The problem asks to explain why the limit $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ exists and compute the limit using an intuitive approach.

Step-by-Step Solution:

1. Simplifying the expression:

The given expression is $\frac{x^3 - 8}{x^2 - 4}$.

We can factor both the numerator and the denominator:

• The numerator $x^3 - 8$ is a difference of cubes: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
• The denominator $x^2 - 4$ is a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$

Now, rewrite the expression: $\frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$

2. Cancelling the common factor:

For $x \neq 2$, we can cancel the $(x - 2)$ term from both the numerator and the denominator: $\frac{x^2 + 2x + 4}{x + 2}$

3. Substituting $x = 2$:

Now, substitute $x = 2$ into the simplified expression: $\frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$

Thus, the limit is $\boxed{3}$.

Why does the limit exist?

The limit exists because, after simplifying the expression by cancelling the removable discontinuity at $x = 2$, the resulting function $\frac{x^2 + 2x + 4}{x + 2}$ is continuous at $x = 2$. Therefore, we can evaluate the limit by direct substitution.

Would you like more details or explanations?

Here are 5 related questions for further practice:

1. How would the solution change if the denominator were $x^2 + 4$ instead of $x^2 - 4$?
2. How can you apply L'Hôpital's Rule to solve this limit?
3. What is the geometrical interpretation of cancelling common factors in limits?
4. Can you find the limit if the expression were $\lim_{x \to -2} \frac{x^3 - 8}{x^2 - 4}$?
5. How would the factorization change if the numerator were $x^3 + 8$ instead of $x^3 - 8$?

Tip: Always check for common factors to simplify limits with indeterminate forms, like $\frac{0}{0}$.