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Let's evaluate the limits one by one from the image you provided.

### 1. Problem (ii):
\[
\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1}
\]

At first glance, if you substitute \( x = 1 \), you get a \( \frac{0}{0} \) indeterminate form. So, let's apply algebraic manipulation to resolve this.

#### Step 1: Rationalize the denominator
We will multiply both the numerator and denominator by the conjugate of the denominator:
\[
\frac{x - 1}{\sqrt{x} - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{(x - 1)(\sqrt{x} + 1)}{(\sqrt{x})^2 - 1^2}
\]

#### Step 2: Simplify
\[
\frac{(x - 1)(\sqrt{x} + 1)}{x - 1}
\]
The \( (x - 1) \) cancels out, leaving:
\[
\lim_{x \to 1} \sqrt{x} + 1
\]

#### Step 3: Substitute \( x = 1 \)
\[
\sqrt{1} + 1 = 1 + 1 = 2
\]

Thus, the value of the limit is:
\[
\boxed{2}
\]

---

### 2. Problem (iii):
\[
\lim_{x \to 1} \frac{(x - 3)(x + 1)}{x^2 - 1}
\]

First, substitute \( x = 1 \) to check for indeterminate form:
\[
\frac{(1 - 3)(1 + 1)}{1^2 - 1} = \frac{(-2)(2)}{0} = \frac{-4}{0}
\]
This is a case of division by zero, so the limit does not exist as a finite number. However, we need to analyze whether it approaches \( +\infty \) or \( -\infty \).

#### Step 1: Factor the denominator
The denominator \( x^2 - 1 \) is a difference of squares:
\[
x^2 - 1 = (x - 1)(x + 1)
\]

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

#### Step 2: Cancel common factors
The \( (x + 1) \) terms cancel out:
\[
\frac{x - 3}{x - 1}
\]

#### Step 3: Substitute \( x = 1 \)
\[
\frac{1 - 3}{1 - 1} = \frac{-2}{0}
\]

#### Step 4: Analyze the sign of the limit near \( x = 1 \)
- As \( x \to 1^+ \) (from the right), \( (x - 1) \) is positive but very small, so \( \frac{-2}{\text{small positive number}} \to -\infty \).
- As \( x \to 1^- \) (from the left), \( (x - 1) \) is negative and very small, so \( \frac{-2}{\text{small negative number}} \to +\infty \).

Thus, the limit does not exist because the function approaches different infinities from either side:
\[
\boxed{\text{DNE (does not exist)}}
\]

---

Would you like any additional details or further clarification?

Here are some related questions for further practice:
1. What is the limit of \(\lim_{x \to 0} \frac{\sin(x)}{x}\)?
2. How would you evaluate \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\)?
3. What happens when you apply L'Hopital's Rule to indeterminate forms like \( \frac{0}{0} \)?
4. Can you factor \( x^3 - 1 \) to find the limit as \( x \to 1 \)?
5. How do you find the one-sided limits as \( x \to 0 \) for \(\frac{1}{x^2}\)?

**Tip:** When you encounter indeterminate forms, try algebraic manipulation (like factoring or rationalizing) before applying more advanced techniques like L'Hopital's Rule.

Learn how to evaluate limits using algebraic manipulation and factorization. This example covers rationalizing the denominator and factorizing quadratic expressions to solve indeterminate forms and limits approaching infinity. Suitable for high school students in Grades 10-12.

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