A.f(x)与x是等价无穷小; B. f(x)与x同阶，但非等价无穷小 C. f(x)是比x高阶的无穷小； D. f(x)是比x低阶的无穷小 证明：
2x?3x?22x?13x?111lim?lim?lim???1x?0x?0x?0xxxln2ln3 ?f(x)与x同阶，但不是等价无穷小三. 计算题 2.设f(x)=limn??x2n?1?ax2?bx,当a,b取何值时，f(x)在（-?,??)上连续
x2n?1当x?1时x?ax2?2n?bx1?2nf(x)?lim?xn??11?2nx?f(x)在(??,?1)和(1,?)是连续的。当x?1时f(x)?ax2?bx?f(x)在(?1,1)是连续的。a?b?1,lim?f(x)??1,lim?f(x)?a?bx??1x??12由f(?1)?lim?f(x)?lim?f(x)f(?1)?x??1x??1?a?b??1a?b?1,lim?f(x)?a?b,lim?f(x)?1x?1x?12由f(1)?lim?f(x)?lim?f(x)f(1)?x?1x?1解：?a?b?1?a?0解得??b?1
3．求下列极限
（1）limn1n?2n???10n;
n??解: ?10?n1n?2n???10n?101010
limn10?1
n??? limnnn??21?2n???10n?1
(2)lim =limln(sin2x?e2)?xln(x?e2xx?0)?2x;
ln(sin2x?e2)?lnexln(x?e22xx?0)?lne2xlnex?sin2xex 22xx?ee2xsin2xex2e2xx =limx?0lnln(1?)=limx?0
)ln(1?sin2xxe=limx?0x2
e2x=1 （3）lim=limx?a?x?ax?a22x?a；
x?a?x?ax?ax?ax?a
?1(x?a)(x?a)=limx?a(x?a)x?ax?a?1x?ax?a
=lim=
1x?a
2a
(4)lim?x?0??1?ax?bx?cx?x3?(a?0,b?0,c?0) ??????
=limex?01?ax?bx?cx?lnx?3?=limex?01?ax?1?bx?1?cx?1??ln?1??x?3??
=e1?ax?1?bx?1?cx?1??limln?1???3x?0x??
?ax?1?bx?1?cx?1??lim??3xx?0??? e=
lna?lnb?lnc3=e
=3abc
(5) lim?sinx?tanx
x??2令y=?x
2?原式=lim??sin(?y)?y?0??tan(?y)?2?2?
=limey?01lncosytany
=e=e
1limln(1?cosy?1)tanyy?01lim(cosy?1)y?0y
y2lim?=ex?02
y=1
(6)设x1?1,xn?2xn?1?3(n?2,3,?),求limxn
n??解：0?xn?3,显然，xn?0,下面用数学归纳法证明xn?3，
当n?1时，x1?1?3, 假设当n?1时，xn?1?3, xn?2xn?1?3?2?3?3?3 所以0?xn?3
222?xn xn?1?2xn?1?3?xn?1?(3?xn?1)(xn?1?1)?0
所的以极限存在 所以xn?xn?1,{xn}是单调递增数列，它
2 设a?limxn,由xn?2xn?1?3，取极限可得，a2?2a?3,
n?? 解得a?3,a??1(舍） 即limxn?3
n??4.设F(x,t)???t?1??1?x?1?x?t((x?t)(t?1)?0,x?t),函数f(x)由下列表达式确定
f(x)?limF(x,t)，试求f(x)的连续区间和间断并点研，究在其间断点
t?x处的左右极限1?x?1?x?t
=lim??1?t?x?1x?t?x?t 解:f(x)?limF(x,t)=lim?t?x?t?1t?x???t?1?
=lim??1?t?x?t?1x?t1x?t?x?tt?1x?t?t?1?=limt?xt?1x?t1?x?t?x?tln?1??t?1x?tt?1??e
t?1x?t1?x?t?x?tlimln?1??t?1x?tt?1t?x??=e t?1x?t1?x?t?x?tlimlimln?1??t?1x?tt?1t?xt?x??=e
1=ex?1
所以x?1是间断点，limf(x)???,limf(x)?0,
x?1?x?1?四．证明题 1．
(1)lim利用及定义证明：
1?6n??3
n??3?2n1?6n1010?3????,成立 3?2n3?2n2n证：??〉0，要使解得，n??5?,取N??? ????51?6n?5???〉0，?N???,?n?N,有?3??,成立
?3?2n??所以lim1?6n??3
n??3?2n4?x2??4
(2)limx?2x2?3x?2??〉0，限定0?x?2?要使4?x2111,则x?1?x?2?1?1?x?2?1??222(2?x)(2?x)?4?3?6x?2??成立(x?2)(x?1)x?1x?2
x2?3x?2?4?解得x?2??1?,所以取??min(,) 6261?4?x2 ??〉0，???min(,),?x,0?x?2??时,有?4??成立 226x?3x?2即lim4?x2??4
x?2x2?3x?2a2.若f(x）在[0,a]上连续（a?0)且f(0)?f(a),证明方程f(x)?f(x?)在（0,a)内至少2有一个实根。
a证：令g(x)?f(x)?f(x?),则g(x)是[0，a]上的连续函数。2ag(0)?f(0)?f(),2aaag()?f()?f(a)?f()?f(0)222
aaa若f(0)?f(),则0,满足f(x)?f(x?)222aaa若f(0)?f(),则g(0)g()?0，根据零点定理，g(x)在(0,)至少存在一个实根222a即f(x)?f(x?)在（0,a)内至少有一个实根。2