Bai tap gioi han ham so 11 (1)

Taøi lieäu Toaùn 11 Naâng cao
MOÄT SOÁ BAØI TAÄP CHÖÔNG IV. GIÔÙI HAÏN
A. GIÔÙI HAÏN DAÕY SOÁ
Baøi taäp 1: Tính caùc giôùi haïn:
2
12
lim/1
+
+
n
n
4
13
lim/2 2
2
+
+
n
n
23
15
lim/3
+
−
n
n
nnn
nn
−+
++
2
2
2
32
lim/4
1
32
lim/5 2
++
+
nn
nn
)3)(23(
)12)(1(
lim/6
++
−+
nn
nn
13
2
lim/7 2
2
++
+
nn
nn
13
2
lim/8 24
3
++ nn
n
)2)(1(
)3)(2(
lim/9
++
+
nn
nnn
1
12
lim/1 2
2
+
−
n
n
2
52
lim/2 2
+−
+
nn
n
23
2
lim/3 2
3
−+
−
nn
nn
( )nnn +−3 32
lim/4
23
12
lim/5 3
2
−
++
n
nn
( )nnn −−3 23
2lim/6
nn
n
32
1
lim/1 2
2
−
+
4
32
)1(
)2()1(
lim/2
−
++
nn
nn
( )1lim/3 22
+−+ nnn
3 32
3lim(/4 nnn −+ )
2
1112
lim/5 2
3
−
+−
n
nn
42
1
lim/6
22
+−+ nn
B. GIÔÙI HAÏN HAØM SOÁ
)32(lim/1
2
+
→
x
x
)432(lim/2 3
2
+−
−→
xx
x
1
14
lim/3 2
2
1 +−
++
→ xx
xx
x
1
21
lim/4
3 +
+−
−→ x
xx
x
)2(lim/5 3
1
xx
x
++
−→ 2
25
lim/6
2
5 +
−
→ x
x
x
Daïng
0
0
1
23
lim/4
4
6
lim/1
23
3
1
2
2
2
+−−
+−
−
−+
→
→
xxx
xx
x
xx
x
x
8
4
lim/5
20
16
lim/2
3
2
2
2
2
4
+
−
−+
−
−→
→
x
x
xx
x
x
x
9
3
lim/6
3
34
lim/3
23
2
3
−
+
−
+−
−→
→
x
x
x
xx
x
x
x
x
x
xx
x
x
x
x
x
2
121
lim/7
4
23
lim/4
2
121
lim/1
0
22
0
−+
−
−−
−+
→
→
→
2
24
lim/8
33
223
lim/5
39
4
lim/2
3
2
1
0
−
−
+
+−+
−+
→
−→
→
x
x
x
xx
x
x
x
x
x
25
32
lim/9
34
472
lim/6
32
372
lim/3
2
3
5
31
1
−
+−
+−
−++
+−
−+
→
→
→
x
x
xx
xx
x
x
x
x
x
-Trang 1-

33
276
lim/7
22
2
lim/4
1
1
lim/1
23
24
3
2
2
2
3
1
+++
−−
−+−
−
−
−
−→
→
→
xxx
xx
xx
x
x
x
x
x
x
33
3 2
0
1
2
23
1
232
11
lim/8
45
32
lim/5
43
42
lim/2
+−+
−−
+−
−+
−−
++−
→
→
−→
xx
x
xx
xx
xx
xxx
x
x
x
314
2
lim/9
23
2423
lim/6
11
lim/3
2
2
2
1
2
0
−+
+−
+−
−−−−
++−+
→
→
→
x
xx
xx
xxx
x
xxx
x
x
x
x
x
xx
xx
xxx
xx
x
xx
x
x
x
x
x
x
x
−−
+−
++
++
++
−+
−
+−
−−
→
−→
→
→
→
51
53
lim/5
62
23
lim/4
)1)(1(
lim/3
3
34
lim/2
11
lim/1
4
2
2
2
23
2
3
2
3
3
0
23
1
lim/10
3
11
lim/9
2
321
lim/8
1
12
lim/7
23
1
lim/6
2
3
1
3
0
4
2
23
1
2
3
1
−+
+
−−
−
−+
−
+−+−
−+
−
−→
→
→
→
→
x
x
x
x
x
x
x
xxx
x
x
x
x
x
x
x
• Tính caùc giôùi haïn baèng caùch theâm, bôùt löôïng lieân hôïp.
3
51
lim/3
11
lim/2
23
7118
lim/1
3
3
3
0
2
3
2
−
+−+
−−+
+−
+−+
→
→
→
x
xx
x
xx
xx
xx
x
x
x
2
122
lim/6
2
66
lim/5
1
39
lim/4
21
2
3
2
3
1
−−
−−+
−+
++−
−
++−
−→
−→
→
xx
xx
xx
xx
x
xx
x
x
x
Daïng ∞
∞
2
3
2
5 2
3
1
1/ lim
2 3
1
2 / lim
2
2 1
3/ lim
1
x
x
x
x
x
x x
x
x x
x
→−∞
→+∞
→∞
+
+
− + +
−
+ +
+
2
2
3
2
4
2 3 1
4 / lim
3 5
( 2)(2 1)(1 4 )
5 / lim
(3 4)
3 8
6 / lim
6 1
x
x
x
x x
x x
x x x
x
x x
x x
→∞
→∞
→∞
+ +
− +
− + −
+
+ −
− +
3
2
2
3 3
2
2
3
4 3 7
7 / lim
3 5
2 3
8 / lim
1
4 1
9 / lim
3 1
2 3
10 / lim
2 1
x
x
x
x
x x
x x
x x
x x
x
x
x
x x
→∞
→∞
→∞
→∞
+ −
− +
+ +
− +
+
−
+
− +
ĐS
1 2 8
1/ ;2 / ;3/ ;4 / ;5 /
2 3 27
− ∞ +∞ −
2
6 / 0;7 / ;8/ 1;9 / ;10 / 0
3
∞ ± ±
-Trang 2-

xx
xxx
x
−++
++++
∞→
214
4132
lim/1
2
2
1
12419
lim/2
22
−
++−++
∞→ x
xxxx
x
ĐS 

−
5
1
/1 


−1
1
/2
Daïng ∞−∞






−
−
−
−+
−−−−
−+
→
∞←
∞→
+∞→
31
2
2
3 23
1
3
1
1
lim/4
)(lim/3
)34412(lim/2
)(lim/1
xx
xxx
xxx
xxx
x
x
x
x






+−
+
+−
++−+−
+−
−+
→
−∞→
+∞→
∞→
65
1
23
1
lim/8
)11(lim/7
)1(lim/6
)3(lim/5
222
22
2
3 32
xxxx
xxxx
xx
xxx
x
x
x
x
ĐS
1/4
2
1
/3
0
/2
3
1
/1
−


 ∞−
2/8
1/7
0/6
1/5
−
Daïng : Tìm giôùi haïn cuûa caùc haøm soá löôïng giaùc:
Cho bieát : 1
sin
lim
0
=
→ x
x
x
Baøi taäp 10: Tính giôùi haïn caùc haøm soá löôïng giaùc sau:
20
0
0
0
2
4cos1
lim/4
sin
2cos1
lim/3
11
2sin
lim/2
2
5sin
lim/1
x
x
xx
x
x
x
x
x
x
x
x
x
−
−
−+
→
→
→
→
20
0
2
2
0
30
6cos1
lim/8
2
3
lim/7
3
sin
lim/6
sin
lim/5
x
x
x
xtg
x
x
x
xtgx
x
x
x
x
−
−
→
→
→
→
x
x
x
xx
xtg
x
x
x
x
x
x
x
cos21
3
sin
lim/12
sin
cossin1
lim/11
cos12
lim/10
5cos1
3cos1
lim/9
3
2
2
0
20
0
−






−
−+
+−
−
−
→
→
→
→
π
π
ÑS:
25
9
/9
2
1
/5
2
5
/1
8
2
/10
9
1
/6
4/2
1/11
2
3
/7
2/3
3
1
/12
18/8
4/4
Daïng 1: Tìm caùc ñieåm giaùn ñoaïn cuûa caùc haøm soá:
Baøi taäp: Tìm caùc ñieåm giaùn ñoaïn cuûa caùc haøm soá sau:
.
23
452
/
.345/
2
2
23
+−
+−
=
−+−=
xx
xx
yb
xxxya
.
2
2sincot
/
.5cos/
xtg
xgx
yd
xtgxyc
+
=
+=
Daïng 2: Xeùt tính lieân tuïc cuûa haøm soá:
-Trang 3-

Baøi taäp 1: Cho haøm soá:






−
+−
−
=
1
23
2
)(
2
2
x
xx
x
xf
)1(
)1(
≥
<
x
x
Xeùt tính lieân tuïc cuûa haøm soá f(x) taïi x0 =
1.





−
−
−
=
2
4
21
)( 2
x
x
x
xf
)2(
)2(
<
≥
x
x
2.







−+
−+
=
11
11
2
3
)(
3
x
x
xf
)0(
)0(
>
≤
x
x
0.





−
−
=
5
1
1
)(
2
x
x
xf )1(
)1(
=
≠
x
x
1.





−
−
+
=
1
1
2
)( 3
x
x
ax
xf
)1(
)1(
<
≥
x
x
Ñònh a ñeå haøm soá f(x) lieân tuïc taïi x0 = 1.





−
−−=
x
xxf
2
321
1
)(
)2(
)2(
≠
=
x
x
2.






+−−
+
−
+
=
x
xx
x
x
a
xf
11
2
4
)(
)0(
)0(
<
≥
x
x
Ñònh a ñeå haøm soá f(x) lieân tuïc taïi x0 = 0.






−
−+
+
=
2
223
4
1
)( 3
x
x
ax
xf
)2(
)2(
>
≤
x
x
Ñònh a ñeå haøm soá f(x) lieân tuïc treân R.






+−
−
+
=
23
24
3
2
)(
2
3
2
xx
x
ax
xf
)2(
)2(
>
≤
x
x
Ñònh a ñeå haøm soá f(x) lieân tuïc treân R.
-Trang 4-





−=
x
xxf cos1
1
)( )0(
)0(
≠
=
x
x
Xeùt tính lieân tuïc cuûa haøm soá treân toaøn
truïc soá.
Daïng 3: Chöùng minh phöông trình coù nghieäm:
Baøi taäp 1: CMR caùc phöông trình sau ñaây coù nghieäm:
010010/
01096/
013/
35
23
4
=+−
=−+−
=+−
xxc
xxxb
xxa
Baøi taäp 2: CMR phöông trình 0162 3
=+− xx coù 3 nghieäm trong khoaûng (-2 ; 2).
Baøi taäp 3: CMR phöông trình 0133
=+− xx coù 3 nghieäm phaân bieät.
Baøi taäp 4: CMR phöông trình 02012643 234
=−+−− xxxx coù ít nhaát hai nghieäm.
Baøi taäp 5: CMR caùc phöông trình sau co ùhai nghieäm phaân bieät:
.0)5()9(/
.032)2)(1(/
2
=−+−
=−+−−
xxxmb
xxxma
-Trang 5-

Bai tap gioi han ham so 11 (1)

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Bai tap gioi han ham so 11 (1)