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The document you uploaded contains practice problems on the limits of trigonometric functions. Let's go through one of the questions as an example:

**Question 1**:
\[
\lim_{x \to \pi} \frac{\sin x}{1 + \cos x}
\]
Options:
- A. \(0\)
- B. \(\frac{1}{2}\)
- C. \(1\)
- D. \(2\)
- E. \(\infty\)

### Solution:
We evaluate the limit by substituting \(x = \pi\):

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

Since \(\sin(\pi) = 0\) and \(\cos(\pi) = -1\), the expression becomes:

\[
\frac{0}{1 - 1} = \frac{0}{0}
\]

This is an indeterminate form, so we apply **L'Hopital's Rule**, which states that for indeterminate forms of \(\frac{0}{0}\), we can differentiate the numerator and denominator separately and then take the limit.

- The derivative of \(\sin x\) is \(\cos x\).
- The derivative of \(1 + \cos x\) is \(-\sin x\).

Thus, the limit becomes:

\[
\lim_{x \to \pi} \frac{\cos x}{-\sin x}
\]

At \(x = \pi\):

\[
\frac{\cos(\pi)}{-\sin(\pi)} = \frac{-1}{0}
\]

This expression suggests the limit approaches infinity (or negative infinity depending on the sign of the sine term), so the answer is:

**E. \(\infty\)**

---

Do you want more detailed explanations or have any other questions?

Here are 5 additional related problems you might want to consider:
1. Evaluate \(\lim_{x \to 0} \frac{\tan x}{x}\).
2. Solve \(\lim_{x \to \pi/2} \frac{1 - \sin x}{\cos^2 x}\).
3. Evaluate \(\lim_{x \to 0} \frac{\sin(2x)}{x}\).
4. Find \(\lim_{x \to \infty} \frac{\sin x}{x}\).
5. Solve \(\lim_{x \to 0} \frac{\tan x - x}{x^3}\).

**Tip**: When solving limits that yield an indeterminate form, L'Hopital's Rule is a powerful technique!

Evaluate the limit: lim_{x \to \pi} \frac{\sin x}{1 + \cos x}.

This page provides a detailed step-by-step solution to the trigonometric limit lim_{x \to \pi} \frac{\sin x}{1 + \cos x}, involving the application of L'Hopital's Rule for an indeterminate form. Key concepts include trigonometric functions and calculus limits, suitable for students in Grades 11-12.

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