Question

use lhôpitals rule to find the following limit.
lim_{x
ightarrow\frac{pi}{2}^{-}}left(\frac{pi}{2}-x
ight)sec x
lim_{x
ightarrow\frac{pi}{2}^{-}}left(\frac{pi}{2}-x
ight)sec x=square \text{ (type an exact answer.)}

Explanation:

Step1: Rewrite the function

Rewrite $(\frac{\pi}{2}-x)\sec x$ as $\frac{\frac{\pi}{2}-x}{\cos x}$. As $x\to\frac{\pi}{2}^{-}$, we have the indeterminate - form $\frac{0}{0}$.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $\frac{\pi}{2}-x$ with respect to $x$ is $- 1$, and the derivative of $\cos x$ with respect to $x$ is $-\sin x$. So, $\lim_{x\to\frac{\pi}{2}^{-}}\frac{\frac{\pi}{2}-x}{\cos x}=\lim_{x\to\frac{\pi}{2}^{-}}\frac{-1}{-\sin x}$.

Step3: Evaluate the limit

Substitute $x = \frac{\pi}{2}$ into $\frac{-1}{-\sin x}$. We know that $\sin\frac{\pi}{2}=1$, so $\lim_{x\to\frac{\pi}{2}^{-}}\frac{-1}{-\sin x}=1$.

Answer:

$1$