To solve the limit [tex]\(\lim_{x \rightarrow 8} \frac{x-8}{\sqrt[3]{x}-2}\)[/tex], let's go through the problem step by step.
1. Understanding the Expression:
We need to find the limit of the function [tex]\(\frac{x-8}{\sqrt[3]{x}-2}\)[/tex] as [tex]\(x\)[/tex] approaches [tex]\(8\)[/tex].
2. Substitute [tex]\(x = 8\)[/tex] if possible:
Before applying any limit laws or techniques, let's try direct substitution to see if we get a determinate form.
Substituting [tex]\(x = 8\)[/tex]:
[tex]\[
\frac{8 - 8}{\sqrt[3]{8} - 2} = \frac{0}{2 - 2} = \frac{0}{0}
\][/tex]
This is an indeterminate form [tex]\(\frac{0}{0}\)[/tex], so we need to use algebraic techniques to resolve this.
3. Algebraic Manipulation:
Let's rewrite the expression in a form that allows us to simplify it:
[tex]\[
\frac{x - 8}{\sqrt[3]{x} - 2}
\][/tex]
Notice that [tex]\(x = 8\)[/tex] is a perfect cube, so we use the fact that [tex]\(8 = 2^3\)[/tex].
4. Rationalizing the Denominator:
To simplify the [tex]\( \frac{0}{0} \)[/tex] indeterminate form, we can make use of the technique of rationalizing the denominator. Let's set [tex]\(u = \sqrt[3]{x} \)[/tex]:
Thus, [tex]\(u^3 = x\)[/tex].
The expression becomes:
[tex]\[
\frac{u^3 - 8}{u - 2}
\][/tex]
5. Factoring the Numerator:
Recognize that [tex]\(u^3 - 8\)[/tex] is a difference of cubes and can be factored using the formula [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex].
Here, [tex]\(a = u\)[/tex] and [tex]\(b = 2\)[/tex]:
[tex]\[
u^3 - 8 = (u - 2)(u^2 + 2u + 4)
\][/tex]
6. Simplifying the Fraction:
Substituting this back in, we get:
[tex]\[
\frac{(u - 2)(u^2 + 2u + 4)}{u - 2}
\][/tex]
Cancel the common factor [tex]\((u - 2)\)[/tex]:
[tex]\[
u^2 + 2u + 4
\][/tex]
7. Substitute Back [tex]\(\sqrt[3]{x}\)[/tex]:
Since [tex]\(u = \sqrt[3]{x}\)[/tex], replace [tex]\(u\)[/tex] back into the expression:
[tex]\[
(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4
\][/tex]
8. Take the Limit as [tex]\(x\)[/tex] Approaches 8:
Now, we take the limit of the simplified expression as [tex]\(x\)[/tex] approaches 8. Recall that [tex]\(\sqrt[3]{8} = 2\)[/tex]:
[tex]\[
\lim_{x \to 8} \left[(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4\right] = (2)^2 + 2(2) + 4 = 4 + 4 + 4 = 12
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{12}
\][/tex]
