一个极限定理的讨论及其应用外文翻译资料

本科生毕业设计（论文）

外文翻译

原文作者 Richard Beals 单位 Yale University

摘要：摘要这篇文章给出了一个能较好地解决一类特殊“和式”的极限问题的极限定理。同时,利用对数函数的特性,又能够用来解决一些“积式”的极限。

关键词：极限，和式，积式

在微积分中，我们经常使用一些特殊的极限来解决和式问题：

但是这个式子是不能直接相加的，也不能转换成函数的积分和的形式。所以很难求出它的极限，为了解决这个问题。这篇文章给出了一个极限定理，能较好地解决这一类特殊“和式”的极限问题。同时,利用对数函数又能够用来解决一些“积式”的极限。

定理1. 令()在时可微且,()在区间内可积,则

其中 :, ，

证明：由条件()可知,对任意的存在,当时有

由条件()可知,这里有存在一个实数,且在时,存在, 当时有

令 ，当时有

（因为）

另外还有

我们注意到到，先前的变量是以为条件的，在的情况中，有我们可以得到：当时

当时

令，则定理1可以变为

这是一个定积分的定义，然后通过对数函数我们可以得到

推论2.如果在时可微且，在区间上可积，则有：

在实际情况下，我们经常将n等分，取

推论3 令在处可微，在上对可积，我们有

1. 如果,我们有

1. 如果,我们有

证明：由定理1和对数函数，我们可得

例1：求下列各式的值

解：

(a)以等价形式进行和的重置：

令 且 则

且

根据定理1得：

（b）以等价形式进行和的重置：

令 且 则

则根据定理1得

（c）令=且则

且

则根据定理1得

（d）令且则且

则根据定理1得

（e）令且则且

根据定理1得

例2：求下列各式的极限

解：

（a）我们可以以等价形式写出积的变换：

令且得

且

则根据推论2得

（b）以等价形式写出积的重置

令且，则

则根据推论2得

例3．求下式极限

解：令，将平均分成份，选择点则

所以，

参考文献：

[1]王寿生等. 微积分解题方法与技巧[M] . 西安:西北工业大学出版社,1990.

[2]林源渠等. 数学分析习题集[M] . 北京:北京大学出版社,1993.

[3][美]波利亚等. 数学分析中的问题与定理[M] . 张奠宙等译. 上海:上海科技出版社,1985.

[4]Loren C Larson. Problem2Solving Through Problems [ M ] . Printed and bound by R. R. Donnelley amp; Sons ,Harrisonburg , Virginia. 175 Fif th Avenue , New York , New York 10010 , U. S. A. Springer Verlag New York Inc. , 1983.

附外文文献原文

A Discussion on a Limit Theorem and Its Application

Abstract: This paper proposes that a limit theoremcan help to solve a specific limit problemof sum formula and that some limit of product formula can also be solved by exploiting the feature of logarithm function.

Keywords: limit theorem; sumformula; product formula

Incalculus,we will usually solve a specific limit problem of sum formula

But this sum formula canrsquo;t sum directly, and it canrsquo;t change into some kinds of functionrsquo;s integral sum. So it is hard to work out its limit , for solve this problem.

This paperrsquo;s proposes is that a limit theorem can help to solve this limit problem of sum formula and that some limit of product formula can also be solved by logarithm function.

Theorem1 Let (a) f be differentiable at x=0 and f (0) =0,(b) g be integrable for xisin;[a, b].

We have

Proof 　By the (a), for every thereis a gt;0 such that implies .

Then by the (b), there exists a real number Mgt;0 such that | g(x)| le;M for xisin;[a,b] and there is a gt;0 such that‖T‖lt;implies

Let ,so when‖T‖lt;delta;, we get

and therefore

We note the preceding argument was based on the assumption that f (0) =0. For the case that f (0) ne;0. We can show that

for f (0) gt;0 and

Let f (x) =x then theorem 1 has become

This is definition of definite integral , and by logarithm function we get

Corollary2 If f be differentiable at x=0 and f (0) =1 and g be integrable for x into [a,b] then we have

In practical is usually divide [0,1] into n parts, and choose (k=1,2, hellip;, n).

Corollary3　Let f be differentiable at x=0 and g be integrable for x into [0,1] , then we have

1. If f (0) =0, we have

1. If f(0) =1, we have

Proof 　By that theorem1 and logarithm function, we get

Example1　Evaluate each of the following:

Solution　(a) Rewrite the sum in the equivalent form

So that by theorem1,

（b）Rewrite the sum in the equivalent form

So that by theorem1,

So that by theorem1,

（d）Let f(x) =sinax and g(x) =x. Then

So that by theorem 1,

So that by theorem 1,

Example2　Evaluate the following limits:

Solution(a) We can change the product intoan equivalent from by writing

Let f(x) = 1 x and g(x) =x. Then

So that by corollary 2,

(b) Rewrite the product in the equivalent from

So that by corollary 2,

Example3　Evaluateof thefollowinglimit

So

[1] 王寿生等.微积分解题方法与技巧[M].西安:西北工业大学出版社,1990.

[2] 林源渠等.数学分析习题集[M].北京:北京大学出版社,1993.

[3] [美]波利亚等.数学分析中的问题与定理[M].张奠宙等译.上海:上海科技出版社,1985.

[4] Loren C Larson. Problem-Solving Through Problems[ M]. Printed and bound by R. R. Donnelley amp;Sons, Harrisonburg, Virginia. 175 Fifth Avenue, NewYork, NewYork10010, U. S. A. Springer Verlag NewYork Inc. , 1983.

剩余内容已隐藏，支付完成后下载完整资料

---

A Discussion on a Limit Theorem and Its Application

Abstract: This paper proposes that a limit theoremcan help to solve a specific limit problemof sum formula and that some limit of product formula can also be solved by exploiting the feature of logarithm function.

Keywords: limit theorem; sumformula; product formula

Incalculus,we will usually solve a specific limit problem of sum formula

But this sum formula canrsquo;t sum directly, and it canrsquo;t change into some kinds of functionrsquo;s integral sum. So it is hard to work out its limit , for solve this problem.

This paperrsquo;s proposes is that a limit theorem can help to solve this limit problem of sum formula and that some limit of product formula can also be solved by logarithm function.

Theorem1 Let (a) f be differentiable at x=0 and f (0) =0,(b) g be integrable for xisin;[a, b].

We have

Proof 　By the (a), for every thereis a gt;0 such that implies .

Then by the (b), there exists a real number Mgt;0 such that | g(x)| le;M for xisin;[a,b] and there is a gt;0 such that‖T‖lt;implies

Let ,so when‖T‖lt;delta;, we get

and therefore

We note the preceding argument was based on the assumption that f (0) =0. For the case that f (0) ne;0. We can show that

for f (0) gt;0 and

Let f (x) =x then theorem 1 has become

This is definition of definite integral , and by logarithm function we get

Corollary2 If f be differentiable at x=0 and f (0) =1 and g be integrable for x into [a,b] then we have

In practical is usually divide [0,1] into n parts, and choose (k=1,2, hellip;, n).

Corollary3　Let f be differentiable at x=0 and g be integrable for x into [0,1] , then we have

1. If f (0) =0, we have

1. If f(0) =1, we have

Proof 　By that theorem1 and logarithm function, we get

Example1　Evaluate each of the following:

Solution　(a) Rewrite the sum in the equivalent form

So that by theorem1,

（b）Rewrite the sum in the equivalent form

So that by theorem1,

So that by theorem1,

（d）Let f(x) =sinax and g(x) =x. Then

So that by theorem 1,

So that by theorem 1,

Example2　Evaluate the following limits:

Solution(a) We can change the product intoan equivalent from by writing

Let f(x) = 1 x and g(x) =x. Then

So that by corollary 2,

(b) Rewrite the product in the equivalent from

So that by corollary 2,

Example3　Evaluateof thefollowinglimit

So

[1] 王寿生等.微积分解题方法与技巧[M].西安:西北工业大学出版社,1990.

[2] 林源渠等.数学分析习题集[M].北京:北京大学出版社,1993.

[3] [美]波利亚等.数学分析中的问题与定理[M].张奠宙等译.上海:上海科技出版社,1985.

[4] Loren C Larson. Problem-Solving Through Problems[ M]. Printed and bound by R. R. Donnelley amp;Sons, Harrisonburg, Virginia. 175 Fifth Avenue, NewYork, NewYork10010, U. S. A. Springer Verlag NewYork Inc. , 1983.

剩余内容已隐藏，支付完成后下载完整资料

---

资料编号：[286829]，资料为PDF文档或Word文档，PDF文档可免费转换为Word