洛必达法则及其证明

洛必达法则内容

定理1：0/0型

若:

1. \(\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = 0\),

2. \(f(x), g(x)\)在\(\mathring{U}(a)\)（点\(a\)的某去心邻域）内可导,

3. \(g^{\prime}(x) \neq 0\),

4. \(\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)（\(A\)为有限数或无穷）.

则\(\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)。

其中\(x \to a\)改为\(x \to a + 0\)或\(x \to a - 0\)结论仍然成立。

定理1证明

证明：如果\(f(x), g(x)\)在\(x = a\)处连续，显然有\(f(a) = 0, g(a) = 0\)。如果\(x = a\)是\(f(x)\)，\(g(x)\)的不连续点，因为极限存在，一定是可去间断点。不妨设\(f(a) = 0, g(a) = 0\)，则可以认为\(f(x)\)与\(g(x)\)在\(x = a\)处连续。设\(x\)为\(a\)的邻域内一点，那么，在以\(x\)及\(a\)为端点的区间上，柯西中值定理的条件均满足，因此有 \[\dfrac{f(x)}{g(x)} = \dfrac{f(x) - f(a)}{g(x) - g(a)} = \dfrac{f^{\prime}(\xi)}{g^{\prime}(\xi)}\] 其中ξ在\(x\)与\(a\)之间，故当\(x \to a\)时，\(\xi \to a\)，对两端取极限得 \[\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(\xi)}{g^{\prime}(\xi)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)}\] 故定理1成立。

定理2：∞/∞型

若:

1. \(\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = \infty\),

2. \(f(x)\)和\(g(x)\)在\(\mathring{U}(a)\)内可导,

3. \(g^{\prime}(x) \neq 0\),

4. \(\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)（\(A\)为有限数或无穷）。

则\(\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)。

其中\(x \to a\)改为\(x \to a + 0\)或\(x \to a - 0\)结论仍然成立。

定理2证明

证明：∵\(\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)，∴对\(\forall \varepsilon > 0 \exists x_{1} \in \mathring{U}(a)\)，对于一切\(x \in (a, x_{1})\)（或\(x \in (x_{1}, a)\)），都有 \[\left| \dfrac{f^{\prime}(x)}{g^{\prime}(x)} - A \right| < \dfrac{\varepsilon}{2}, \tag{3}\]

对\(f(x)\)与\(g(x)\)在\([x, x_{1}]\)上应用柯西中值定理，则有 \[\left| \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} - A \right| = \dfrac{f^{\prime}(\xi)}{g^{\prime}(\xi)} \tag{4}\]

(4)式中\(\xi \in (x, x_{1}) \subset (a, x_{1})\).

由(3)，(4)可得 \[\left| \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} - A \right| < \dfrac{\varepsilon}{2}, \tag{5}\]

又因为，

\(\left| \dfrac{f(x)}{g(x)} - \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right|\)\(= \left| \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right| \cdot \left| \dfrac{g(x_{1}) - g(x) \cdot f(x)}{[f(x_{1}) - f(x)] \cdot g(x)} - 1 \right|\)\(= \left| \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right| \cdot \left| \dfrac{g(x_{1}) / g(x) - 1}{f(x_{1}) / f(x) - 1} - 1 \right|\),

由(5)式可知，当\(x \rightarrow a\)时，\(\dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)}\)是有界量，又由\(\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = \infty\)，当\(x \rightarrow a\)时， \(\dfrac{g(x_{1}) / g(x) - 1}{f(x_{1}) / f(x) - 1} - 1\)是无穷小量，进而有，当\(x \rightarrow a\)时，\(\left| \dfrac{f(x)}{g(x)} - \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right|\)也是无穷小量，即\(∃δ > 0\)，使得当\(a < x < a + \delta\)时有 \(\left| \dfrac{f(x)}{g(x)} - \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right| < \dfrac{\varepsilon}{2}\).

进而对一切\(a < x < a + \delta\)，由(5)式可得\(\left| \dfrac{f(x)}{g(x)} - A \right|\)\(= \left| \dfrac{f(x)}{g(x)} - \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} + \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} - A \right|\)\(\leq \left| \dfrac{f(x)}{g(x)} - \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} \right| + \left| \dfrac{f(x_{1}) - f(x)}{g(x_{1}) - g(x)} - A \right|\)\(< \varepsilon\).

故有\(\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\).

在定理2中，若对条件进行弱化，洛必达法则仍然成立，即不必验证分子\(f(x)\)是否为无穷大，也可利用洛必达法则求解极限，进而得到洛必达法则的一般情形.

定理3：广义∞/∞型

若：

1. \(\lim\limits_{x \to a} g(x) = \infty\),

2. \(f(x), g(x)\)在\(\mathring{U}(a)\)内可导,

3. \(g^{\prime}(x) \neq 0\)

4. \(\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)（\(A\)为有限数或无穷）.

则\(\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)。

其中\(x \to a\)改为\(x \to a + 0\)或\(x \to a - 0\)结论仍然成立。

定理3证明

此证明\(A\)是有限的情况，∵\(\lim\limits_{x \to a} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A\)，所以，对于\(\forall \varepsilon > 0, \exists \delta_{1} > 0\)，使得当\(x \in \mathring{U}(a, \delta_{1})\)时，则有 \[\left| \dfrac{f^{\prime}(x)}{g^{\prime}(x)} - A \right| < \dfrac{\varepsilon}{2}.\]

∴\(g^{\prime}(x) \neq 0\)

由柯西中值定理可知，对\(x_{1} \in \mathring{U}(a, \delta_{1}), \exists \xi \in (x_{1}, a)\)或\((a, x_{1})\)，使得 \[\dfrac{f(x) - f(a)}{g(x) - g(a)} = \dfrac{f^{\prime}(\xi)}{g^{\prime}(\xi)},\]

因此 \[\left| \dfrac{f(x) - f(a)}{g(x) - g(a)} - A \right| = \left| \dfrac{f^{\prime}(\xi)}{g^{\prime}(\xi)} - A \right| < \dfrac{\varepsilon}{2}.\]

于是，当\(x \neq a\)时，进而有 \[\dfrac{f(x)}{g(x)} = \dfrac{f(x) - f(a)}{g(x)} + \dfrac{f(a)}{g(x)} = \dfrac{g(x) - g(a)}{g(x)} \cdot \dfrac{f(x) - f(a)}{g(x) - g(a)} + \dfrac{f(a)}{g(x)} = \left(1 - \dfrac{g(a)}{g(x)}\right) \cdot \dfrac{f(x) - f(a)}{g(x) - g(a)} + \dfrac{f(a)}{g(x)} \tag{7}\]

因此根据(7)式及基本不等式有 \[\left| \dfrac{f(x)}{g(x)} - A \right| = \left| (1 - \dfrac{g(a)}{g(x)}) \cdot \dfrac{f(x) - f(a)}{g(x) - g(a)} + \dfrac{f(a)}{g(x)} - A \right| \leq \left| 1 - \dfrac{g(a)}{g(x)} \right| \cdot \left| \dfrac{f(x) - f(a)}{g(x) - g(a)} - A \right| + \left| \dfrac{f(a) - Ag(a)}{g(x)} \right| \tag{8},\]

又因为\(\lim\limits_{x \to a} g(x) = \infty\)，所以，\(\exists \delta > 0\)，使得当\(x \in \mathring{U}(a, \delta)\)时，有 \[0 < 1 - \dfrac{g(a)}{g(x)} < 1, \left| \dfrac{f(a) - Ag(a)}{g(x)} \right| < \dfrac{\varepsilon}{2}, \tag{9}\]

因此，根据(8)，(9)式，对于\(\forall \varepsilon > 0, \exists \delta > 0,\) 当\(x \in \mathring{U}(a, \delta)\)时，有 \[\left| \dfrac{f(x)}{g(x)} - A \right| \leq \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} = \varepsilon, \tag{10}\] 进而根据(10)式有\(\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = A\)，故定理3成立。

洛必达法则推论

推论1

设\(f(x)\)在区间\((a, +\infty)\)内可导，若极限\(\lim\limits_{x \to +\infty} f(x)\)与\(\lim\limits_{x \to +\infty} f^{\prime}(x)\)都存在，则有\(\lim\limits_{x \to +\infty} f(x) = 0\)。

推论1证明

假设\(\lim\limits_{x \to +\infty} f^{\prime}(x) = A, \lim\limits_{x \to +\infty} f(x) = B\)，根据定理3，则有 \[\lim\limits_{x \to +\infty} \dfrac{f(x)}{x} = \lim\limits_{x \to +\infty} \dfrac{f^{\prime}(x)}{1} = \lim\limits_{x \to +\infty} f^{\prime}(x) = A.\]

又因为\(\lim\limits_{x \to +\infty} f(x)\)存在，所以 \[\lim\limits_{x \to +\infty} \dfrac{f(x)}{x} = \lim\limits_{x \to +\infty} \dfrac{1}{x}\cdot\lim\limits_{ x \rightarrow + \infty } f ( x )= 0 \times B = 0 ,\]

即 \[\lim\limits_{x \to +\infty} f^{\prime}(x) = 0.\]

推论2

设\(f(x)\)在\((a_{1}, +\infty)\)内\(n\)阶导数存在，若\(\lim\limits_{x \to +\infty} f(x), \lim\limits_{x \to +\infty} f^{(n)}(x)\)都存在，则有\(\lim\limits_{x \to +\infty} f^{(k)}(x) = 0\)\((k = 1, 2, \cdots, n).\)

推论2证明

根据定理3，则有 \[\lim\limits_{x \to +\infty} \dfrac{f(x)}{x^{k}} = \lim\limits_{x \to +\infty} \dfrac{f^{\prime}(x)}{k x^{(k-1)}} = \lim\limits_{x \to +\infty} \dfrac{f^{\prime}(x)}{k(k-1) x^{(k-2)}} = \cdots = \lim\limits_{x \to +\infty} \dfrac{f^{(k)}(x)}{k!} = \dfrac{1}{k!} \lim\limits_{x \to +\infty} f^{(k)}(x)\]

由\(\lim\limits_{x \to +\infty} f^{(n)}(x)\)存在，所以 \[\lim\limits_{x \to +\infty} \dfrac{f(x)}{x^{k}} = 0,\] 即 \[\lim\limits_{x \to +\infty} f^{(k)}(x) = 0 (k = 1, 2, \cdots, n).\]

推论3

设\(f(x)\)在\((a, +\infty)\)内可导，且 \(\lim\limits_{x \to +\infty} \left[ f(x) + f^{\prime}(x) \right] = A\)（\(A\)为有限数或无穷），则有 \(\lim\limits_{x \to +\infty} f^{\prime}(x) = 0.\)

推论3证明

根据定理3有， \[\lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} \dfrac{e^{x} f(x)}{e^{x}} = \lim\limits_{x \to +\infty} \dfrac{\left[ e^{x} f(x) \right]^{\prime}}{(e^{x})^{\prime}} = \lim\limits_{x \to +\infty} \dfrac{e^{x} f(x) + e^{x} f^{\prime}(x)}{e^{x}} = \lim\limits_{x \to +\infty} \left[ f(x) + f^{\prime}(x) \right] = A,\] ∴ \[\lim\limits_{x \to +\infty} f(x) = A,\] 又∵ \[\lim\limits_{x \to +\infty} \left[ f(x) + f^{\prime}(x) \right] = A,\] 即 \[\lim\limits_{x \to +\infty} f^{\prime}(x) = 0.\]

推论4

设\(f(x)\)在区间\((a, +\infty)\)内可导，若\(\lim\limits_{x \to +\infty} [k_{1} f(x) + k_{2} f^{\prime}(x)] = A\)（其中\(k_{1}\)，\(k_{2}\)为正数），则有\(\lim\limits_{x \to +\infty} f(x) = \dfrac{A}{k_{1}}, \lim\limits_{x \to +\infty} f^{\prime}(x) = 0.\)

推论4证明

根据定理3，则有\(\lim\limits_{x \to +\infty} f(x)\)\(= \lim\limits_{x \to +\infty} \dfrac{e^{(k_{1}/k_{2})x} f(x)}{e^{(k_{1}/k_{2})x}}\)\(= \lim\limits_{x \to +\infty} \dfrac{(k_{1}/k_{2}) e^{(k_{1}/k_{2})x} f(x) + e^{(k_{1}/k_{2})x} f^{\prime}(x)}{(k_{1}/k_{2}) e^{(k_{1}/k_{2})x}}\)\(= \lim\limits_{x \to +\infty} \dfrac{(k_{1}/k_{2}) f(x) + f'(x)}{(k_{1}/k_{2})}\)\(= \lim\limits_{x \to +\infty} \left[ f(x) + \dfrac{k_{2}}{k_{1}} f^{\prime}(x) \right]\)\(= \lim\limits_{x \to +\infty} \dfrac{1}{k_{1}} \left[ k_{1} f(x) + k_{2} f^{\prime}(x) \right]\)\(= \dfrac{A}{k_{1}},\)

所以，\(\dfrac{k_{2}}{k_{1}} \lim\limits_{x \to +\infty} f^{\prime}(x)\)\(= \lim\limits_{x \to +\infty} \left[ f(x) + \dfrac{k_{2}}{k_{1}} f^{\prime}(x) - f(x) \right]\)\(= \dfrac{A}{k_{1}} - \dfrac{A}{k_{1}} = 0,\)

故\(\lim\limits_{x \to +\infty} f^{\prime}(x) = 0.\)

注意事项及错误案例

注意事项

1. 每次使用洛必达法则前，要注意验证条件的成立，若条件成立可用，否则不能用；
2. 若\(\lim\limits_{x \to a} \dfrac{f(x)}{g^{\prime}(x)}\)既不存在也不是∞，且通过应用洛必达法则求导后出现循环式的情况，即通过几次求导后的极限式又回到原式极限，则洛必达法则失效。

1. \(\lim\limits_{x \to \infty} \dfrac{x + \sin x}{x - \sin x}\)；

   错误解

   因为\(\lim\limits_{x \to \infty} \dfrac{x + \sin x}{x - \sin x} = \lim\limits_{x \to \infty} \dfrac{1 + \cos x}{1 - \cos x}\)，此时，发现右端极限不存在，所以判断左端极限也不存在，即原极限不存在。

   分析：这种解法错误，因为洛必达法则指明，在适当条件下，\(\lim\limits_{x \to \infty} \dfrac{f^{\prime}(x)}{g^{\prime}(x)} = A \Rightarrow \lim\limits_{x \to \infty} \dfrac{f(x)}{g(x)} = A\)，而这里\(\lim\limits_{x \to \infty} \dfrac{f^{\prime}(x)}{g^{\prime}(x)}\)不存在，也不是∞，不具备用洛必达法则的条件，即\(\lim\limits_{x \to \infty} \dfrac{f^{\prime}(x)}{g^{\prime}(x)}\)不存在也不是∞，不能断定原极限不存在，只是说明应用洛必达法则求解失效了，应想办法采用其他方法求解。

   正确解

   应用极限的四则运算法则及有界函数乘以无穷小仍然是无穷小进行求解，即

   \(\lim\limits_{x \to \infty} \dfrac{x + \sin x}{x - \sin x}\)\(= \lim\limits_{x \to \infty} \dfrac{1 + \sin x / x}{1 - \sin x / x}\)\(= \lim\limits_{x \to \infty} \dfrac{1 + (1 / x) \cdot \sin x}{1 - (1 / x) \cdot \sin x}\)\(= \dfrac{1 + 0}{1 - 0} = 1.\)

2. \(\lim\limits_{x \to 0} \dfrac{x + \cos x}{\sin x}\)。

   错误解

   \(\lim\limits_{x \to 0} \dfrac{x + \cos x}{\sin x}\)\(= \lim\limits_{x \to 0} \dfrac{1 - \sin x}{\cos x}\)\(= 1.\)

   分析：洛必达法则是求\(\dfrac{0}{0}\)型或\(\dfrac{\infty}{\infty}\)型极限的一种方法，此极限既不是\(\dfrac{0}{0}\)型，也不是\(\dfrac{\infty}{\infty}\)型，不能应用洛必达法则求解。应注意洛必达法则不是求极限的万能公式，应用是有前提条件的，每用一次必须验证条件是否成立，只有条件成立才能应用。

   正确解

   应用无穷大与无穷小的关系求解，因为\(\lim\limits_{x \to 0} \dfrac{\sin x}{1 + \cos x}\)\(= \dfrac{0}{2} = 0,\)

   所以\(\lim\limits_{x \to 0} \dfrac{1 + \cos x}{\sin x}\)\(= \infty.\)