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.