Let's analyze and solve the given problem step-by-step.

The problem is:

$\lim_{x \to \pi} \frac{x + \pi \sec x}{x^2 - \pi^2}$

Step 1: Analyze the expression

The denominator $x^2 - \pi^2$ can be factored as: $x^2 - \pi^2 = (x - \pi)(x + \pi).$

Substitute $x = \pi$:

• In the numerator: $(x + \pi \sec x)$ involves $\sec(\pi)$, which equals $-1$, so we get $x + \pi(-1) = \pi - \pi = 0$.
• In the denominator: $(x - \pi)(x + \pi)$ becomes $0 \cdot 2\pi = 0$.

Thus, both the numerator and denominator approach 0 as $x \to \pi$. This indicates that L'Hôpital's Rule is applicable.

Step 2: Apply L'Hôpital's Rule

We differentiate the numerator and denominator with respect to $x$:

• Numerator: $x + \pi \sec x$ $\frac{d}{dx} \left(x + \pi \sec x\right) = 1 + \pi \sec x \tan x.$

• Denominator: $x^2 - \pi^2$ $\frac{d}{dx} \left(x^2 - \pi^2\right) = 2x.$

So, the limit becomes: $\lim_{x \to \pi} \frac{1 + \pi \sec x \tan x}{2x}.$

Step 3: Evaluate the new limit as $x \to \pi$

Substitute $x = \pi$ into the simplified expression:

• Recall that:
  • $\sec(\pi) = -1$,
  • $\tan(\pi) = 0$.

Thus, the numerator becomes: $1 + \pi \sec(\pi) \tan(\pi) = 1 + \pi(-1)(0) = 1.$

The denominator becomes: $2x = 2\pi.$

Therefore, the limit is: $\frac{1}{2\pi}.$

Final Answer:

The correct choice is: $\boxed{C \, \frac{1}{2\pi}}$

Would you like a deeper explanation of any step? Let me know!

Here are 5 related questions to explore further:

1. How is L'Hôpital's Rule derived and when can it be used?
2. What are the trigonometric values of $\sec x$ and $\tan x$ for key angles like $\pi$?
3. What happens if we apply Taylor expansion to approximate functions near a point?
4. Can you find the limit of similar expressions involving secant and tangent?
5. What are the conditions for the existence of a limit?

Tip: Always check if both the numerator and denominator approach 0 or $\infty$ before using L'Hôpital's Rule.