高数上-2012年9月27日条子题

b1p40-9另一方面,固定自然数m，并设 nm，nnen11n

两边令n, 11nnxnke

1k1nn1⋯nk1

k kkk k

1n1n⋯1 n1,ene,∑m1nn1⋯nk11

1nn1⋯nk1 得kk kmk ee211⋯1xm ∑1nn1⋯nk1

k

再 m 得 e 2111⋯11112⋯ b1p53-

xn1

limn1x1limt11 x1tn 解原

x1

n

其 lim1xx2⋯ n推 n1x 1n

x2x 的根,limna 只要 limn1x11

n1ր

nnnn1

tn1x x0tn1x1

t

先求 tn根据b1p52-3(10),即

x2x1 n 则有limn

b

不妨假 则

nnn1nn1b

5 52

nnnn1n 5,limn

nn∵n1nn

e1ln1bnlimn

limnn

ln1

bb,b

lim1ln1blimbnlimnn

nn

n n

15 15

limnn

limn1于n

1 43 53 9

lim

a1

(x2x

的正根再回答问题 n1nnnnn1n 事实上,我们前面已

一目了然了吧 头缩小提供本题另一解法并引

1 ,显

1

n1

1

bn n1,n

1 n1b 1nnnn1这时,约去, 就n1bnn1bnn1

n1 n1b n1 n1从此U形不等式串的两端即1 n1b1b ○limx23 4x9这 解注意limn1 1,limnnn

x2

1

就更显然了 lim33x201 4x9 求limm1xn1x

32

3334limx23x201132

4x9 m1xn1x n1xm1x1n1x

limx

x2

x

2x

xxtx1xx x22x x2tx1xx x22x x2x112t21tt2limm1xn1x1 用Taylor212t1t1t2Ot321t11t1t2Ot3 解法2 21 12t21t14t2Ot3

因为limx12使

1 所以X即得

1 1

34lim x22x4b1p52-

x2xx1

只 于

X

12 lim mb 1 1x3 ⇒lim1x

axmaxm1 b0xnb1xn1⋯bn1

12

12mlima0xma1xm1⋯am1xamamtx0b0xnb1xn1⋯bn1 t

解法二令x1 则b1p65-

112xx

tt

1

1,xt0t0

1 t12 12 t 1 12

1lnxxx ∵lim1lnt1

xln1 x1 x1,xt0 t1lntlime

t2

xlim1xx12

x

x x

b1p65-

x1

1 x

b1p16-6可以用来证明n1 x11

a 时,

a1,对x1x1

xlim1tt

0a 怎么

回答:b1p16-6n0na1a1n

a 求证证根据等比数列求和1xn1xn2⋯11x1xn1x1xn1xn2⋯在上式中,令xna 则因 a所

回过头来,对 0aa b1 ba用前limnalimn x1⇒xn1xn2⋯1故 a1nnan0na1a1⇒limnan

lim nnb1p76-x,x1fx 1e10,x

limn

f(0)limf(x)f x001e1f(0)依题f(x)f(ax()0⇒()(aaa因x001