Let's solve each of the limits step by step:

(a) $\lim_{x \to 3} \frac{x^3 - 5x^2 + 3x + 9}{x - 3}$

The given function is indeterminate at $x = 3$, so we can factorize the numerator: $x^3 - 5x^2 + 3x + 9$ Factorizing it: $x^3 - 5x^2 + 3x + 9 = (x - 3)(x^2 - 2x - 3)$ $x^2 - 2x - 3 = (x - 3)(x + 1)$ So: $x^3 - 5x^2 + 3x + 9 = (x - 3)^2 (x + 1)$ Thus, the limit becomes: $\lim_{x \to 3} \frac{(x - 3)^2 (x + 1)}{x - 3}$ Cancel $x - 3$: $\lim_{x \to 3} (x - 3)(x + 1)$ Evaluate at $x = 3$: $(3 - 3)(3 + 1) = 0$ Answer: $\boxed{0}$

(b) $\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1}$

This is indeterminate at $x = 1$, so we rationalize the denominator: $\sqrt{x} - 1 \to \frac{1}{\sqrt{x} + 1}$ Rewrite: from sympy import symbols, limit, sqrt

Define the variable

x = symbols('x')

(b) Expression and Limit

expr_b = (x - 1) / (sqrt(x) - 1) limit_b = limit(expr_b, x, 1) limit_b22successThe limit for (b) is $\boxed{2}$.

(c) $\lim_{x \to 2} \frac{\sqrt{5x - 1} - 3}{x - 2}$

This is indeterminate at $x = 2$, so we rationalize the numerator: $\sqrt{5x - 1} - 3 \to \frac{5x - 1 - 9}{\sqrt{5x - 1} + 3}$ Let's calculate:

$\lim_{x \to 2} \frac{\sqrt{5x - 1} - 3}{x - 2}$

(c) Expression and Limit

expr_c = (sqrt(5*x - 1) - 3) / (x - 2) limit_c = limit(expr_c, x, 2) limit_c5/65/6successThe limit for (c) is $\boxed{\frac{5}{6}}$.

Final Answers:

(a) $\boxed{0}$
(b) $\boxed{2}$
(c) $\boxed{\frac{5}{6}}$

Would you like detailed explanations for any of these steps, or do you have additional questions?

Here are five related questions to explore:

1. How can you apply L'Hôpital's rule to these types of limits?
2. How do you factorize polynomials to simplify limits?
3. What is the role of rationalization in solving indeterminate limits?
4. How can you identify when a function is indeterminate?
5. What are the differences between one-sided and two-sided limits?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ when solving limits. These forms guide you to use techniques like factoring or rationalization.