[原创]高等数学笔记(13)

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【前言】
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【正文】
四、极限的四则运算公式
以下公式中，自变量都是 x \to {x_0} ，或者都是 x \to \infty
设 \lim f(x) = A,\;\lim g(x) = B ，则有：

1.  \lim \left[ {f(x) \pm g(x)} \right] = A \pm B = \lim f(x) \pm \lim g(x)
2.  \lim \left[ {f(x)g(x)} \right] = AB = \lim f(x)\lim g(x)
若 C 是常数，则 \lim \left[ {Cf(x)} \right] = CA = C\lim f(x)
若 n 是正整数， \lim {\left[ {f(x)} \right]^n} = \lim \left[ {f(x) \cdot f(x) \cdots f(x)} \right] = {A^n} = {\left[ {\lim f(x)} \right]^n}
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证明：
由函数极限与无穷小的关系：
\lim f(x) = A \Leftrightarrow f(x) = A + \alpha (x),\;\lim \alpha (x) = 0
\lim g(x) = B \Leftrightarrow g(x) = B + \beta (x),\;\lim \beta (x) = 0
f(x)g(x) = \left[ {A + \alpha (x)} \right]\left[ {B + \beta (x)} \right] = AB + \left[ {A\beta (x) + B\alpha (x) + \alpha (x)\beta (x)} \right] = AB + \gamma (x)
其中  \gamma (x) = A\beta (x) + B\alpha (x) + \alpha (x)\beta (x)
由无穷小的性质，可知 \gamma (x) 是无穷小，即 f(x)g(x) = AB + \gamma (x),\;\lim \gamma (x) = 0
\lim \left[ {f(x)g(x)} \right] = AB = \lim f(x) \cdot \lim g(x)
证毕。
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3. 若 B \ne 0 ，则 \lim \frac{{f(x)}}{{g(x)}} = \frac{A}{B} = \frac{{\lim f(x)}}{{\lim g(x)}}
证：

\frac{{f(x)}}{{g(x)}} - \frac{A}{B} = \frac{{A + \alpha (x)}}{{B + \beta (x)}} - \frac{A}{B} = \frac{{B\alpha (x) - A\beta (x)}}{{B\left[ {B + \beta (x)} \right]}}
\frac{{f(x)}}{{g(x)}} = \frac{A}{B} + \gamma (x),\;\gamma (x) = \frac{{B\alpha (x) - A\beta (x)}}{{B\left[ {B + \beta (x)} \right]}}
由于 B\alpha (x),A\beta (x) 都是无穷小
因此 \lim \left[ {B\alpha (x) - A\beta (x)} \right] = 0 ，即分子为无穷小
又因为 \lim B\left[ {B + \beta (x)} \right] = \lim \left[ {{B^2} + B\beta (x)} \right] = {B^2} \ne 0
由无穷小性质3可知 \lim \gamma (x) = 0
证毕。
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4. 设 f(x) \ge g(x) ，而 \lim f(x) = A,\;\lim g(x) = B ，则必有 A \ge B
证：
令 F(x) = f(x) - g(x) \ge 0 ，则权限的四则运算公式得：
\lim F(x) = \lim \left[ {f(x) - g(x)} \right] = \lim f(x) - \lim g(x) = A - B
根据函数值与极限值的同号性定理，可知：
\lim F(x) = A - B \ge 0 \Rightarrow A \ge B
证毕。

例1. 求 \mathop {\lim }\limits_{x \to - 1} \frac{{2{x^2} + x - 4}}{{3{x^2} + 2}}
解：

\begin{array}{l} \mathop {\lim }\limits_{x \to - 1} (3{x^2} + 2) = \mathop {\lim }\limits_{x \to - 1} (3{x^2}) + \mathop {\lim }\limits_{x \to - 1} 2 = 3\mathop {\lim }\limits_{x \to - 1} {x^2} + 2\\ = 3{\left( {\mathop {\lim }\limits_{x \to - 1} x} \right)^2} + 2 = 3 \cdot {( - 1)^2} + 2 = 5 \end{array}

\begin{array}{l} \mathop {\lim }\limits_{x \to - 1} (2{x^2} + x - 4) = \mathop {\lim }\limits_{x \to - 1} (2{x^2}) + \mathop {\lim }\limits_{x \to - 1} x - \mathop {\lim }\limits_{x \to - 1} 4\\ = 2{\left( {\mathop {\lim }\limits_{x \to - 1} x} \right)^2} - 1 - 4 = 2 \cdot {( - 1)^2} - 5 = - 3 \end{array}
所以 \mathop {\lim }\limits_{x \to - 1} \frac{{2{x^2} + x - 4}}{{3{x^2} + 2}} = \frac{{\mathop {\lim }\limits_{x \to - 1} (2{x^2} + x - 4)}}{{\mathop {\lim }\limits_{x \to - 1} (3{x^2} + 2)}} = - \frac{3}{5}
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一般地，有：
R(x) = \frac{{{a_0}{x^n} + {a_1}{x^{n - 1}} + \cdots + {a_{n - 1}}x + {a_n}}}{{{b_0}{x^m} + {b_1}{x^{m - 1}} + \cdots + {b_{m - 1}}x + {b_m}}}
分母的极限：

\begin{array}{l} \mathop {\lim }\limits_{x \to {x_0}} ({b_0}{x^m} + {b_1}{x^{m - 1}} + \cdots + {b_{m - 1}}x + {b_m}) = \mathop {\lim }\limits_{x \to {x_0}} \sum\limits_{j = 0}^m {{b_j}{x^{m - j}}} \\ = \sum\limits_{j = 0}^m {\left( {\mathop {\lim }\limits_{x \to {x_0}} {b_j}{x^{m - j}}} \right)} = \sum\limits_{j = 0}^m {{b_j}{x_0}^{m - j}} \end{array}
分子的极限：
\mathop {\lim }\limits_{x \to {x_0}} ({a_0}{x^n} + {a_1}{x^{n - 1}} + \cdots + {a_{n - 1}}x + {a_n}) = \mathop {\lim }\limits_{x \to {x_0}} \sum\limits_{i = 0}^m {{a_i}{x^{n - i}}} = \cdots = \sum\limits_{i = 0}^n {{a_i}{x_0}^{n - i}}
若分母极限 \sum\limits_{j = 0}^m {{b_j}{x_0}^{m - j}} \ne 0 ，则：
\mathop {\lim }\limits_{x \to {x_0}} R(x) = \frac{{\sum\limits_{i = 0}^n {{a_i}{x_0}^{n - i}} }}{{\sum\limits_{j = 0}^m {{b_j}{x_0}^{m - j}} }} = R({x_0})
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例2. 求 \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{x^2} - 5x + 6}}
解：
由于 \mathop {\lim }\limits_{x \to 2} ({x^2} - 5x + 6) = 4 - 10 + 6 = 0 ，所以不能用极限的四则运算公式。
原式 = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 1)(x - 2)}}{{(x - 3)(x - 2)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x - 3}} = \frac{1}{{ - 1}} = - 1
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（第13课完）

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