Here is a breakdown of the given math problems and their solutions:

29. Limits

1. 29.2 $\lim_{x \to 0} \frac{2x + 2}{2x - 1}$: Substituting $x = 0$: $\frac{2(0) + 2}{2(0) - 1} = \frac{2}{-1} = -2.$ The limit is not given in the choices. Likely an error in options.

2. 29.2 $\lim_{x \to 0} \frac{2^x - 1}{x + 2}$: Using the known limit property $\lim_{x \to 0} \frac{2^x - 1}{x} = \ln(2)$: Approximate $\ln(2) \approx 0.693$, and the denominator is slightly above $2$. Thus, this problem requires simplification: The correct answer is approximately $0$.

3. 29.3 $\lim_{x \to 1} \frac{3x^2 + x}{x^2 + 1}$: Substituting $x = 1$: $\frac{3(1)^2 + 1}{1^2 + 1} = \frac{3 + 1}{1 + 1} = \frac{4}{2} = 2.$ The answer is C. 2.

30. Extreme Values

4. 30.1 $\lim_{x \to -2} \frac{x}{\sqrt{x^3 + 1}}$: Substitute $x = -2$ in the denominator: $x^3 + 1 = (-2)^3 + 1 = -8 + 1 = -7.$ So: $\frac{-2}{\sqrt{-7}} \quad (\text{undefined since square root of a negative number}).$ The answer is D. 不存在 (Does not exist).

5. 30.2 $f(x) = x^2 - 4x + 3$ - Minimum Value: Rewrite $f(x) = x^2 - 4x + 3$ in vertex form: $f(x) = (x - 2)^2 - 1.$ The minimum occurs at $x = 2$ with $f(2) = -1$. The answer is A. -1.

6. 30.2 $f(x) = \sin x + \cos x$ on $[0, 2\pi]$: The maximum of $\sin x + \cos x$ is $\sqrt{2}$, found by rewriting as: $R = \sqrt{\sin^2 x + \cos^2 x} = \sqrt{2}.$ The answer is C. $\sqrt{2}$.

7. 30.3 $f(x) = x^2 - 4x + 3$: Critical Point $x$: Find $f'(x) = 2x - 4$. Set $f'(x) = 0$: $2x - 4 = 0 \implies x = 2.$ The answer is C. 2.

8. 30.4 $f(x) = x^2 + 2x - 5$: Maximum Value: Vertex form: $f(x) = (x + 1)^2 - 6.$ Maximum value does not exist since the parabola opens upwards ($a > 0$). The answer is D. 不存在 (Does not exist).

31. Derivatives

9. 31.1 $f(x) = 3x^2 + 5x - 2$: Differentiate: $f'(x) = 6x + 5.$ The answer is A. 6x + 5.

10. 31.2 $f(x) = e^x \sin(x)$: Using product rule: $f'(x) = e^x \cdot \sin(x) + e^x \cdot \cos(x) = e^x (\sin x + \cos x).$ The answer is C. $e^x [\sin(x) + \cos(x)]$.

11. 31.3 $f(x) = \tan(x)$: Derivative of $\tan(x)$ is: $f'(x) = \sec^2(x).$ The answer is A. $\sec^2 x$.

Summary of Answers:

1. 29.2 (First): None of the above.
2. 29.2 (Second): A. 0.
3. 29.3: C. 2.
4. 30.1: D. 不存在.
5. 30.2 (First): A. -1.
6. 30.2 (Second): C. $\sqrt{2}$.
7. 30.3: C. 2.
8. 30.4: D. 不存在.
9. 31.1: A. $6x + 5$.
10. 31.2: C. $e^x [\sin x + \cos x]$.
11. 31.3: A. $\sec^2 x$.

Follow-Up Questions:

1. How do you convert a quadratic function into vertex form?
2. What is the product rule for derivatives?
3. How do you evaluate limits involving square roots or indeterminate forms?
4. Why does the derivative of $\tan(x)$ equal $\sec^2(x)$?
5. How can you determine the maximum or minimum values of trigonometric functions?

Tip: Always check for undefined limits when a square root involves a negative number.