函数极限的定义及运算

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函数极限的定义

函数极限的定义有几种等价表述, 这里介绍柯西 (Cauchy) 提出的 \varepsilon-\delta 式定义.

设 x_0, A \in \mathbb{R}, 并设函数 f(x) 在 x_0 附近有定义. 如果对任意 \varepsilon > 0, 存在 \delta > 0, 使得只要 0 < |x - x_0| < \delta, 就有

|f(x) - A| < \varepsilon,

则称 x \to a 时函数 f(x) 的极限是 A, 记作

\lim_{x \to x_0} f(x) = A.

函数极限的运算

在以下的讨论中, 我们约定: 设 f(x) 和 g(x) 在 x_{0} 处极限存在, 即 \displaystyle \lim_{x \to x_{0}} f(x) = A, \lim_{x \to x_{0}} g(x) = B.

定理 1

\lim_{x \to x_0} kf(x) = k \lim_{x \to x_0} f(x).

证明

由 \displaystyle \lim_{x \to x_0} f(x) = A \Rightarrow \forall \frac{\varepsilon}{|k|} > 0, \exists \delta > 0, 使得 0 < |x - x_0| < \delta, 均有

|f(x) - A| < \frac{\varepsilon}{|k|},

因此 \forall \varepsilon > 0, \exists \delta > 0, 使得 0 < |x - x_0| < \delta, 均有

|k| \cdot |f(x) - A| < \varepsilon \Rightarrow |kf(x)-kA|<\varepsilon
\Rightarrow \lim_{x \to x_0} kf(x) = kA = k \lim_{x \to x_0} f(x).

定理 2

\lim_{x \to x_0} \left[ f(x) \pm g(x) \right] = \lim_{x \to x_0} f(x) \pm \lim_{x \to x_0} g(x).

证明

这里展示 \displaystyle \lim_{x \to x_0} \left[ f(x) + g(x) \right] = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x), 的证明, 对于 \displaystyle \lim_{x \to x_0} \left[ f(x) - g(x) \right] = \lim_{x \to x_0} f(x) - \lim_{x \to x_0} g(x), 读者可尝试自行证明.

由 \displaystyle \lim_{x \to x_0} f(x) = A \Rightarrow \forall \frac{\varepsilon}{2} > 0, \exists \delta_1 > 0, 使得 0 < |x - x_0| < \delta_1, 均有

|f(x) - A| < \frac{\varepsilon}{2};

由 \displaystyle \lim_{x \to x_0} g(x) = B \Rightarrow \forall \frac{\varepsilon}{2} > 0, \exists \delta_2 > 0, 使得 0 < |x - x_0| < \delta_2, 均有

|g(x) - B| < \frac{\varepsilon}{2},

取 \delta = \min \{ \delta_1, \delta_2 \}, 则 \forall \varepsilon > 0, \exists \delta > 0, 使得 0 < |x - x_0| < \delta, 均有

|f(x) + g(x) - A - B| \leq |f(x) - A| + |g(x) - B| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
\Rightarrow \lim_{x \to x_0} \left[ f(x) + g(x) \right] = A + B = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x).

定理 3

\lim_{x \to x_0} f(x) \cdot g(x) = \lim_{x \to x_0} f(x) \cdot \lim_{x \to x_0} g(x);
\lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x \to x_0} f(x)}{\displaystyle \lim_{x \to x_0} g(x)}.

证明

这里展示 \displaystyle \lim_{x \to x_0} f(x) \cdot g(x) = \lim_{x \to x_0} f(x) \cdot \lim_{x \to x_0} g(x), 的证明, 对于 \displaystyle \lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{\displaystyle \lim_{x \to x_0} f(x)}{\displaystyle \lim_{x \to x_0} g(x)}, 读者可尝试自行证明.

由 \displaystyle \lim_{x \to x_0} g(x) = B 存在, 设 M \in \mathbb{R} 且 M > 0, 满足 \exists \delta_1 > 0, 使得 0 < |x - x_0| < \delta_1, 均有

|g(x)| < M,

由 \displaystyle \lim_{x \to x_0} f(x) = A \Rightarrow \forall \frac{\varepsilon}{2M} > 0, \exists \delta_2 > 0, 使得 0 < |x - x_0| < \delta_2, 均有

|f(x) - A| < \frac{\varepsilon}{2M};

由 \displaystyle \lim_{x \to x_0} g(x) = B \Rightarrow \forall \frac{\varepsilon}{2|A|} > 0, \exists \delta_3 > 0, 使得 0 < |x - x_0| < \delta_3, 均有

|g(x) - B| < \frac{\varepsilon}{2|A|},

取 \delta = \min \{ \delta_1, \delta_2, \delta_3 \}, 则 \forall \varepsilon > 0, \exists \delta > 0, 使得 0 < |x - x_0| < \delta, 均有

\begin{align*} |f(x) \cdot g(x) - A \cdot B| &= |f(x) \cdot g(x) + A \cdot g(x) - A \cdot g(x) - A \cdot B| \\ &= |[f(x) - A] \cdot g(x) + A \cdot [g(x) - B]| \\ &\leq |f(x) - A| \cdot |g(x)| + |A| \cdot |g(x) - B| \\ &< \frac{\varepsilon}{2M} \cdot M + \frac{\varepsilon}{2|A|} \cdot |A| \\ &= \varepsilon \end{align*}
\Rightarrow \lim_{x \to x_0} f(x) \cdot g(x) = A \cdot B = \lim_{x \to x_0} f(x) \cdot \lim_{x \to x_0} g(x).