1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{x^r - \lfloor x \rfloor}{x - \sqrt{x}}$$ where $\lfloor x \rfloor$ is the greatest integer function (floor function). 2. **Analyze the floor function:** Since $x \to 1$, and for values of $x$ close to but less than 1, $\lfloor x \rfloor = 0$, and at $x=1$, $\lfloor 1 \rfloor = 1$. The limit from the left and right might differ, so consider the behavior near 1. 3. **Evaluate numerator and denominator at $x=1$:** - Numerator: $1^r - \lfloor 1 \rfloor = 1 - 1 = 0$ - Denominator: $1 - \sqrt{1} = 1 - 1 = 0$ This is an indeterminate form $\frac{0}{0}$, so we can apply L'Hôpital's Rule. 4. **Apply L'Hôpital's Rule:** Differentiate numerator and denominator with respect to $x$: - Numerator derivative: $\frac{d}{dx}(x^r - \lfloor x \rfloor) = r x^{r-1} - 0 = r x^{r-1}$ (since floor function is constant almost everywhere except at integers, derivative is zero except at discontinuities) - Denominator derivative: $\frac{d}{dx}(x - \sqrt{x}) = 1 - \frac{1}{2 \sqrt{x}}$ 5. **Evaluate derivatives at $x=1$:** - Numerator derivative at 1: $r \cdot 1^{r-1} = r$ - Denominator derivative at 1: $1 - \frac{1}{2 \cdot 1} = 1 - \frac{1}{2} = \frac{1}{2}$ 6. **Compute the limit:** $$\lim_{x \to 1} \frac{x^r - \lfloor x \rfloor}{x - \sqrt{x}} = \frac{r}{\frac{1}{2}} = 2r$$ **Final answer:** $$\boxed{2r}$$