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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2012 年 02月 10日A Discussion on a Limit Theorem and Its ApplicationAbstract: This paper proposes that a limit theoremcan help to solve aspecific 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 formulaIncalculus,we will usually solve a specific limit problem of sum formulaBut this sum formula can’t sum directly, and it can’t change into some kinds of function’s integral sum. So it is hard to work out its limit , for solve this problem. This paper’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 beintegrable for x ∈[a, b]. We haveProofBy the (a), for every 0>ε thereis a 1δ>0 such that 1δ<ximplies x x o f x f ε<'-)()(.Then by the (b), there exists a real number M>0 such that | g(x)| ≤M for x ∈[a,b] and there is a 2δ>0 such that ‖T ‖<2δimpliesLet }Mmin{21δδδ，=,so when ‖T ‖<δ, we getand thereforeWe note the preceding argument was based on the assumption that f (0) =0. For the case that f (0) ≠0. We can show thatfor f (0) >0 andLet 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 be integrable for xinto [a,b] then we haveIn practical is usually divide [0,1] into n parts, and choose nk =k ξ(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 haveProofBy that theorem1 and logarithm function, 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 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 thefollowinglimitSo[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 &Sons, Harrisonburg, Virginia. 175 Fifth Avenue, NewY ork, NewY ork10010, U. S. A. Springer V erlag NewY ork 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：23223232211nn nn n Xn++++++=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(a )f 在0=x 时可微且0)0(=f ,(b )g 在[]b a x ,∈区间内可积,则()[]⎰∑'=∆=→bank k k T dx x g f x g f )()0(lim1ξ其中 T :b x x x a n =<<<= 21, ],[1k k k x x -∈ξ，证明：由条件(a )可知,对任意的0>ε存在01>δ,当1δ<x 时有x x o f x f ε<'-)()(由条件(b )可知,这里有存在一个实数0>M ,且在[]b a x ,∈时M x g ≤)(,存在02>δ,当2δ<T 时有εξ<-∆⎰∑ba i i dxx g x g )()(令 }Mmin{21δδδ，=，当δ<T 时有⎰∑'-∆=bani i i dx x g f x f f )()0(])([1ξ∑∑-=∆'-≤ni i i ni i x g f g f 11)()0()]([ξξ+)0(f 'dxx g x g ni ba i i ∑⎰=-∆1)()(ξ∑=∆'-∆+'≤ni i i i i x g f x g f f 1)()0(])([)0(ξξε∑=∆+'≤ni i i x g f 1)()0(ξεε（因为i i i x g δξ<∆)(）)]()0([)0(1a b M f x Mf ni i-+'=∆+'≤∑=εεε另外还有dx x g f x g f bani i i T ⎰∑'=∆=→)()0(])([lim1ξ我们注意到到，先前的变量是以0)0(=f 为条件的，在0)0(≠f 的情况中，有我们可以得到：当0)0(>f 时+∞=∆∑=→ni i i T x g f 1])([limξ当0)0(<f 时令x x f =)(，则定理1可以变为⎰∑=∆=→ba ni ii T dx x g x g )()(lim1ξ-∞=∆∑-→ni i i T x g f 1])([limξ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f 在0=x 时可微且1)0(=f ，g 在[]b a x ,∈区间上可积，则有： {}⎰∏'=∆=→bank k k T dx x g f x g f )()0(exp ])([lim1ξ在实际情况下，我们经常将]1,0[n 等分，取nk k =ξ)3,2,1(n k =推论3 令f 在0=x 处可微，g 在]1,0[上对x 可积，我们有(a)如果0)0(=f ,我们有⎰∑'==∞→11)()0(]1[limdx x g f nk gnf nk n(b)如果1)0(=f ,我们有{}⎰∏'==∞→11)()0(exp ]1[lim dx x g f nk gnf nk n证明：由定理1和对数函数，我们可得⎭⎬⎫⎩⎨⎧∆=∆∑∏=→=→n k k k T nk k k T x g f x g f 101])([ln exp lim ]()[limξξ{}⎰∑==→'=⎭⎬⎫⎩⎨⎧∆=b ax nk k k T dx x g x f x g f )(])([ln exp ])([ln lim exp 010ξ{}⎰⎰'=⎭⎬⎫⎩⎨⎧'=b ab a dx x g f dx x g f f )()0(exp )()0()0(exp例1：求下列各式的值∑=∞→∞→∞→⎥⎦⎤⎢⎣⎡-+⎥⎦⎤⎢⎣⎡++++++⎥⎦⎤⎢⎣⎡++++++nk n n n n kc n n nn n b n n nn n a 13223223223224224224211lim )(2211lim )(2211lim )( ∑∑=∞→=∞→-⎥⎦⎤⎢⎣⎡nk n kn nk n a e n ka d 112)1(lim)(sin lim)(2解：(a)以等价形式进行和的重置：⎥⎥⎦⎤⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡=+∑∑==21212421.11.n n k n n k kn knk nk 令 221)(xxx f +=且 x x g =)( 则0)0(='f 且⎰=121)(dx x g根据定理1得：⎰∑='=⎥⎥⎦⎤⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡++++++=∞→12122422422420)()0(1.11.lim 2211lim dx x g f nn k n n k n n n n n nk n（b ）以等价形式进行和的重置：∑∑==⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛=+nk nk n n k n n k kn k1221232111令 xx x f +=1)(且2)(x x g = 则31)(1)0(121==='⎰⎰dx x dx x g f 且则根据定理1得312211lim 232232232=⎥⎦⎤⎢⎣⎡++++++∞→n n n n n n（c ）令)(x f =113-+x 且x x g =)(则31)0(='f 且 21)(1=⎰dx x g则根据定理1得（d ）令ax x f sin )(=且x x g =)(则a f =')0(且21)(10=⎰dx x g则根据定理1得⎰∑='=⎥⎦⎤⎢⎣⎡=∞→101261)()0(sin limdx x g f n ka nk n（e ）令1)(-=xa x f 且x x g =)(则a f ln )0(='且21)(1=⎰dx x g根据定理1得∑=∞→=-nk n kn a a 12ln )1(lim2例2：求下列各式的极限∏=∞→⎥⎦⎤⎢⎣⎡+nk n n k a 121lim )( ∏=∞→-+n k n kn b 122n k lim )(解：（a）我们可以以等价形式写出积的变换：令x x f +=1)(且x x g =)(得1)0(='f 且 21)(10=⎰dx x g则根据推论2得21121lim e n k nk n =⎪⎭⎫ ⎝⎛+∏=∞→ （b）以等价形式写出积的重置令xx x f +-=11)(且x x g =)(，则则根据推论2得e kn k n nk n =-+∏=∞→122lim例3．求下式极限⎪⎪⎭⎫⎝⎛++++++∞→n n n n n 22212111lim解：令11)(,0)(2+==n x g x f ，将[]1,0平均分成2n 份，选择点2nk k =ξ则211222111122nk nn k g nf nk nk +=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛∑∑==)12(2)()0(111lim 1lim121122222-='=+=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛⎰∑∑==∞→∞→dx x g f nk n n k g nf nk nk n n所以，)(11112112222∞→∞→+=+∑∑-=n nk nn kn nk nk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 & 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）。