Certainly! Let's find the limit of the function [tex]\( g(x) = x^3 \)[/tex] as [tex]\( x \)[/tex] approaches 2 for the given expression:
[tex]\[
\lim _{x \rightarrow 2} \frac{g(x)-g(2)}{x-2}
\][/tex]
First, let's evaluate the function [tex]\( g(x) \)[/tex] at [tex]\( x = 2 \)[/tex]:
[tex]\[
g(2) = 2^3 = 8
\][/tex]
Now, substitute [tex]\( g(x) = x^3 \)[/tex] and [tex]\( g(2) = 8 \)[/tex] into the limit expression:
[tex]\[
\lim _{x \rightarrow 2} \frac{x^3 - 8}{x - 2}
\][/tex]
Next, we need to simplify the expression inside the limit. Notice that [tex]\( x^3 - 8 \)[/tex] is a difference of cubes, which can be factored as follows:
[tex]\[
x^3 - 8 = (x - 2)(x^2 + 2x + 4)
\][/tex]
So, the limit expression becomes:
[tex]\[
\lim _{x \rightarrow 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}
\][/tex]
Since [tex]\( x \neq 2 \)[/tex] in the limit (we are approaching 2 but not equal to 2), we can cancel out [tex]\( (x - 2) \)[/tex] in the numerator and the denominator:
[tex]\[
\lim _{x \rightarrow 2} (x^2 + 2x + 4)
\][/tex]
Now, we simply evaluate the remaining polynomial [tex]\( x^2 + 2x + 4 \)[/tex] at [tex]\( x = 2 \)[/tex]:
[tex]\[
2^2 + 2(2) + 4 = 4 + 4 + 4 = 12
\][/tex]
Thus, the limit is:
[tex]\[
\lim _{x \rightarrow 2} \frac{g(x) - g(2)}{x - 2} = 12
\][/tex]
So, the limit of the given expression is [tex]\( 12 \)[/tex]. Additionally, we found that:
[tex]\[
g(2) = 8
\][/tex]
Putting these together:
[tex]\[
(g(2), \lim _{x \rightarrow 2} \frac{g(x) - g(2)}{x - 2}) = (8, 12)
\][/tex]
Therefore, the limit and the value of the function at [tex]\( x = 2 \)[/tex] are 8 and 12, respectively.
