1. We are asked to find the limit $$\lim_{x \to 1^+} (x^x - 1)^h$$ where $h$ is a constant. 2. First, evaluate the inner expression $x^x - 1$ as $x$ approaches 1 from the right. 3. Recall that $x^x = e^{x \ln x}$. When $x \to 1$, $x \ln x \to 1 \cdot 0 = 0$, so $x^x \to e^0 = 1$. 4. Therefore, $x^x - 1 \to 0$ as $x \to 1^+$. 5. To understand the behavior of $(x^x - 1)^h$, we need the rate at which $x^x - 1$ approaches 0. 6. Use the expansion for $x^x$ near $x=1$: $$x^x = e^{x \ln x} = e^{(1 + (x-1)) \ln(1 + (x-1))}$$ 7. For small $t = x-1$, $\ln(1+t) \approx t - \frac{t^2}{2}$, so $$x \ln x = (1 + t)(t - \frac{t^2}{2}) = t + t^2 - \frac{t^2}{2} - \frac{t^3}{2} = t + \frac{t^2}{2} + O(t^3)$$ 8. Then, $$x^x = e^{t + \frac{t^2}{2} + O(t^3)} = e^t e^{\frac{t^2}{2} + O(t^3)} = (1 + t + \frac{t^2}{2} + O(t^3))(1 + \frac{t^2}{2} + O(t^3)) = 1 + t + t^2 + O(t^3)$$ 9. So, $$x^x - 1 = t + t^2 + O(t^3) = (x-1) + (x-1)^2 + O((x-1)^3)$$ 10. As $x \to 1^+$, $x^x - 1 \sim (x-1)$. 11. Therefore, $$(x^x - 1)^h \sim (x-1)^h$$ 12. The limit depends on $h$: - If $h > 0$, then $(x-1)^h \to 0$. - If $h = 0$, then limit is 1. - If $h < 0$, then $(x-1)^h \to +\infty$. 13. Final answer: $$\lim_{x \to 1^+} (x^x - 1)^h = \begin{cases} 0, & h > 0 \\ 1, & h = 0 \\ +\infty, & h < 0 \end{cases}$$