To solve the limit:

$\lim_{{x \to 1}} \frac{\sqrt[3]{x} - 1}{\sqrt{x} - 1}$

we'll analyze it by using the Taylor expansions or by simplifying each term around $x = 1$.

Step-by-Step Solution

1. Recognize Indeterminate Form: As $x \to 1$, both the numerator and denominator go to 0. This is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule: Differentiate the numerator and the denominator with respect to $x$.

   • Numerator: $\frac{d}{dx} \left( \sqrt[3]{x} - 1 \right) = \frac{1}{3} x^{-\frac{2}{3}}$
   • Denominator: $\frac{d}{dx} \left( \sqrt{x} - 1 \right) = \frac{1}{2} x^{-\frac{1}{2}}$

   So the limit becomes:

   $\lim_{{x \to 1}} \frac{\frac{1}{3} x^{-\frac{2}{3}}}{\frac{1}{2} x^{-\frac{1}{2}}}$

3. Simplify the Expression: Rewrite the expression by dividing the fractions:

   $= \lim_{{x \to 1}} \frac{\frac{1}{3} x^{-\frac{2}{3}}}{\frac{1}{2} x^{-\frac{1}{2}}} = \lim_{{x \to 1}} \frac{2}{3} x^{\frac{1}{6}}$

4. Evaluate the Limit: Now substitute $x = 1$ directly:

   $= \frac{2}{3} \cdot 1^{\frac{1}{6}} = \frac{2}{3}$

Final Answer

$\lim_{{x \to 1}} \frac{\sqrt[3]{x} - 1}{\sqrt{x} - 1} = \frac{2}{3}$

Would you like more details, or do you have any questions?

1. How does L'Hôpital's Rule apply to different types of indeterminate forms?
2. What are alternative methods for solving limits involving roots?
3. How does the behavior of fractional exponents affect limits near specific points?
4. What are Taylor expansions and how are they useful in limits?
5. How can we generalize this limit for other roots?

Tip: Always check for the simplest method first; sometimes rewriting expressions avoids more complex rules like L'Hôpital's Rule.