极限的一个定理及其应用-外文翻译

1、本科生毕业设计论文外文翻译学生姓名： 张朋宇学 号： 408114010113 专业班级： 数学与应用数学指导教师： 梁海燕老师 2021 年 02月 10日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 e

2、xploiting the feature of logarithm function.Keywords: limit theorem; sumformula; product formulaIncalculus,we will usually solve a specific limit problem of sum formulaBut this sum formula cant sum directly, and it cant change into some kinds of functions integral sum. So it is hard to work out its

3、limit , for solve this problem. This papers 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 xa, b.We

4、haveProof By the (a), for every thereis a 0 such that implies .Then by the (b), there exists a real number M0 such that | g(x)| M for xa,b and there is a 0 such thatTimpliesLet ,so whenT0 and Let f (x) =x then theorem 1 has becomeThis is definition of definite integral , and by logarithm function we

5、 getCorollary2 If f be differentiable at x=0 and f (0) =1 and g be integrable for x into a,b then we haveIn practical is usually divide 0,1 into n parts, and choose (k=1,2, , n).Corollary3Let f be differentiable at x=0 and g be integrable for x into 0,1 , then we have(a) If f (0) =0, we have(b)(c) I

6、f f(0) =1, we haveProof By that theorem1 and logarithm function, we getExample1Evaluate each of the following:Solution(a) Rewrite the sum in the equivalent formSo that by theorem1, bRewrite the sum in the equivalent formSo that by theorem1,So that by theorem1,dLet f(x) =sinax and g(x) =x. ThenSo tha

7、t by theorem 1,So that by theorem 1,Example2Evaluate the following limits:Solution(a) We can change the product intoan equivalent from by writingLet f(x) = 1+x and g(x) =x. ThenSo that by corollary 2,(b) Rewrite the product in the equivalent fromSo that by corollary 2,Example3Evaluateof thefollowing

8、limitSo1 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 &Sons, Harrisonburg, Virginia. 175 Fifth Avenue, NewYork, NewYork10010, U. S. A. Springer

9、Verlag NewYork Inc. , 1983.极限的一个定理及其应用摘要：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,那么其中 :, ， 证明：由条件()可知,对任意的

10、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，那么定理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，那么有：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(a) 如果,我们有(b) 如果,我们有证明：由定理1和对数函数，我们可得例1：求以下各式的值解：(a)以等价形式进行和的重置：令 且 那么且根据定理1得： b以等价形式进行和的重置：令 且 那么那么根据定理1得c令=且那么 且 那么根据定理1得d令且那么且那么根据定理1得e令且那么且根据定理1得例2：求以下各式的极限解：a我们可以以等价形式写出积的变换：令且得 且 那么根据推论2得 b以等价形式写出积的重置令且，那么那么根据推论2得例3求下式极限解：令，将平均分成份，选择点那么所以， 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,1985.4Loren C Larson. Problem2Solving Through Prob