Certainly! Let's dive into the problem step-by-step to solve it:
### Problem Statement
Given:
[tex]\[ \sqrt[3]{3} \][/tex]
We need to calculate:
[tex]\[ \lim _{x \rightarrow 3} \frac{x^3-27}{x-3} \][/tex]
### Step-by-Step Solution
1. Evaluate the expression [tex]\(\sqrt[3]{3}\)[/tex]:
The cube root of 3, [tex]\(\sqrt[3]{3}\)[/tex], is a constant value. Its approximate value is:
[tex]\[ \sqrt[3]{3} \approx 1.4422495703074083 \][/tex]
2. Evaluate the limit [tex]\(\lim _{x \rightarrow 3} \frac{x^3-27}{x-3}\)[/tex]:
First, observe that the denominator [tex]\(x - 3\)[/tex] causes a direct substitution of [tex]\(x = 3\)[/tex] to result in an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. Hence, we need to simplify the expression before evaluating the limit.
Notice that [tex]\(x^3 - 27\)[/tex] can be factored as a difference of cubes:
[tex]\[
x^3 - 27 = (x - 3)(x^2 + 3x + 9)
\][/tex]
Substituting this back into the limit expression gives:
[tex]\[
\lim_{x \rightarrow 3} \frac{x^3 - 27}{x - 3} = \lim_{x \rightarrow 3} \frac{(x - 3)(x^2 + 3x + 9)}{x - 3}
\][/tex]
We can cancel the common factor [tex]\(x - 3\)[/tex] in the numerator and the denominator:
[tex]\[
\lim_{x \rightarrow 3} (x^2 + 3x + 9)
\][/tex]
Now, we substitute [tex]\(x = 3\)[/tex] into the simplified expression:
[tex]\[
(3)^2 + 3(3) + 9 = 9 + 9 + 9 = 27
\][/tex]
Therefore, we find that:
[tex]\[
\lim _{x \rightarrow 3} \frac{x^3-27}{x-3} = 27
\][/tex]
### Conclusion
The evaluated values are:
[tex]\[ \sqrt[3]{3} \approx 1.4422495703074083 \][/tex]
[tex]\[ \lim _{x \rightarrow 3} \frac{x^3-27}{x-3} = 27 \][/tex]
