Certainly! Let's go through the steps to find the limit of [tex]\(\lim _{x \rightarrow 4} \frac{x^3-64}{x^2-16}\)[/tex] in detail.
### Step 1: Factor the Numerator and the Denominator
First, observe the given expression:
[tex]\[
\frac{x^3-64}{x^2-16}
\][/tex]
Numerator: [tex]\(x^3 - 64\)[/tex]
Notice that [tex]\(x^3 - 64\)[/tex] is a difference of cubes, which can be factored using the formula:
[tex]\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = 4\)[/tex]. Thus:
[tex]\[
x^3 - 64 = (x - 4)(x^2 + 4x + 16)
\][/tex]
Denominator: [tex]\(x^2 - 16\)[/tex]
Notice that [tex]\(x^2 - 16\)[/tex] is a difference of squares, which can be factored using the formula:
[tex]\[
a^2 - b^2 = (a - b)(a + b)
\][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = 4\)[/tex]. Thus:
[tex]\[
x^2 - 16 = (x - 4)(x + 4)
\][/tex]
### Step 2: Simplify the Expression
Now substitute the factored forms of the numerator and the denominator into the original expression:
[tex]\[
\frac{x^3 - 64}{x^2 - 16} = \frac{(x - 4)(x^2 + 4x + 16)}{(x - 4)(x + 4)}
\][/tex]
We can cancel out the common factor [tex]\((x - 4)\)[/tex] from the numerator and the denominator:
[tex]\[
\frac{(x - 4)(x^2 + 4x + 16)}{(x - 4)(x + 4)} = \frac{x^2 + 4x + 16}{x + 4}
\][/tex]
### Step 3: Evaluate the Limit
Now, we need to find the limit of the simplified expression as [tex]\(x\)[/tex] approaches 4:
[tex]\[
\lim_{x \to 4} \frac{x^2 + 4x + 16}{x + 4}
\][/tex]
To do this, simply substitute [tex]\(x = 4\)[/tex] into the simplified expression:
[tex]\[
\frac{4^2 + 4 \cdot 4 + 16}{4 + 4} = \frac{16 + 16 + 16}{8} = \frac{48}{8} = 6
\][/tex]
### Final Answer
Thus, the limit is:
[tex]\[
\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16} = 6
\][/tex]
