To determine the limit of the function [tex]\( g(x) = \frac{x^3 - 8}{x^2 - 4} \)[/tex] as [tex]\( x \)[/tex] approaches 2, we will go through the following steps.
First, let's rewrite the function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{x^3 - 8}{x^2 - 4} \][/tex]
Notice that both the numerator and the denominator can be factored.
1. Factor the numerator [tex]\( x^3 - 8 \)[/tex]:
[tex]\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \][/tex]
2. Factor the denominator [tex]\( x^2 - 4 \)[/tex]:
[tex]\[ x^2 - 4 = (x - 2)(x + 2) \][/tex]
Now, rewrite the function using these factorizations:
[tex]\[ g(x) = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} \][/tex]
For [tex]\( x \neq 2 \)[/tex], we can cancel the [tex]\((x - 2)\)[/tex] term in both the numerator and the denominator:
[tex]\[ g(x) = \frac{x^2 + 2x + 4}{x + 2} \][/tex]
Now, we need to find the limit as [tex]\( x \)[/tex] approaches 2:
[tex]\[ \lim_{x \to 2} g(x) = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} \][/tex]
Substitute [tex]\( x = 2 \)[/tex] into the simplified function:
[tex]\[ g(2) = \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3 \][/tex]
Therefore, the limit of [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches 2 is:
[tex]\[ \boxed{3} \][/tex]
Thus, the correct answer is:
b. 3
