To solve the limit [tex]\(\lim_{x \rightarrow 1} \frac{1 - x^{-1/3}}{1 - x^{-2/3}}\)[/tex], let's proceed step-by-step:
1. Substitute [tex]\(x = 1\)[/tex]:
First, we substitute [tex]\(x = 1\)[/tex] into the expression to see if it yields a determinate form.
[tex]\[
\frac{1 - 1^{-1/3}}{1 - 1^{-2/3}} = \frac{1 - 1}{1 - 1} = \frac{0}{0}
\][/tex]
Since the form is [tex]\(\frac{0}{0}\)[/tex], it is an indeterminate form, suggesting we may need to apply L'Hospital's Rule or simplify the expression further.
2. Using series expansion (optional but insightful):
Another method to handle indeterminate forms involves using a series expansion around [tex]\(x = 1\)[/tex]. However, for the sake of clarity, we can look at the expressions more conceptually.
3. L'Hospital's Rule:
Since we have a [tex]\(\frac{0}{0}\)[/tex] form, applying L'Hospital's Rule is a valid approach. This rule indicates that for limits of the form [tex]\(\frac{0}{0}\)[/tex], you can take the derivative of the numerator and the denominator separately.
Define [tex]\( f(x) = 1 - x^{-1/3} \)[/tex] and [tex]\( g(x) = 1 - x^{-2/3} \)[/tex].
[tex]\[
f(x) = 1 - x^{-1/3} \quad \text{and} \quad g(x) = 1 - x^{-2/3}
\][/tex]
Compute the derivatives:
[tex]\[
f'(x) = \frac{d}{dx}(1 - x^{-1/3}) = \frac{1}{3} x^{-4/3},
\][/tex]
[tex]\[
g'(x) = \frac{d}{dx}(1 - x^{-2/3}) = \frac{2}{3} x^{-5/3}.
\][/tex]
Now, applying L'Hospital's Rule to the limit:
[tex]\[
\lim_{x \to 1} \frac{1 - x^{-1/3}}{1 - x^{-2/3}} = \lim_{x \to 1} \frac{\frac{d}{dx}(1 - x^{-1/3})}{\frac{d}{dx}(1 - x^{-2/3})} = \lim_{x \to 1} \frac{\frac{1}{3} x^{-4/3}}{\frac{2}{3} x^{-5/3}}
\][/tex]
4. Simplification:
Simplify the limit expression:
[tex]\[
\lim_{x \to 1} \frac{\frac{1}{3} x^{-4/3}}{\frac{2}{3} x^{-5/3}} = \lim_{x \to 1} \frac{\frac{1}{3} x^{-4/3} \cdot 3}{2 x^{-5/3}} = \lim_{x \to 1} \frac{x^{-4/3}}{2 x^{-5/3}}
\][/tex]
[tex]\[
= \lim_{x \to 1} \frac{x^{-4/3}}{2 x^{-5/3}} = \lim_{x \to 1} \frac{1}{2} x^{1/3} = \frac{1}{2} \cdot 1 = \frac{1}{2}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{\frac{1}{2}}
\][/tex]
