ce Test \( 1(4.2,4.5-4.9)< \) Question 1, *4.7.57-Setup \& Solve rrade \[ \lim _{x \rightarrow \frac{\pi}{2}_{2}^{+}}\left(\frac{\pi}{2}-x\right) \sec x \]

To evaluate the limit: \[ \lim _{x \rightarrow \frac{\pi}{2}^{+}}\left(\frac{\pi}{2}-x\right) \sec x \] **Step 1: Understand the Behavior Near \(\frac{\pi}{2}\)** - **Approaching \(\frac{\pi}{2}\) from the Right (\(x \to \frac{\pi}{2}^{+}\)):** - \(x > \frac{\pi}{2}\), so \(\frac{\pi}{2} - x\) is negative and approaching 0. - \(\sec x = \frac{1}{\cos x}\). As \(x\) approaches \(\frac{\pi}{2}\), \(\cos x\) approaches 0. - Specifically, from the right, \(\cos x\) approaches 0 **from the negative side** (since cosine is negative just after \(\frac{\pi}{2}\)), making \(\sec x\) approach \(-\infty\). **Step 2: Substitute and Simplify** Let’s perform a substitution to simplify the limit: - Let \(h = x - \frac{\pi}{2}\), where \(h \to 0^{+}\) as \(x \to \frac{\pi}{2}^{+}\). Rewriting the expression in terms of \(h\): \[ \frac{\pi}{2} - x = -h \] \[ \cos x = \cos\left(\frac{\pi}{2} + h\right) = -\sin h \] \[ \sec x = \frac{1}{\cos x} = \frac{1}{-\sin h} = -\csc h \] Now, the original limit becomes: \[ \lim_{h \to 0^{+}} (-h) \cdot (-\csc h) = \lim_{h \to 0^{+}} \frac{h}{\sin h} \] **Step 3: Evaluate the Simplified Limit** We know from standard calculus limits that: \[ \lim_{h \to 0} \frac{h}{\sin h} = 1 \] **Conclusion** Therefore, the limit is: \[ \lim _{x \rightarrow \frac{\pi}{2}^{+}}\left(\frac{\pi}{2}-x\right) \sec x = 1 \]