【答案】C

【解析】由题易知，\(x \to 0\)时，\(f(x)\)是\(g(x)\)高阶无穷小.

则有\(\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = 0\)及\(\lim\limits_{x \to 0}f(x) = 0\)，\(\lim\limits_{x \to 0}g(x) = 0\).

又\(f(x)\)，\(g(x)\)在\(x = 0\)某去心邻域内有定义且不恒等于\(0\).

故对于A选项，等式两端同除\(g(x)\)得：

\(\frac{f(x)}{g(x)} + 1 = \frac{o[g(x)]}{g(x)}\)

取极限得

\(\lim\limits_{x \to 0}(\frac{f(x)}{g(x)} + 1) = \lim\limits_{x \to 0}\frac{o[g(x)]}{g(x)}\)

即\(0 + 1 = 0\)，显然A不成立.

对于B选项，等式两端同除\(f^2(x)\)得

\(\frac{g(x)}{f(x)} = \frac{o[f^2(x)]}{f^2(x)}\)

两端取极限得\(\lim\limits_{x \to 0}\frac{g(x)}{f(x)} = \lim\limits_{x \to 0}\frac{o[f^2(x)]}{f^2(x)}\)，即\(\infty = 0\)，显然不成立.

对于C选项，等式两端同除\(g(x)\)得

\(\frac{f(x)}{g(x)} = \frac{o[\mathrm{e}^{g(x)} - 1]}{g(x)}\)

取极限得\(\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = \lim\limits_{x \to 0}\frac{o[\mathrm{e}^{g(x)} - 1]}{g(x)} = \lim\limits_{x \to 0}\frac{o[g(x)]}{g(x)}\)

显然有\(0 = 0\)，故C正确.

对于D等式两端同除\(g(x)\)得

\(\frac{f(x)}{g^2(x)} = \frac{o[g^2(x)]}{g^2(x)}\)

取极限得\(\lim\limits_{x \to 0}\frac{f(x)}{g^2(x)} = \lim\limits_{x \to 0}\frac{o[g^2(x)]}{g^2(x)}\)，显然不成立.

综上选C.