ce Test $$ 1(4.2,4.5-4.9)

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

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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:

$$
\lim _{x \rightarrow \frac{\pi}{2}^{+}}\left(\frac{\pi}{2}-x\right) \sec x = 1
$$
The limit is 1.

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

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