1. **Problem Statement:** Evaluate the limit $$\lim_{x \to 0^+} \sin(x)^x$$ which is an indeterminate form of type $0^0$. 2. **Approach:** To evaluate limits of the form $f(x)^{g(x)}$, especially indeterminate forms like $0^0$, $1^\infty$, or $\infty^0$, we use the logarithm transformation: $$\lim_{x \to a} f(x)^{g(x)} = \lim_{x \to a} e^{g(x) \ln(f(x))} = e^{\lim_{x \to a} g(x) \ln(f(x))}$$ This reduces the problem to evaluating the limit of the product $g(x) \ln(f(x))$. 3. **Apply to the problem:** Here, $f(x) = \sin(x)$ and $g(x) = x$, so we consider: $$\lim_{x \to 0^+} x \ln(\sin(x))$$ 4. **Simplify inside the logarithm:** As $x \to 0^+$, $\sin(x) \approx x$ (using the small angle approximation), so: $$\ln(\sin(x)) \approx \ln(x)$$ Thus, $$\lim_{x \to 0^+} x \ln(\sin(x)) \approx \lim_{x \to 0^+} x \ln(x)$$ 5. **Evaluate $\lim_{x \to 0^+} x \ln(x)$:** Rewrite as: $$\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}$$ Apply L'Hôpital's Rule since numerator and denominator both approach $-\infty$ and $\infty$ respectively: $$\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$$ 6. **Conclusion:** Since $$\lim_{x \to 0^+} x \ln(\sin(x)) = 0,$$ then $$\lim_{x \to 0^+} \sin(x)^x = e^0 = 1.$$