借助极限与无穷小的关系求解极限

一、题目

已知 $\lim_{x \rightarrow 0} \frac{\sin 6x + x f(x)}{x^{3}} = 0$, 则 $\lim_{x \rightarrow 0} \frac{6 + f(x)}{x^{2}} = ?$

二、解析

由于：

$$
\lim_{x \rightarrow 0} \frac{\sin 6x + x f(x)}{x^{3}} = 0
$$

因此，根据极限与无穷小的关系可知：

$$
\begin{aligned}
\frac{\sin 6x + x f(x)}{x^{3}} & = 0 + \alpha(x) \\ \\
& = \alpha(x)
\end{aligned}
$$

其中 $\alpha(x)$ 为 $x \rightarrow 0$ 时的无穷小量.

于是可知：

$$
\begin{aligned}
& \frac{\sin 6x + x f(x)}{x^{3}} = \alpha(x) \\ \\
\leadsto \ & \sin 6x + x f(x) = x^{3} \alpha (x) \\ \\
\leadsto \ & x f(x) = x^{3} \alpha (x) – \sin 6x \\ \\
\leadsto \ & \textcolor{lightgreen}{ f(x) = x^{2} \alpha(x) – \frac{\sin 6x}{x} }
\end{aligned}
$$

于是：

$$
\begin{aligned}
& \lim_{x \rightarrow 0} \frac{6 + \textcolor{lightgreen}{ f(x) }}{x^{2}} \\ \\
& = \lim_{x \rightarrow 0} \frac{6 + \textcolor{lightgreen}{ x^{2} \alpha(x) – \frac{\sin 6x}{x} } }{x^{2}} \\ \\
& = \lim_{x \rightarrow 0} \frac{6}{x^{2}} + \lim_{x \rightarrow 0} \alpha(x) + \lim_{x \rightarrow 0} \frac{6x – \sin 6x}{x^{3}} \\ \\
& = 0 + 0 + \lim_{x \rightarrow 0} \frac{6 – 6 \cos 6x}{3x^{2}} \\ \\
& = 2 \lim_{x \rightarrow 0} \frac{1 – \cos 6x}{x^{2}} \\ \\
& \leadsto \textcolor{gray}{\text{洛必达运算}} \\ \\
& = 2 \cdot \lim_{x \rightarrow 0} \frac{6 \sin 6x}{2x} \\ \\
& = \textcolor{green}{\boldsymbol{36}}
\end{aligned}
$$

综上可知，本 题 应 选 A

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