This is a limit problem. You are asked to compute:

$\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 3x + 2}.$

Step 1: Check for indeterminate form

When $x = 2$, substitute into both the numerator and denominator:

• Numerator: $x^3 - 8 = 2^3 - 8 = 0$,
• Denominator: $x^2 - 3x + 2 = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0$.

Since both the numerator and denominator are 0, this is an indeterminate form ($0/0$). We need to simplify the expression.

Step 2: Factorize the numerator and denominator

Numerator: $x^3 - 8$

This is a difference of cubes: $x^3 - 8 = (x - 2)(x^2 + 2x + 4).$

Denominator: $x^2 - 3x + 2$

This is a quadratic trinomial that factors as: $x^2 - 3x + 2 = (x - 1)(x - 2).$

Step 3: Simplify the expression

The original limit becomes: $\frac{x^3 - 8}{x^2 - 3x + 2} = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 1)(x - 2)}.$

Cancel the common factor $(x - 2)$ (valid as long as $x \neq 2$): $\frac{x^2 + 2x + 4}{x - 1}.$

Step 4: Evaluate the limit

Now, substitute $x = 2$ into the simplified expression: $\frac{x^2 + 2x + 4}{x - 1} = \frac{2^2 + 2(2) + 4}{2 - 1} = \frac{4 + 4 + 4}{1} = \frac{12}{1} = 12.$

Final Answer:

$\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 3x + 2} = 12.$

Would you like further clarifications or additional examples?

5 Related Questions:

1. How do you handle limits with indeterminate forms like $\frac{0}{0}$?
2. What are the steps to factor a difference of cubes like $x^3 - a^3$?
3. How do you factor quadratic expressions like $x^2 - 3x + 2$?
4. What happens if there are higher-order indeterminate forms (e.g., $\frac{\infty}{\infty}$)?
5. How does the cancellation of terms affect the domain of the function?

Tip:

Always check for possible factorization in limit problems to resolve indeterminate forms.