Evaluate the following limits, if they exist, using the limit laws:
a. lim
𝑥𝑥→3
𝑥𝑥 2 −9
𝑥𝑥 2 −2𝑥𝑥−3
b. lim
𝑥𝑥→−1
(2𝑥𝑥−1)2 −9
𝑥𝑥+1
c. lim
ℎ→0
1
5+ℎ−
1
5
ℎ

Let's evaluate each of the given limits step by step.

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

**Step 1: Simplify the expression**
- The numerator is a difference of squares: \(x^2 - 9 = (x - 3)(x + 3)\).
- The denominator is a quadratic expression: \(x^2 - 2x - 3 = (x - 3)(x + 1)\) (factoring).

So, the expression becomes:
\[
\frac{(x - 3)(x + 3)}{(x - 3)(x + 1)}
\]

**Step 2: Cancel the common factor \((x - 3)\)**
For \(x \neq 3\), we can cancel \((x - 3)\) in the numerator and denominator:
\[
\frac{x + 3}{x + 1}
\]

**Step 3: Substitute \(x = 3\)**
Now, substitute \(x = 3\) into the simplified expression:
\[
\frac{3 + 3}{3 + 1} = \frac{6}{4} = \frac{3}{2}
\]

Thus, the limit is:
\[
\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3} = \frac{3}{2}
\]

---

### b. \(\lim_{x \to -1} \frac{(2x - 1)^2 - 9}{x + 1}\)

**Step 1: Simplify the expression**
- The numerator \((2x - 1)^2 - 9\) is a difference of squares:
\[
(2x - 1)^2 - 9 = \left((2x - 1) - 3\right)\left((2x - 1) + 3\right)
\]
Simplifying both terms:
\[
(2x - 1) - 3 = 2x - 4 = 2(x - 2)
\]
\[
(2x - 1) + 3 = 2x + 2 = 2(x + 1)
\]

Thus, the numerator becomes:
\[
(2x - 1)^2 - 9 = 2(x - 2) \cdot 2(x + 1) = 4(x - 2)(x + 1)
\]

**Step 2: Cancel the common factor \((x + 1)\)**
For \(x \neq -1\), we can cancel \((x + 1)\) in the numerator and denominator:
\[
\frac{4(x - 2)(x + 1)}{x + 1} = 4(x - 2)
\]

**Step 3: Substitute \(x = -1\)**
Now, substitute \(x = -1\) into the simplified expression:
\[
4((-1) - 2) = 4(-3) = -12
\]

Thus, the limit is:
\[
\lim_{x \to -1} \frac{(2x - 1)^2 - 9}{x + 1} = -12
\]

---

### c. \(\lim_{h \to 0} \frac{\frac{1}{5+h} - \frac{1}{5}}{h}\)

**Step 1: Combine the fractions in the numerator**
- The expression becomes:
\[
\frac{\frac{1}{5 + h} - \frac{1}{5}}{h} = \frac{\frac{5 - (5 + h)}{5(5 + h)}}{h} = \frac{\frac{-h}{5(5 + h)}}{h}
\]

**Step 2: Simplify the expression**
- Cancel the \(h\) in the numerator and denominator:
\[
\frac{-h}{h} \cdot \frac{1}{5(5 + h)} = \frac{-1}{5(5 + h)}
\]

**Step 3: Substitute \(h = 0\)**
Now, substitute \(h = 0\) into the simplified expression:
\[
\frac{-1}{5(5 + 0)} = \frac{-1}{25}
\]

Thus, the limit is:
\[
\lim_{h \to 0} \frac{\frac{1}{5 + h} - \frac{1}{5}}{h} = \frac{-1}{25}
\]

---

### Summary of Results:
a. \(\frac{3}{2}\)  
b. \(-12\)  
c. \(\frac{-1}{25}\)

Do you have any questions or need further explanation on any of these steps?

---

**Relative questions:**
1. How do you factor quadratic expressions efficiently?
2. What are the limit laws, and how do they apply to rational functions?
3. How does the difference of squares formula help simplify limits?
4. Why is it important to cancel terms before evaluating limits?
5. Can limits always be evaluated using algebraic simplifications?

**Tip:** Always check if direct substitution works before factoring when evaluating limits.

Evaluate the following limits, if they exist, using the limit laws:
a. lim x→3 (x^2 −9) / (x^2 −2x−3)
b. lim x→−1 ((2x−1)^2 −9) / (x+1)
c. lim h→0 (1 / (5+h) − 1 / 5) / h

Explore how to evaluate limits using factoring and limit laws. This problem involves simplifying rational functions, applying the difference of squares formula, and using limit laws for rational functions. Detailed solutions provided for limits as x approaches 3 and -1, and as h approaches 0. Suitable for Grades 11-12.

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