find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

Explanation:

Step1: Check the form

As $x\to(\frac{\pi}{2})^+$, $\cos x\to0$ and $1 - \sin x\to0$. So, it is in $\frac{0}{0}$ - form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $\cos x$ is $-\sin x$, and the derivative of $1-\sin x$ is $-\cos x$. So, the limit becomes $\lim_{x\to(\frac{\pi}{2})^+}\frac{-\sin x}{-\cos x}=\lim_{x\to(\frac{\pi}{2})^+}\tan x$.

Step3: Evaluate the new - limit

As $x\to(\frac{\pi}{2})^+$, $\tan x\to-\infty$.

Answer:

$-\infty$