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Let's go over the questions one by one:

### 1. \(\lim_{x \to 2} \frac{1 - \frac{2}{x}}{x^2 - x - 2}\)

First, we factor the denominator:
\[
x^2 - x - 2 = (x - 2)(x + 1)
\]

Next, simplify the numerator:
\[
1 - \frac{2}{x} = \frac{x - 2}{x}
\]

Thus, the expression becomes:
\[
\lim_{x \to 2} \frac{\frac{x - 2}{x}}{(x - 2)(x + 1)}
\]
Canceling \((x - 2)\) from numerator and denominator:
\[
\lim_{x \to 2} \frac{1}{x(x + 1)}
\]
Substituting \(x = 2\):
\[
\frac{1}{2(2 + 1)} = \frac{1}{6}
\]
**Answer**: \(\frac{1}{6}\)

### 2. \(\lim_{x \to 0} \frac{4 - 2x + x^2}{2 + x^2}\)

Substitute \(x = 0\) directly:
\[
\frac{4 - 2(0) + (0)^2}{2 + (0)^2} = \frac{4}{2} = 2
\]
**Answer**: \(2\)

### 3. \(\lim_{x \to 0} \frac{1 - \cos(2x)}{x \tan(x)}\)

First, recall that \(\cos(2x) \approx 1 - 2x^2\) for small \(x\), and \(\tan(x) \approx x\) for small \(x\).

This simplifies the expression to:
\[
\frac{1 - (1 - 2x^2)}{x \cdot x} = \frac{2x^2}{x^2} = 2
\]
**Answer**: \(2\)

### 4. Derivatives

a. \(\text{cot}(x)\)

The derivative of \(\text{cot}(x)\) is:
\[
\frac{d}{dx}[\text{cot}(x)] = -\csc^2(x)
\]

b. \(\cos(x) \tan^2(x)\)

Use the product rule:
\[
\frac{d}{dx} [\cos(x) \tan^2(x)] = \frac{d}{dx} [\cos(x)] \cdot \tan^2(x) + \cos(x) \cdot \frac{d}{dx} [\tan^2(x)]
\]
\[
= -\sin(x) \cdot \tan^2(x) + \cos(x) \cdot 2\tan(x) \sec^2(x)
\]

c. \(\cot^2(x)\)

Using the chain rule:
\[
\frac{d}{dx} [\cot^2(x)] = 2\cot(x) \cdot \frac{d}{dx} [\cot(x)] = 2\cot(x)(-\csc^2(x)) = -2\cot(x)\csc^2(x)
\]

d. \(\sec^2(x)\)

The derivative of \(\sec^2(x)\) is:
\[
\frac{d}{dx}[\sec^2(x)] = 2\sec(x) \cdot \sec(x) \tan(x) = 2\sec^2(x) \tan(x)
\]

e. \(\sqrt{x - x\sin^2(x)}\)

Use the chain rule and simplify:
\[
\frac{d}{dx} \left( \sqrt{x - x \sin^2(x)} \right) = \frac{1}{2\sqrt{x - x \sin^2(x)}} \cdot \left( 1 - (\sin^2(x) + 2x\sin(x)\cos(x)) \right)
\]
This requires further simplification, but the general form follows the chain rule.

Would you like any further details or explanations?

### Follow-up questions:
1. How does the limit of \(\frac{1 - \cos(2x)}{x\tan(x)}\) change with different trigonometric approximations?
2. Could you expand on how the chain rule applies to \(\cot^2(x)\)?
3. Why is it necessary to use the product rule for \(\cos(x)\tan^2(x)\)?
4. What happens if the denominator in the first limit cannot be factored?
5. How do small-angle approximations help in solving limits like in question 3?

**Tip**: Always check for factorization or small-angle approximations when handling limits with trigonometric expressions.

1. \lim_{x \to 2} \frac{1 - \frac{2}{x}}{x^2 - x - 2}, 2. \lim_{x \to 0} \frac{4 - 2x + x^2}{2 + x^2}, 3. \lim_{x \to 0} \frac{1 - \cos(2x)}{x \tan(x)}, 4. Determine the derivatives of: a. Cot(x), b. Cos(x)tan^2(x), c. cotan^2(x), d. sec^2(x), e. \sqrt{x - x\sin^2(x)}.

Explore solutions for limits and derivatives involving trigonometric functions such as cotangent, secant, cosine, and small-angle approximations. This includes applying L'Hopital's rule, the product rule, and chain rule for advanced calculus problems. Suitable for Grades 11-12 and early college students.

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