高數(shù)上-2012年9月27日條子題

b1p40-9另一方面,固定自然數(shù)m，并設(shè) nm，nnen11n

兩邊令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

n1?

nnnn1

tn1x x0tn1x1

t

先求 tn根據(jù)b1p52-3(10),即

x2x1 n 則有l(wèi)imn

b

不妨假 則

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

的正根再回答問題 n1nnnnn1n 事實上,我們前面已

一目了然了吧 頭縮小提供本題另一解法并引

1 ,顯

1

n1

1

bn n1,n

1 n1b 1nnnn1這時,約去, 就n1bnn1bnn1

n1 n1b n1 n1從此U形不等式串的兩端即1 n1b1b ○limx23 4x9這 解注意limn1 1,limnnn

x2

1

就更顯然了 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

因為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 則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可以用來證明n1 x11

a 時,

a1,對x1x1

xlim1tt

0a 怎么

回答:b1p16-6n0na1a1n

a 求證證根據(jù)等比數(shù)列求和1xn1xn2?11x1xn1x1xn1xn2?在上式中,令xna 則因 a所

回過頭來,對 0aa b1 ba用前l(fā)imnalimn x1?xn1xn2?1故 a1nnan0na1a1?limnan

lim nnb1p76-x,x1fx 1e10,x

limn

f(0)limf(x)f x001e1f(0)依題f(x)f(ax()0?()(aaa因x001