To solve the limit [tex]\(\lim_{x \to -2} \frac{x^3 + 8}{x + 2}\)[/tex], we need to handle the potential indeterminate form that arises when substituting [tex]\( x = -2 \)[/tex]. Here is a detailed, step-by-step approach to finding this limit:
1. Identify the form at [tex]\( x = -2 \)[/tex]:
When [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex]:
[tex]\[
x + 2 \to 0 \quad \text{and} \quad x^3 + 8 \to (-2)^3 + 8 = -8 + 8 = 0.
\][/tex]
This gives us the indeterminate form [tex]\(\frac{0}{0}\)[/tex].
2. Factor the numerator:
We recognize that [tex]\( x^3 + 8 \)[/tex] can be factored using the sum of cubes:
[tex]\[
x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4).
\][/tex]
Therefore, the expression becomes:
[tex]\[
\frac{x^3 + 8}{x + 2} = \frac{(x + 2)(x^2 - 2x + 4)}{x + 2}.
\][/tex]
3. Simplify the expression:
Provided [tex]\( x \neq -2 \)[/tex], the [tex]\( (x + 2) \)[/tex] terms can be canceled out:
[tex]\[
\frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4.
\][/tex]
4. Evaluate the limit:
Now, substitute [tex]\( x = -2 \)[/tex] into the simplified expression:
[tex]\[
x^2 - 2x + 4 \bigg|_{x = -2} = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12.
\][/tex]
Therefore, we have:
[tex]\[
\lim_{x \to -2} \frac{x^3 + 8}{x + 2} = 12.
\][/tex]
That completes our calculation, and the limit as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex] of the given expression is indeed [tex]\( 12 \)[/tex].
