一元函数一点处极限的运算之相乘运算

一、运算关系

设函数 $f\left(x\right)$ 和函数 $g\left(x\right)$ 在点 $x_{0}$ 处的极限分别为 $A$ 和 $B$, 则下式成立：

$$
\lim_{x \to x_{0}} \left(f\left(x\right) \cdot g\left(x\right)\right) = A \cdot B
$$

二、运算关系的证明

已知 $\displaystyle \lim_{x \to x_{0}} f\left(x\right) = A$, $\lim_{x \to x_{0}} g\left(x\right) = B$，因此，$\forall \varepsilon > 0$:

• $\exists \delta_{2} > 0,\ \forall x \left(0 < \left|x – x_{0}\right| < \delta_{2}\right)$, 使得 $\left|f\left(x\right) – A\right| < \varepsilon$；
• $\exists \delta_{3} > 0,\ \forall x \left(0 < \left|x – x_{0}\right| < \delta_{3}\right)$, 使得 $\left|g\left(x\right) – B\right| < \varepsilon$.

取 $\delta = \min \left\{\delta_{1}, \delta_{2}\right\}$，由三角不等式，当 $0 < \left|x – x_{0}\right| < \delta$ 时，有：

$$
\begin{aligned}
& \ \left|f\left(x\right) g\left(x\right) – AB\right| \\ \
= & \ \left|f\left(x\right)\left(g\left(x\right) – B\right) + B\left(f\left(x\right) – A\right)\right| \\ \
\leqslant & \ \left|f\left(x\right)\right| \cdot \left|g\left(x\right) – B\right| + \left|B\right| \cdot \left| f \left( x \right) – A \right| \\ \
= & \ \left| A \right| \cdot \left|g\left(x\right) – B\right| + \left|B\right| \cdot \left| f \left( x \right) – A \right| \\ \
< & \ \left( \left| A \right| + \left| B \right| \right) \cdot \epsilon \to 0 \\ \
\textcolor{lightgreen}{ \leadsto } & \ \left|f\left(x\right) g\left(x\right) – AB\right| \to 0 \\ \
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ f\left(x\right) g\left(x\right) = AB }
\end{aligned}
$$

综上可知：

$$
\textcolor{lightgreen}{
\lim_{x \to x_{0}} \left(f \left( x \right) \cdot g\left(x\right)\right) = A \cdot B
}
$$

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