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

(2)使用羅必達法則

\(\lim\limits_{x \to 0} \frac{{\sin x}}{x}\mathop = \limits^L \lim\limits_{x \to 0} \frac{{{{\left( {\sin x} \right)}^\prime }}}{{{{\left( x \right)}^\prime }}} = \lim\limits_{x \to 0} \frac{{\cos x}}{1} = \frac{{\cos 0}}{1} = 1\)

故選(A)