(1 sin x )(1 3 sin x ) (1 n sin x )
例1.13 lim
(1 sin x)n1
x
2
: / = 3 - t
t -X X =
,
(1-sm) (l-smt) (1-
"SuCE
T . ...
Um
=
+
[1-SuCE-t)"
(1-lost) (1 -Most (1- "cost)
· ...
In
=
To
+
Cl-cost"
I I I
Z 2 5 t in th
t
um Z - .... . I
= 2 +
2 [xyx
x
Ito = ... =
N ni
⑳th gu+1
f (x) x f (x)
例1.14 已知lim 1 x e3,lim
x x2
x0 x0
f()
In(
: +x
+ e
e
↓ =
to
+
In Cit
M
=>
3
=
*
X
f)
In 1
+
x+
FigABX
35-
:
X+O
#J 3+ d(x + 0)
, =
, Y
+ x(3 d)
=> In (1 + x ) = x (3 + 2) = H+ x+ +
e
=
(3+
ex
=>
1 X
= - -m M-X eXB
=>
M
= = /
-
M X
= =
+
35 Mum(+
= : = AMX
3 =
=
*
X
In Ux
=
M
=>
xt
Im
=> I
= 3
=
*
Mo
=
2
=1 x
例1.15 设函数 f (x) lim , 则 f (x)（ ）.
1 nx2n
n
(A) 在 x 1, x 1处都连续 (B) 在 x 1
D
处连续, x 1处不连续
(C) 在 x 1, x 1处都不连续 (D) 在 x 1处不连续, x 1处连续
X (x*
"She
* C X31 8
[ =
= = , .
=
# DECIAF Y x
: new -0 n - 0
, , , .
fix
: 1+X
=
Y (X fix
② EX2 1 At nto n = n . + 0 0
=
, , ,
. " fill Ms it
③ EX = 1 #J n = n , = = O
,
fil
EX
④ = -AJ = o
,X
i
1xx
f(x ((+x. =
:
=
X1X : -1 >
0
.
f(l 0 fit = ME (X) = 0 fH) = 0 : + *
= . ,
f
f 0
(i M 2 =
=
=例1.16 设函数 f (x)在 x 0处可导，且 f (0) 1, f (0) 3,求数列极限
1
1 1
n(1cos )
I lim f ( ) .
n
n
n I
fitl
In
fast
M
not
1 &
= =
fint-1]
flu)-1
In [l
+
m
me
= e n.til
=
n+
fin)-fiol
eafin
2. 6
I
C = e
= I =
- 0
i2 2 2
1 2 n
例1.17 limln n 1 1 1 等于( B ).
n n n
n
2 2 2 2
(A) ln2 xdx (B)2 ln xdx (C)2 ln(1 x)dx (D) ln2(1 x)dx
1 1 1 1
m(h]
(H)
|(H+E
7. Uns[ In
+ + ... +
=
In
[mC
2M +
=
=I
m
InCl
x]
+ dx
= 2
t = HX 2/2 me /2 mx
de 2 dx
=
I例1.18 设 x 0, x 1 e x n ,(n 1,2, ).(1)证明lim x 存在，并求此极
1 n1 n
n
限；
SERA
X 7 0
(1) : ,
,
XI
-
1-e <
: 07Xz =
*
: xX 1 e
= -
i
i 17 SIA 35
It
o < Xn < /1-EM
35 Xnt-Xn Xn
- : = -
*
1 (0(X 1)
9m 1 -e - x - 91 %
= 0
= ,
*
gin -
1
= e -
-X0
gix) 9 V
0
: <
,
9I 910)
: <
= 0
Xnt1 < Xn
[x] NATO
:e-X fix 1 - eX
/2
1 (0xx()
3 = = Xn = - , =
ex
fix
> 0
=
* 1137 A P
[x]
:
.
.
P
:H3) X a
,
Ve M(-eX
Xnt ,
1 =
nico
a
-
1 e
a -
=
: a = 0
nu
Xn
0
im =
Hesx x
(2)求lim n1 n .
x 2
n
n
-
Ml-e -x et
An Xnt-X2 Im 1 t
- -
=
Yu
Xu
nico to0 th
1)
(m (e - (-4) - Etti
- My-
=
-
=
Eto -2 to =
2
-
例1.19 设 f x 是区间 0, 上单调减少且非负的连续函数，
n
n
a f k f x dx n 1,2, ，证明数列 a 的极限存在.
n n
1
iK1
fin)-S
"fix
fill fill
G EDA
:
an
=
+ +...+ ax
,
fill fill fini finti) 14t fixdx
Any = + +... + +
-
"
1
finti fixax 9" fix
Any-an = - + ex
:
, ,
firs (frfixax-f"f(xax)
= -
.
fintil In fix
ax
= -
fint1)-flul
EFEY SuE(n
(i5 : IRIE Gut-C ntl)
= .
,
and
fix
V . Ant-an
: 10.
-C
*
(*1ax-fr*
(35 Ant-an fint fixax final fixdx
== = - =
Crfint-fix] Gut An)
<o <
= ax
,
And Th
:
fin)-1
fill fill "funax
: an = + -... +
,
fil f(zl f) [ ? fixax- 12 fix ax - Max
= + +... +
-
...
fill f(si)
fizl f(n) fEr)- fint)
= + +... +
- - ... -
, 5 EC 2) E2E (23)
: En E (n n)
, . ... +,
+
fin)
: an > > 0: anTTB0 : M CTP
-
.x4 1
例1.20 曲线 y arctan 渐近线的条数为( B ).
x2 1 x
.
(A) 2 (B) 3 (C) 4 (D) 5
T M
: DKE Mr aritmt
: · = = 0
,
/nTd7-
:
EX
② E X = - 1 X= 1 x0
: , ,
.
xY
I
m
ariem -
·
+)X
1
x -
TEE-
X + 5 X =1
: =
-⑪t o 1
↳
m an = 0 + 0... X =0 RAY FRE
↓
#
-
14 ↑
③
x3
avcint = I
Am
=
1
=
-
#
manm-x]
b - Ef
= acctit -
to Ez-1
concent My concert-tc-ti E
m) E] : : Y X
=
= =
-
to
t +Y E0
(1 t))
- t
-
t -to(E)
Pe constant + + A t
- Mr
= - +
=
#or 2 = O
+
e2ln x
例1.21 设函数 f (x) sin x，则 f (x)有( ).
A
x 1
(A) 1 个可去间断点，1 个跳跃间断点
(B) 1 个跳跃间断点，1 个无穷间断点
(C) 2 个可去间断点 (D)2 个无穷间断点
EX
X=
15X =0
.
lux
Ink
Un It
jun
smx
sml SmM
=
=
x1(x 11
-1x 1) -
-
+
sml/m x f(i) filt
sml
= = - sm/
, =
x (x 1)
-
1 -**
:Ink
(n(x
jun
smx MX =0
= .
&
01x-11
+
* 33
X= 0 - --1 1 1 1
例1.22 设 f (x) , x ,1 , 试补充定义使得
x sinx (1 x) 2
1
f (x)在 ,1 上连续.
=
2 P fill
I I
fit
=
J
+
T M -
:
smax
x( x)
-
I
r(smax
(
=+ -
acces
I
t 1-X smae
= m] 1) sit)
+ - = + -
tot Smirk-t)
Xt
+
-(x
Smit
7t-
pr +
mm =
= +
= +
tot 7t Smit z to thax ln(1 x)
例1.23 设函数 f (x) 的可去间断点为 x 0，则a,b满足
x bsin x
( ).
(A)a 1,b 任意数 (B)a 1,b 任意数
(C)a任意数 b 1 (D) a
D
任意数 b 1
P
05
#
x=
,
ax-InCITN)
nu * E = X+ o Afax-InCI + X) Bil > X + bsmX
.
bSmX
# **
o() 2
(x
E
=> Max - - + im Ga-l . X + + oix
=
x0x o(xi)
b(X + x *0
+
- + (1 + b) x -X o()
- +
a
b
1
: +
,
