一元函数一点处极限的性质之夹逼性

一、性质

设函数 $g\left(x\right)$ 和函数 $h\left(x\right)$ 在点 $x_{0}$ 处的极限均为 $A$, 若存在 $\rho > 0$, 使得当 $0 < \left|x – x_{0}\right| < \rho$ 时，$g\left(x\right) \leqslant f\left(x\right) \leqslant h\left(x\right)$ 成立，则有：

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

二、性质的证明

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

• $\exists \delta_{1} > 0, , \forall x\left(0 < \left|x – x_{0}\right| < \delta_{1}\right)$, 使得 $\left|g\left(x\right) – A\right| < \epsilon$, 即：

$$
\begin{aligned}
& \ \left|g\left(x\right) – A\right| < \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ – \epsilon < g\left(x\right) – A < \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{g\left(x\right) > A – \epsilon}
\end{aligned}
$$

• $\exists \delta_{2} > 0, , \forall x\left(0 < \left|x – x_{0}\right| < \delta_{2}\right)$, 使得 $\left|h\left(x\right) – A\right| < \epsilon$, 即：

$$
\begin{aligned}
& \ \left|h\left(x\right) – A\right| < \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ – \epsilon < h\left(x\right) – A < \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{h\left(x\right) < A + \epsilon}
\end{aligned}
$$

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

$$
\begin{aligned}
& \ A – \epsilon < g\left(x\right) \leqslant f\left(x\right) \leqslant h\left(x\right) < A + \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ A – \epsilon \leqslant f\left(x\right) \leqslant A + \epsilon \\ \
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{\displaystyle \lim_{x \to x_{0}} f\left(x\right) = A}
\end{aligned}
$$

综上可知，性质得证.

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