To evaluate the limit [tex]\(\lim _{x \rightarrow 0} \frac{(x+8)^{\frac{1}{3}}-2}{\mid x \mid}\)[/tex], we need to carefully analyze the behavior of the function near [tex]\(x = 0\)[/tex]. The problem involves a combination of a cubic root function and an absolute value, which makes it non-trivial, but let's break it down step by step.
1. Original Function:
First, we look at the function we need to take the limit of:
[tex]\[
\frac{(x+8)^{\frac{1}{3}}-2}{\mid x \mid}
\][/tex]
2. Simplify with Substitution:
To simplify the expression, let's substitute [tex]\(u = x + 8\)[/tex]. As [tex]\(x \rightarrow 0\)[/tex], [tex]\(u \)[/tex] approaches [tex]\(8\)[/tex]. Under this substitution, the expression becomes:
[tex]\[
\frac{u^{\frac{1}{3}} - 2}{|u - 8|}
\][/tex]
3. Analyze the Limit:
Now we need to evaluate the limit of this function as [tex]\(u \to 8\)[/tex].
Consider the Taylor expansion of [tex]\(u^{\frac{1}{3}}\)[/tex] around [tex]\(u = 8\)[/tex]:
[tex]\[
u^{\frac{1}{3}} = (8 + (u-8))^{\frac{1}{3}} \approx 2 + \frac{1}{12}(u - 8) \quad \text{for } u \text{ near } 8
\][/tex]
4. Apply the Taylor Expansion:
Substitute the approximation back into the expression:
[tex]\[
\frac{(2 + \frac{1}{12}(u - 8)) - 2}{|u - 8|}
\][/tex]
Simplify the numerator:
[tex]\[
\frac{\frac{1}{12}(u - 8)}{|u - 8|} = \frac{1}{12} \frac{u - 8}{|u - 8|}
\][/tex]
5. Evaluate the Limit:
Consider the quotient [tex]\(\frac{u - 8}{|u - 8|}\)[/tex]:
- When [tex]\(u > 8\)[/tex], [tex]\(|u-8| = u-8\)[/tex], so [tex]\(\frac{u - 8}{|u - 8|} = 1\)[/tex]
- When [tex]\(u < 8\)[/tex], [tex]\(|u-8| = -(u-8)\)[/tex], so [tex]\(\frac{u - 8}{|u - 8|} = -1\)[/tex]
As [tex]\(u\)[/tex] approaches [tex]\(8\)[/tex] from either side, this quotient simplifies to:
[tex]\[
\hat{\lim_{u \to 8}} \frac{u - 8}{|u - 8|} = 1
\][/tex] or [tex]\(\hat{\lim_{u \to 8}} \frac{u - 8}{|u - 8|} = -1\)[/tex]
Plugging back into the initial expansion:
[tex]\[
\frac{1}{12} \lim_{u \to 8} \frac{u - 8}{|u - 8|} = \frac{1}{12}
\][/tex]
Hence the limit is:
[tex]\[
\boxed{\frac{1}{12}}
\][/tex]
