}$. That means, $$A=\lim_{x\to(\pi/2)^-}\frac{\cos x}{\cos 5x}$$ $\lim_{x\to(\pi/2)^-}(\cos x)=\cos(\frac{\pi}{2})=0$ and $\lim_{x\to(\pi/2)^-}(\cos 5x)=\cos(\frac{5\pi}{2})=0.$ We have here an indeterminate form of $\frac{0}{0}$, which can be solved by L'Hospital's Rule: $$A=\lim_{x\to(\pi/2)^-}\frac{(\cos x)'}{(\cos 5x)'}$$ $$A=\lim_{x\to(\pi/2)^-}\frac{-\sin x}{-5\sin 5x}$$ $$A=\lim_{x\to(\pi/2)^-}\frac{\sin x}{5\sin 5x}$$ $$A=\frac{\sin(\pi/2)}{5\sin(\frac{5\pi}{2})}$$ $$A=\frac{1}{5\times1}$$ $$A=\frac{1}{5}$$