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Let's solve the limits step by step.

### (a) \(\lim_{x \to \infty} \frac{1 + 2x^2 + 3x^9}{1 - x - x^9}\)

#### Step 1: Analyze the degrees of the polynomials.
- The highest degree term in the numerator is \(3x^9\).
- The highest degree term in the denominator is \(-x^9\).

#### Step 2: Divide both the numerator and denominator by \(x^9\).
We do this to simplify the expression and focus on the dominant terms as \(x \to \infty\):
\[
\frac{1 + 2x^2 + 3x^9}{1 - x - x^9} = \frac{\frac{1}{x^9} + \frac{2}{x^7} + 3}{\frac{1}{x^9} - \frac{1}{x^8} - 1}
\]
As \(x \to \infty\), the terms with \(x\) in the denominator (\(\frac{1}{x^n}\)) approach 0, so we are left with:
\[
\lim_{x \to \infty} \frac{3}{-1} = -3
\]

Thus, the limit for part (a) is:
\[
\boxed{-3}
\]

---

### (b) \(\lim_{x \to 0} \frac{\sin^2 x + x^2}{x \sin x}\)

#### Step 1: Analyze the limit at \(x = 0\).
This expression looks like an indeterminate form \( \frac{0}{0} \), so we should try to simplify it.

#### Step 2: Apply approximations for small \(x\).
For small \(x\), we can use the approximations:
- \(\sin x \approx x\)

Using this approximation in the expression:
\[
\frac{\sin^2 x + x^2}{x \sin x} \approx \frac{x^2 + x^2}{x^2} = \frac{2x^2}{x^2} = 2
\]

Thus, the limit for part (b) is:
\[
\boxed{2}
\]

---

### Do you want more details or have any other questions?

Here are 5 related questions to expand your understanding:
1. How do you handle limits at infinity when the degrees of the numerator and denominator are the same?
2. What happens if the highest degree term in the denominator has a higher power than in the numerator?
3. How can L'Hopital's Rule be used to evaluate limits involving indeterminate forms?
4. What are the typical approximations for \(\sin x\), \(\cos x\), and \(\tan x\) as \(x\) approaches 0?
5. How would you deal with limits involving logarithmic or exponential functions?

**Tip:** When solving limits, always check if factoring, division by the highest degree, or approximation methods can help simplify the problem before applying more advanced techniques like L'Hopital's Rule.

Compute the following limits:
(a) \( \lim_{x \to \infty} \frac{1 + 2x^2 + 3x^9}{1 - x - x^9} \)
(b) \( \lim_{x \to 0} \frac{\sin^2 x + x^2}{x \sin x} \)

Learn how to compute limits at infinity and evaluate indeterminate forms. This problem demonstrates polynomial division and small-angle approximations for two limit problems, ideal for Grades 11-12 or early college students.

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