Certainly! Let's find the limit [tex]\(\lim_{x \rightarrow 1} \left( \frac{1 - x^{-\frac{1}{3}}}{1 - x^{-\frac{2}{3}}} \right)\)[/tex]. Here's the detailed, step-by-step solution:
1. Identify the problem: We need to find the limit of the function as [tex]\( x \)[/tex] approaches 1.
[tex]\[
\lim_{x \to 1} \left( \frac{1 - x^{-\frac{1}{3}}}{1 - x^{-\frac{2}{3}}} \right)
\][/tex]
2. Analyze the behavior at [tex]\( x = 1 \)[/tex]: Substitute [tex]\( x = 1 \)[/tex] into the function inside the limit.
[tex]\[
\text{Substitute } x = 1: \frac{1 - 1^{-\frac{1}{3}}}{1 - 1^{-\frac{2}{3}}} = \frac{1 - 1}{1 - 1} = \frac{0}{0}
\][/tex]
We get an indeterminate form [tex]\(\frac{0}{0}\)[/tex], which means we need to use algebraic manipulation or apply L'Hôpital's Rule.
3. Algebraic Manipulation: Rewrite the expressions inside the numerator and the denominator.
Let [tex]\( y = x^{-\frac{1}{3}} \)[/tex].
When [tex]\( x \to 1 \)[/tex], [tex]\( y \to 1 \text{ as well} \)[/tex].
Substitute [tex]\( y \)[/tex] back into the limit:
[tex]\[
\lim_{y \to 1} \left( \frac{1 - y}{1 - y^2} \right)
\][/tex]
4. Factor the Denominator: Use the difference of squares to factorize [tex]\( 1 - y^2 \)[/tex]:
[tex]\[
1 - y^2 = (1 - y)(1 + y)
\][/tex]
Rewrite the limit with the factorization:
[tex]\[
\lim_{y \to 1} \left( \frac{1 - y}{(1 - y)(1 + y)} \right)
\][/tex]
5. Simplify the Expression: Cancel the common factor [tex]\(1 - y\)[/tex] in the numerator and denominator:
[tex]\[
\lim_{y \to 1} \left( \frac{1 - y}{(1 - y)(1 + y)} \right) = \lim_{y \to 1} \left( \frac{1}{1 + y} \right)
\][/tex]
6. Substitute the Limit: Now, substitute [tex]\( y = 1 \)[/tex] back into the simplified expression:
[tex]\[
\frac{1}{1 + 1} = \frac{1}{2}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \rightarrow 1}\left(\frac{1 - x^{-\frac{1}{3}}}{1 - x^{-\frac{2}{3}}}\right) = \frac{1}{2}
\][/tex]
