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1、本科生毕业设计（论文）外文翻译 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 the
2、orem; 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 limit , for solve this problem. This papers proposes is that a
3、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 haveProof By the (a), for every thereis a 0 such that implies .
4、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 getCorollary2 If f be differentiable at x=0 and f (0) =1 and g
5、 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) If f(0) =1, we haveProof By that theorem1 and logarithm function
6、, we getExample1Evaluate each of the following:Solution(a) Rewrite the sum in the equivalent formSo that by theorem1, （b）Rewrite the sum in the equivalent formSo that by theorem1,So that by theorem1,（d）Let f(x) =sinax and g(x) =x. ThenSo that by theorem 1,So that by theorem 1,Example2Evaluate the fo
7、llowing 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 thefollowinglimitSo1 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
8、.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 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
9、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，
10、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(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
11、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,1985.4Loren C Larson. Problem2Solving Through Problems M . Printed and bound by R. R. Donnelley & Sons ,Harrisonburg , Virginia. 175 Fif th Avenue , New York , New York 10010 , U. S. A. Springer Verlag New York Inc. , 1983.（此处为翻译的中文名）（原著作者名Times New Roman字体）
