一、题目
已知，当 $x \rightarrow 0$ 时：
$$
\begin{aligned}
\alpha(x) & = \tan x-\sin x \\ \\
\beta(x) & = \sqrt{1+x^{2}}-\sqrt{1-x^{2}} \\ \\
\gamma(x) & = \int_{0}^{1-\cos x} \sin t \mathrm{~d} t
\end{aligned}
$$
都是无穷小，将它们关于 $x$ 的阶数从低到高排列，正确的顺序为（ ）
(A). $\alpha(x)$, $\beta(x)$, $\gamma(x)$
(B). $\alpha(x)$, $\gamma(x)$, $\beta(x)$
(C). $\beta(x)$, $\alpha(x)$, $\gamma(x)$
(D). $\gamma(x)$, $\alpha(x)$, $\beta(x)$
难度评级：
二、解析 
当 $x \rightarrow 0$ 时，由题可知：
$$
\begin{aligned}
\textcolor{springgreen}{\beta(x)} \\ \\
& = \sqrt{1+x^{2}}-\sqrt{1-x^{2}} \\ \\
& = \frac{2 x^{2}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \\ \\
& = \frac{2x^{2}}{2} \sim \textcolor{springgreen}{x^{2}}
\end{aligned}
$$
又可知：
$$
\begin{aligned}
\textcolor{springgreen}{\alpha(x)} \\ \\
& = \tan x – \sin x \\ \\
& = \frac{\sin x}{\cos x} – \cos x \cdot \frac{\sin x}{\cos x} \\ \\
& = \tan x – \cos x \tan x \\ \\
& = (1-\cos x) \tan x \\ \\
& = \frac{x^{2}}{2} \cdot x \sim \textcolor{springgreen}{\frac{x^{3}}{2}}
\end{aligned}
$$
还可知
$$
\begin{aligned}
\textcolor{springgreen}{\gamma(x)} \\ \\
& =\int_{0}^{1-\cos x} \sin t \mathrm{~d} t \\ \\
& = -\cos t \Big|_{0} ^{1-\cos x} \\ \\
& =1-\cos (1-\cos x) \\ \\
& = \frac{(\frac{x^{2}}{2})^{2}}{2} \\ \\
& = \frac{(1-\cos x)^{2}}{2} \sim \textcolor{springgreen}{\frac{x^{4}}{8}}
\end{aligned}
$$
如果按照 $x$ 的次幂从低到高排列，就是：
$$
x^{2}, \quad \frac{x^{3}}{2}, \quad \frac{x^{4}}{8} \Rightarrow
$$
$$
\beta(x), \quad \alpha(x), \quad \gamma(x)
$$
综上可知， 本 题 应 选 C 