两个函数$f(x)$与$F(x)$当$x→a$(或$x→∞$)时都趋于零或无穷大，那么极限

\[\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to a} \\ {(x \to \infty )} \end{array}} \frac{{f(x)}}{{F(x)}}\]

可能存在、也可能不存在.通常把这种极限叫做未定式（Indeterminate form），并分别记作$\frac{0}{0},\frac{{\infty }}{{\infty }}.$这类极限之所以称为未定式，原因在于它们的极限（有可能存在，比如0，1、也可能不存在，比如无穷大）是不确定的，未定的，需要根据$f(x)$与$F(x)$的实际表达式来具体分析确定.实际上，未定式的极限可能是任何实数.

比如：

$$\mathop {\lim }\limits_{x \to 0} \frac{x}{{{x^3}}} = {\infty },\mathop {\lim }\limits_{x \to 0} \frac{x}{x} = 1,\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{x} = 0,\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$ $$\mathop {\lim }\limits_{x \to {\infty }} \frac{{3{x^3} + 4{x^2} + 2}}{{7{x^3} + 5{x^2} - 3}} = \frac{3}{7}$$

很多情况下，我们可以用代数消元（约分）、洛必达法则等方法作用于未定式的表达式上使得未定式可以求出.

In many cases，algebraic elimination，${L}'{H}o {\text{pital}}'{\text{s}}$ rule，or other methods can be used to manipulate the expression of indeterminate forms so that the limit can be evaluated.

下面来介绍洛必达法则.我们着重讨论$x→a$时的未定式$\frac{0}{0}$的情形.

定理1 设

（1）当$x→a$时，函数$f(x)$及$F(x)$都趋于零；

（2）在点$a$的某个去心邻域内，$f'(x)$及$F'(x)$都存在且$F'(x)≠0$；

（3）$\mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$存在（或为无穷大），

那么

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

证明：求有理分式函数当$x→a$的极限时，若$\mathop {\lim }\limits_{x \to a} F\left( a \right) \ne 0$，根据商的极限等于极限的商的极限运算法则有：

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \frac{{\mathop {{\text{lim}}}\limits_{x \to a} f\left( x \right)}}{{\mathop {{\text{lim}}}\limits_{x \to a} F\left( x \right)}}$$

现在$\mathop {{\text{lim}}}\limits_{x \to a} F\left( x \right) = 0$，所以不能根据商的极限等于极限的商的极限运算法则来求. 前面我们用消元法求解过不定式，因此，可以消元必定可以增元. 当$x→a$时，未定式的分子分母同时乘以$\frac{1}{{x - a}}$

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right)}}{{x - a}}}}{{\frac{{F\left( x \right)}}{{x - a}}}}$$

上式中分子分母类似于函数$f(x)$和$F(x)$在$x=a$点的导数的定义式，即

$$\eqalign{ & f'\left( a \right) = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr & F'\left( a \right) = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{F\left( x \right) - F\left( a \right)}}{{x - a}} \cr} $$

若$f(a)=F(a)=0$，则

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right)}}{{x - a}}}}{{\frac{{F\left( x \right)}}{{x - a}}}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right) - 0}}{{x - a}}}}{{\frac{{F\left( x \right) - 0}}{{x - a}}}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right) - f(a)}}{{x - a}}}}{{\frac{{F\left( x \right) - F\left( a \right)}}{{x - a}}}}$$

根据已知条件：当$x→a$时，函数$f(x)$及$F(x)$都趋于零.

若$f(a)=F(a)=0$，则$f(x)$和$F(x)$正好在$x=a$点连续.

在证明复合函数的求导法则时，推导过程中要求$\Delta u \ne 0$，而$\Delta u \ne 0$导致函数$y=f(u)$在$u$点不连续.这与函数$y=f(u)$在$u$点可导（进而连续）的条件相矛盾，我们后来若设$\Delta u = 0$时，$α=0$来解决这个矛盾.

现在我们假设$f(a)=F(a)=0$并没有与已知条件产生矛盾，则

$$\eqalign{ & \mathop {{\text{lim}}}\limits_{x \to a} \frac{{f\left( x \right) - f(a)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} f'\left( x \right) \cr & \mathop {{\text{lim}}}\limits_{x \to a} \frac{{F\left( x \right) - F(a)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} F'\left( x \right) \cr} $$

这就要求定理中的条件（2）（3）成立，则有

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right)}}{{x - a}}}}{{\frac{{F\left( x \right)}}{{x - a}}}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right) - 0}}{{x - a}}}}{{\frac{{F\left( x \right) - 0}}{{x - a}}}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{\frac{{f\left( x \right) - f(a)}}{{x - a}}}}{{\frac{{F\left( x \right) - F\left( a \right)}}{{x - a}}}} $$ $$ = \frac{{f'\left( a \right)}}{{F'\left( a \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

前提是$F'(x)≠0$，若$F'(x)=0$，则针对

$$\frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

继续重复上述过程.

上述讨论过程的优点是通俗易懂，逻辑简单，我们假定$f(x)$及$F(x)$在$x=a$处连续.

下面介绍运用拉格朗日中值定理的证明.

证明：设$f(a)=F(a)=0$，这个假设与已知不矛盾.

根据已知条件，$f(x)$和$F(x)$ 在$[x,a]$或$[a,x]$上连续，在开区间$(x,a)$或$(a,x)$内可导（$x$在$a$的某个邻域内），根据拉格朗日中值定理

$$\frac{{f\left( x \right)}}{{F\left( x \right)}} = \frac{{f\left( x \right) - 0}}{{F\left( x \right) - 0}} = \frac{{f\left( x \right) - f(a)}}{{F\left( x \right) - F\left( a \right)}} = \frac{{f'\left( \xi \right)}}{{F'\left( \xi \right)}},(ξ在a与x之间)$$

则

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{f'\left( \xi \right)}}{{F'\left( \xi \right)}} = \mathop {{\text{lim}}}\limits_{x \to a} \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

定理2 设

(1) 当$x→∞$时，函数$f(x)$及$F(x)$都趋于零；

(2) 当$|x|>N$时$f' (x)$及$F' (x)$都存在，且$F' (x)≠0;$

(3)$\mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$存在（或为无穷大）;

那么

$$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{{F\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

证明：设$x = \frac{1}{t},(t \ne 0)$，当$x→∞$时，$t→0$，当$|x|>N$时，$\left| t \right| < \frac{1}{N}$，即$t \in \mathop U\limits^0 \left( {0,{1 \over N}} \right)$，此时$f' (x)$及$F' (x)$都存在，且$F' (x)≠0$，则

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{t \to 0} f\left( {\frac{1}{t}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to \infty } F\left( x \right) = \mathop {\lim }\limits_{t \to 0} F\left( {\frac{1}{t}} \right) = 0 \cr} $$

即(1’)当$t→0$时，函数$f(x)$及$F(x)$都趋于零；

$$\left[ {f\left( {\frac{1}{t}} \right)} \right]' = f'\left( {\frac{1}{t}} \right) \cdot \frac{{ - 1}}{{{t^2}}} = - \frac{1}{{{t^2}}}f'(x)存在$$ $$\left[ {F\left( {\frac{1}{t}} \right)} \right]' = {F'}\left( {\frac{1}{t}} \right) \cdot \frac{{ - 1}}{{{t^2}}} = - \frac{1}{{{t^2}}}F'(x) \ne 0$$

所以(2’)当$t \in \mathop U\limits^0 \left( {0,{1 \over N}} \right),f\left( {\frac{1}{t}} \right),F\left( {\frac{1}{t}} \right)$都可导且$F'\left( {\frac{1}{t}} \right) \ne 0,$

$$\mathop {\lim }\limits_{t \to 0} \frac{{\left[ {f\left( {\frac{1}{t}} \right)} \right]'}}{{\left[ {F\left( {\frac{1}{t}} \right)} \right]'}} = \mathop {\lim }\limits_{t \to 0} \frac{{f'\left( {\frac{1}{t}} \right) \cdot \frac{{ - 1}}{{{t^2}}}}}{{F'\left( {\frac{1}{t}} \right) \cdot \frac{{ - 1}}{{{t^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{F'\left( x \right)}}$$

所以

$$(3{\text{')}}\mathop {\lim }\limits_{t \to 0} \frac{{\left[ {f\left( {\frac{1}{t}} \right)} \right]'}}{{\left[ {F\left( {\frac{1}{t}} \right)} \right]'}} = \mathop {\lim }\limits_{{\text{x}} \to {\infty }} \frac{{f'\left( x \right)}}{{F'\left( x \right)}}都存在（或为无穷大）$$

根据条件（1’）（2’）（3’），运用证明$x→a$时的$\frac{0}{0}$型不定式的洛必达法则的证明方法证明即可.

一些$0 \cdot \infty, \infty - \infty, {0^0}{1^\infty },{\infty ^0}$型的未定式，也可以通过$\frac{0}{0}$或$\frac{\infty }{\infty }$型的未定式来计算.

例 求

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\sec x - \tan x)$$

解

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\sec x - \tan x) = \mathop {{\text{lim}}}\limits_{x \to \frac{\pi }{2}} \left( {\frac{1}{{\cos x}} - \frac{{\sin x}}{{\cos x}}} \right) = \mathop {{\text{lim}}}\limits_{x \to \frac{\pi }{2}} \frac{{1 - \sin x}}{{\cos x}}$$是未定式$\frac{0}{0}$，判断洛必达法则中的第2，第3个条件并应用洛必达法则，得 $$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\sec x - \tan x) = \mathop {{\text{lim}}}\limits_{x \to \frac{\pi }{2}} \frac{{1 - \sin x}}{{\cos x}} = \mathop {{\text{lim}}}\limits_{x \to \frac{\pi }{2}} \frac{{ - \cos x}}{{ - \sin x}} = 0$$

例 求

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^n}\ln x(n > 0)$$

解 这是未定式$0 \cdot \infty $. 从这里可以看出，未定式$0 \cdot \infty ,\infty - \infty ,{0^0},{1^\infty },{\infty ^0},\frac{0}{0},\frac{\infty }{\infty }$中的$\infty $可以指趋于$+\infty $，$-\infty $或同时趋于正负无穷.

$$\eqalign{ & {x^n}\ln x = \frac{{\ln x}}{{\frac{1}{{{x^n}}}}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {x^n}\ln x = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{{x^n}}}}} = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} \frac{{\frac{1}{x}}}{{ - n{x^{ - n - 1}}}} = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} \left( {\frac{{ - {x^n}}}{n}} \right) = 0 \cr} $$

例 求

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^x}$$

解：$x^x$是幂指函数，设$y=x^x$，两边取对数得

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^x}$$ $$\mathop {{\text{lim}}}\limits_{x \to {0^ + }} \ln y = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} x\ln x = 0$$

设$\ln y = ? \to {{\text{e}}^?} = y \to {{\text{e}}^{\ln y}} = y$

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^x} = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} y = \mathop {{\text{lim}}}\limits_{x \to {0^ + }} {{\text{e}}^{\ln y}} = {{\text{e}}^{\mathop {{\text{lim}}}\limits_{x \to {0^ + }} \ln y}} = {{\text{e}}^0} = 1.$$

概念上，洛必达法则的意思是：具有相同自变量和定义域的函数$f(x)$相对于函数$F(x)$的在$x=x_0$点或$x→∞$时的变化率

\[\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to {x_0}} \\ {(x \to \infty )} \end{array}} \frac{{f(x)}}{{F\left( x \right)}}\]

等于两个函数相对于$x$的变化率的比

\[\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to {x_0}} \\ {(x \to \infty )} \end{array}} \frac{{f'(x)}}{{F'\left( x \right)}},(F'\left( x \right) \ne 0)\]

从概念上讲，洛必达法则的逻辑是清晰易懂的，这就是研究洛必达法则的出发点.