1. **State the problem:** We want to find the limit $$\lim_{x \to 1^-} \frac{x^y - \lfloor x \rfloor}{x - \sqrt{x}}$$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. 2. **Analyze the floor function near 1 from the left:** For $x$ approaching 1 from the left, $x < 1$, so $\lfloor x \rfloor = 0$ because the greatest integer less than or equal to any number just less than 1 is 0. 3. **Rewrite the limit using this:** $$\lim_{x \to 1^-} \frac{x^y - 0}{x - \sqrt{x}} = \lim_{x \to 1^-} \frac{x^y}{x - \sqrt{x}}$$ 4. **Evaluate the denominator near 1:** $$x - \sqrt{x} = x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$ As $x \to 1$, $x^{1/2} \to 1$, so denominator $\to 1 \cdot (1 - 1) = 0$. 5. **Evaluate the numerator near 1:** $$x^y \to 1^y = 1$$ 6. **We have an indeterminate form $\frac{1}{0}$, but since denominator approaches 0, we check the sign:** For $x < 1$, $x^{1/2} < 1$, so $x^{1/2} - 1 < 0$, thus denominator $x - \sqrt{x} < 0$. 7. **Therefore, numerator $\to 1$, denominator $\to 0^-$, so the fraction tends to $-\infty$ if $y$ is finite and constant.** **Final answer:** $$\lim_{x \to 1^-} \frac{x^y - \lfloor x \rfloor}{x - \sqrt{x}} = -\infty$$