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The given limit problem is:

\[
\lim_{x \to \frac{\pi}{2}} \frac{\cos(2x) + 1}{2x + \pi}
\]

### Step-by-Step Solution:

1. **Substitute \( x = \frac{\pi}{2} \)**:
   - First, let's evaluate the expression by direct substitution of \( x = \frac{\pi}{2} \).
   
   \[
   \cos(2 \times \frac{\pi}{2}) + 1 = \cos(\pi) + 1 = -1 + 1 = 0
   \]
   
   For the denominator:
   \[
   2 \times \frac{\pi}{2} + \pi = \pi + \pi = 2\pi
   \]

   So, after substitution, we get:
   \[
   \frac{0}{2\pi} = 0
   \]

Thus, the limit evaluates to **0**.

### Conclusion:
\[
\lim_{x \to \frac{\pi}{2}} \frac{\cos(2x) + 1}{2x + \pi} = 0
\]

Would you like more details on the steps, or do you have further questions?

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Here are five related questions for practice:
1. How do you evaluate limits involving trigonometric functions?
2. What is the limit of \( \lim_{x \to 0} \frac{\sin(x)}{x} \)?
3. How do you apply L'Hopital's rule to solve indeterminate limits?
4. What is the behavior of cosine and sine as \( x \to \infty \)?
5. How do you deal with limits that result in the form \( \frac{0}{0} \)?

**Tip**: Always check if direct substitution works for limits before using other methods like L'Hopital's Rule!

Learn how to solve the limit problem \( \lim_{x \to \frac{\pi}{2}} \frac{\cos(2x) + 1}{2x + \pi} \) with a detailed step-by-step solution using direct substitution method. This problem covers key calculus and trigonometric limit concepts, suitable for undergraduate students.

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