詳解：(1)發現\(\lim \limits_{x \to {0^ + }} {x^x}\) 為 \({0^0}\) 不定型

(2)改造表達方式以便使用羅必達法則

\(\lim \limits_{x \to {0^ + }} {x^x} = \lim\limits_{x \to {0^ + }} {e^{\ln {x^x}}} = \lim \limits_{x \to {0^ + }} {e^{x\ln x}} = {e^{ \lim \limits_{x \to {0^ + }} x\ln x}}\)

(3)考慮\(\lim \limits_{x \to {0^ + }} x\ln x\) 為 \(0 \cdot ( - \infty )\) 不定型

(4)將乘法改為除法以便使用羅必達法則

\(\lim \limits_{x \to {0^+}}x \ln x =\lim \limits_{x \to {0^+}} \frac { \ln x }{ ( \frac{1}{x}) } \) 為 \(\frac{-\infty}{\infty}\) 不定型

(5) 使用羅必達法則

\(\lim \limits_{x \to {0^+}} \frac {\ln x}{(\frac {1}{x})} \mathop = \limits^L = \lim \limits_{x \to {0^+}} \frac { ( \frac{1}{x})} {-( \frac {1}{x^2} )} = \lim \limits_{x \to {0^+}}(-x)=0\)

 (6)此結果代回(2)式

\(\lim \limits_{x \to {0^ + }} {x^x} = {e^{\lim \limits_{x \to {0^ + }} x\ln x}} = {e^0} = 1\)

故選(A)