From the uploaded image, the question is asking which equations are correct among the given limits.

Let us carefully analyze the options:

Option 1:

$\lim_{x \to 8} \frac{x - 8}{x - 8} = \lim_{x \to 8} (x - 8)$ This is incorrect because the left-hand side simplifies to $1$ (as long as $x \neq 8$), whereas the right-hand side is $0$ when $x = 8$.

Option 2:

$\lim_{x \to 8} \frac{x - 8}{x + 8} = \lim_{x \to 8} \frac{x - 8}{x + 8}$ This is correct, as both sides are the same and well-defined for $x \to 8$.

Option 3:

$\lim_{x \to 1} \frac{x^3}{x - 1} = \lim_{x \to 1} \frac{x}{x - 1}$ This is incorrect because $\frac{x^3}{x-1} \neq \frac{x}{x-1}$.

Option 4:

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1)$ This is correct because $\frac{x^2 - 1}{x - 1}$ simplifies to $x + 1$ using factorization of the numerator.

Option 5:

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \sqrt{\lim_{x \to 1} \frac{x^2 - 1}{x - 1}}$ This is incorrect because the square root of a limit would not match the original expression unless additional conditions are satisfied.

Conclusion:

The correct equations are:

• Option 2
• Option 4

Would you like a detailed explanation for any of these steps?

5 Related Questions:

1. What are the conditions for applying L'Hopital's Rule to solve indeterminate limits?
2. How do you simplify rational expressions when dealing with limits?
3. Why does factoring help evaluate limits where direct substitution fails?
4. What are the common pitfalls in evaluating piecewise limits?
5. How does the square root affect the calculation of limits?

Tip:

Always check if a limit leads to an indeterminate form (e.g., $0/0$) before simplifying or applying rules like L'Hopital’s.