Let’s solve each of the limits systematically.

Problem 5

$\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$

Solution:

1. Factorization via rationalization: Multiply numerator and denominator by the conjugate of the numerator: $\frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$

2. Simplify the numerator: $(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9$

   The expression becomes: $\frac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3}$

3. Take the limit as $x \to 9$: $\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}$

Answer: $\boxed{\frac{1}{6}}$

Problem 6

$\lim_{h \to 0} \frac{(1 + h)^2 - 1^2}{h}$

Solution:

1. Expand the numerator: $(1 + h)^2 - 1^2 = 1 + 2h + h^2 - 1 = 2h + h^2$

   The expression becomes: $\frac{2h + h^2}{h}$

2. Simplify by canceling $h$: $\frac{2h + h^2}{h} = 2 + h$

3. Take the limit as $h \to 0$: $\lim_{h \to 0} (2 + h) = 2$

Answer: $\boxed{2}$

Problem 7

$\lim_{h \to 0} \frac{(3 + h)^2 - 3^2}{h}$

Solution:

1. Expand the numerator: $(3 + h)^2 - 3^2 = 9 + 6h + h^2 - 9 = 6h + h^2$

   The expression becomes: $\frac{6h + h^2}{h}$

2. Simplify by canceling $h$: $\frac{6h + h^2}{h} = 6 + h$

3. Take the limit as $h \to 0$: $\lim_{h \to 0} (6 + h) = 6$

Answer: $\boxed{6}$

Summary of Answers:

1. Problem 5: $\frac{1}{6}$
2. Problem 6: $2$
3. Problem 7: $6$

Would you like further explanation, or assistance with similar problems?

Here are some related questions to explore:

1. What is the importance of rationalization in solving limits involving square roots?
2. How does the difference quotient relate to derivatives?
3. What happens if the denominator of a limit cannot be canceled or rationalized?
4. How do you compute limits approaching infinity instead of a specific point?
5. Why does expanding terms often simplify limit calculations?

Tip: Always check if direct substitution into the limit simplifies the problem before applying algebraic techniques.