1. **State the problem:** Evaluate the limit $$\lim_{x \to 1} \frac{\sqrt{x} - \sqrt[3]{x}}{x - 1}$$. 2. **Recognize the indeterminate form:** Substituting $x=1$ gives $$\frac{\sqrt{1} - \sqrt[3]{1}}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0}$$, which is indeterminate. So, we can apply L'Hôpital's Rule. 3. **Recall L'Hôpital's Rule:** If $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 4. **Differentiate numerator and denominator:** - Numerator: $$f(x) = \sqrt{x} - \sqrt[3]{x} = x^{\frac{1}{2}} - x^{\frac{1}{3}}$$ - Derivative: $$f'(x) = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{2}{3}}$$ - Denominator: $$g(x) = x - 1$$ - Derivative: $$g'(x) = 1$$ 5. **Apply L'Hôpital's Rule:** $$\lim_{x \to 1} \frac{\sqrt{x} - \sqrt[3]{x}}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{2}{3}}}{1} = \frac{1}{2} \cdot 1^{-\frac{1}{2}} - \frac{1}{3} \cdot 1^{-\frac{2}{3}}$$ 6. **Evaluate the limit:** Since $1^{-\frac{1}{2}} = 1$ and $1^{-\frac{2}{3}} = 1$, we get $$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$ **Final answer:** $$\boxed{\frac{1}{6}}$$