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.