﻿ 关于∞ / ∞未定式L’Hospital法则的一个注记 A Note on L’Hospital Rule of Indeterminate Form ∞ / ∞

Pure Mathematics
Vol. 09  No. 08 ( 2019 ), Article ID: 32608 , 6 pages
10.12677/PM.2019.98119

A Note on L’Hospital Rule of Indeterminate Form $\frac{\infty }{\infty }$

Dan Liu, Liying Wang, Kai Mao

Naval Aviation University, Yantai Shandong

Received: Sep. 28th, 2019; accepted: Oct. 16th, 2019; published: Oct. 23rd, 2019

ABSTRACT

L’Hospital rule is a classical and effective method to solve indeterminate form in limit problem. This paper studies a class of extended indeterminate form $\frac{\ast }{\infty }$ based on common indeterminate form $\frac{\infty }{\infty }$ , generalizes L’Hospital rule under certain conditions and obtains generalized L’Hospital rule.

Keywords:Indeterminate Form $\frac{\infty }{\infty }$ , L’Hospital Rule, Indeterminate Form $\frac{\ast }{\infty }$ , Generalized L’Hospital Rule

L’Hospital法则是求解极限问题中未定式的一种经典而有效的方法。本文从未定式常见类型 $\frac{\infty }{\infty }$ 出发，研究了一类拓展型未定式 $\frac{\ast }{\infty }$ ，在一定条件下将L’Hospital法则进行了推广，给出了广义L’Hospital法则。

1. 引言

(1) 当 $x\to a$ 时，函数 $f\left(x\right)$$g\left(x\right)$ 均趋于无穷大；

(2) 在点a的某去心邻域内， ${f}^{\prime }\left(x\right)$${g}^{\prime }\left(x\right)$ 都存在且 ${g}^{\prime }\left(x\right)\ne 0$

(3) $\underset{x\to a}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$ 存在(或为无穷大)，

$\underset{x\to a}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

$\begin{array}{c}{\int }_{0}^{x}{\text{e}}^{t}f\left(t\right)\text{d}t={\int }_{0}^{{X}_{0}}{\text{e}}^{t}f\left(t\right)\text{d}t+{\int }_{{X}_{0}}^{x}{\text{e}}^{t}f\left(t\right)\text{d}t\ge {\int }_{0}^{{X}_{0}}{\text{e}}^{t}f\left(t\right)\text{d}t+{\int }_{{X}_{0}}^{x}\frac{1}{2}{\text{e}}^{{X}_{0}}\text{d}t\\ ={\int }_{0}^{{X}_{0}}{\text{e}}^{t}f\left(t\right)\text{d}t+\frac{1}{2}{\text{e}}^{{X}_{0}}\left(x-{X}_{0}\right)\end{array}$

$\underset{x\to +\infty }{\mathrm{lim}}\frac{{\int }_{0}^{x}{\text{e}}^{t}f\left(t\right)\text{d}t}{{\text{e}}^{x}}=\underset{x\to +\infty }{\mathrm{lim}}\frac{{\text{e}}^{x}f\left(x\right)}{{\text{e}}^{x}}=\underset{x\to +\infty }{\mathrm{lim}}f\left(x\right)=1$

L’Hospital法则 [1] 。

2. $\frac{\ast }{\infty }$ 未定式的广义L’Hospital法则

${g}^{\prime }\left(x\right)\ne 0,x\in \left(a,b\right)$ ；(2) $\underset{x\to {a}^{+}}{\mathrm{lim}}g\left(x\right)=+\infty$ ；(3) $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}=A$ (A为有限数或 $-\infty$ , $+\infty$ )，则

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

(证法一) 由 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}=A$ ，则对 $\forall \epsilon >0,\text{\hspace{0.17em}}\exists \delta >0$ ，使得当 $x\in \left(a,a+\delta \right)$ 时，有

$|\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}-A|<\epsilon$

$A-\epsilon <\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}

$\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}=\frac{{f}^{\prime }\left(\xi \right)}{{g}^{\prime }\left(\xi \right)}$ (1)

$A-\epsilon <\frac{\frac{f\left(x\right)}{g\left(x\right)}-\frac{f\left({x}_{0}\right)}{g\left(x\right)}}{1-\frac{g\left({x}_{0}\right)}{g\left(x\right)}}=\frac{{f}^{\prime }\left(\xi \right)}{{g}^{\prime }\left(\xi \right)}

$\left(A-\epsilon \right)\left(1-\frac{g\left({x}_{0}\right)}{g\left(x\right)}\right)+\frac{f\left({x}_{0}\right)}{g\left(x\right)}<\frac{f\left(x\right)}{g\left(x\right)}<\left(A+\epsilon \right)\left(1-\frac{g\left({x}_{0}\right)}{g\left(x\right)}\right)+\frac{f\left({x}_{0}\right)}{g\left(x\right)}$ (2)

$A-\epsilon \le \underset{x\to {a}^{+}}{\underset{_}{\mathrm{lim}}}\frac{f\left(x\right)}{g\left(x\right)}\le \underset{x\to {a}^{+}}{\stackrel{¯}{\mathrm{lim}}}\frac{f\left(x\right)}{g\left(x\right)}\le A+\epsilon$

$A\le \underset{x\to {a}^{+}}{\underset{_}{\mathrm{lim}}}\frac{f\left(x\right)}{g\left(x\right)}\le \underset{x\to {a}^{+}}{\stackrel{¯}{\mathrm{lim}}}\frac{f\left(x\right)}{g\left(x\right)}\le A$

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=A$ ，即

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

(证法二) 由 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}=A$ ，则对 $\forall \epsilon >0,\text{\hspace{0.17em}}\exists \delta >0$ ，使得当 $x\in \left(a,a+\delta \right)$ 时，有

$|\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}-A|<\epsilon$

$A-\epsilon <\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}

$\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}=\frac{{f}^{\prime }\left(\xi \right)}{{g}^{\prime }\left(\xi \right)}$

$A-\epsilon <\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}

$\frac{f\left(x\right)}{g\left(x\right)}-A=\left(1-\frac{g\left({x}_{0}\right)}{g\left(x\right)}\right)\cdot \left(\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}-A\right)+\frac{f\left({x}_{0}\right)-Ag\left({x}_{0}\right)}{g\left(x\right)}$

$|\frac{f\left(x\right)}{g\left(x\right)}-A|\le |1-\frac{g\left({x}_{0}\right)}{g\left(x\right)}|\cdot |\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}-A|+|\frac{f\left({x}_{0}\right)-Ag\left({x}_{0}\right)}{g\left(x\right)}|<\epsilon +\epsilon =2\epsilon$

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=A$ ,从而 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

(ii) 当A为 $+\infty$ 时。

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}=+\infty$ ，则对 $\exists \delta >0$ ，使得当 $x\in \left(a,a+\delta \right)$ 时，有 $\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}>3M$ ，进一步地，

$\forall {x}_{0},x\in \left(a,a+\delta \right)$${x}_{0} ，由Cauchy中值定理， $\exists \xi \in \left({x}_{0},x\right)\subset \left(a,a+\delta \right)$ 使得 [3]

$\frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}=\frac{{f}^{\prime }\left(\xi \right)}{{g}^{\prime }\left(\xi \right)}>3M$

$\frac{g\left({x}_{0}\right)}{g\left(x\right)}<\frac{1}{2}$$\frac{f\left({x}_{0}\right)}{g\left(x\right)}>-\frac{1}{2}M$$x\in \left(a,a+\delta \right)$

$\frac{f\left(x\right)}{g\left(x\right)}=\frac{g\left(x\right)-g\left({x}_{0}\right)}{g\left(x\right)}\cdot \frac{f\left(x\right)-f\left({x}_{0}\right)}{g\left(x\right)-g\left({x}_{0}\right)}+\frac{f\left({x}_{0}\right)}{g\left(x\right)}>\frac{1}{2}\cdot 3M-\frac{1}{2}M=M$

$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=+\infty$ ，从而 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

(iii) 当A为 $-\infty$ 时。

${f}_{1}\left(x\right)=-f\left(x\right)$ 。由 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}=-\infty$$\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{{f}^{\prime }}_{1}\left(x\right)}{{g}^{\prime }\left(x\right)}=+\infty$ 。进一步地，由(ii)之结论可得 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}_{1}\left(x\right)}{g\left(x\right)}=+\infty$ ，即 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{-f\left(x\right)}{g\left(x\right)}=+\infty$ ，故 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=-\infty$ 。从而 $\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to {a}^{+}}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

3. 结论

$\underset{x\to +\infty }{\mathrm{lim}}\frac{{\int }_{0}^{x}{\text{e}}^{t}f\left(t\right)\text{d}t}{{\text{e}}^{x}}=\underset{x\to +\infty }{\mathrm{lim}}\frac{{\text{e}}^{x}f\left(x\right)}{{\text{e}}^{x}}=1$

A Note on L’Hospital Rule of Indeterminate Form ∞ / ∞[J]. 理论数学, 2019, 09(08): 922-927. https://doi.org/10.12677/PM.2019.98119

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