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Bài giảng Giải Tích 1: Giới Hạn và Liên Tục của Hàm Số. Môn Giải tích 1 (GT01) | Trường Đại học Giao thông Vận tải.

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Bài giảng Giải Tích 1: Giới Hạn và Liên Tục của Hàm Số. Môn Giải tích 1 (GT01) | Trường Đại học Giao thông Vận tải.

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BÀI TẬP: GIẢI TÍCH I
CHƯƠNG III: GIỚI HẠN VÀ SỰ LIÊN TỤC CỦA HÀM SỐ
GIẢI BÀI TOÁN TÌM GIỚI HẠN
Bài 1: CMR khi x 0→
1) 1−cosax a x2 2 (a 0)
2
2) sinax+ sin bx2 ax (a 0)
3) ax −1 xlna (1 a 0)
4) ln 1( + ax) ax a( 0)
5) (1+kx)α −1 kαx (kα 0 )
6) axα + a x1 α 1+ + +... a xn α n+ axα (α 0,a 0)
Hướng dẫn giải
1) Ta có: 1−cosax = 2sin
2
ax
2.
ax
2
=
a x
2 2
khi x 0→ , đpcm.
2 2 2
2) limx 0 sinax+sin bx
2
= limx 0
sinax + sin bx
2
= limx 0
→
sinaxax +limx 0
→
sin bxax
2
= +1
limx 0
→
(bxax)
2
= + =1 0 1
→
ax
→
ax ax
sinax+sin bx
2
ax, đpcm.
3) limx 0
→
axlna 0x −1 0
=L limx 0
→
a lnaxlna = limax 0
→
x =1 đpcm.
4) limx 0→ ( ) =L limx 0→ ( a ) = limx 0→ 1+1ax = 1 đpcm. ln 1+ ax 0
a / 1+ ax ax 0
1+kx −1
5) limx 0 ( )α
0 =L limx 0→ α 1( +kakx)α 1− k = lim 1 kxx 0→ ( +
)α 1− =1 đpcm.
→ kαx 0
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Thầy Phạm Ngọc Lam Trường 1
α
α1
x0
0
lim
0
ax
a
−
→
=
α
α
α1
αn
1
n
ax
ax
...
ax
ax
+
+
+
++
,
đpcm
.
x
0
lim
x
x
x
→
=
+
2
7
/
4
1
/
4
sinx
x/2
x
x
6) limx 0 1 n 1( ) n( ) = lim 1( + o x( ))= 1
axα + a xα 1+ + +...a xα n+ L aαxα 1− + a α+1 xα + +... a α+n xα n 1+ −
→ x 0→
Bài 2: Khi x 0→ , cặp VCB sau có tương đương không?
1) α x( )= +x x , β x( )= e
sinx
−cosx
2) α x( )= ln cosx( ), β x( )=−
3) α x( )= arctan sin2x( ) , β x( )= etanx −cos2x
4) α x( )= 3 x− x , β x( )= cosx−1
Hướng dẫn giải
1) limx→0 esinx −cosx esinx − + −1 1 cosx = limx→0 esinx1/4−1+1−cosx1/4 = limx→0
esinxx1/4−1+limx→0 1−xcosx1/4
x x
= lim 1/4 +lx 0im = limx 0 1/4 +lx 0im = limxx 0
3/4
+limx 0
x
7/4
= 0 β
x( )= o(α( )x ) x 0→ x → x → x → 2 → → 2
ln 1 2sin 2 x 2 x
ln cosx ln cosx − −2sin x/ 2
2) lim (
2
) =−2lim ( 2 ) =−2limx 0 2 2 =−2limx 0 2 2 = 4limx 0 ( 2 )2 =1 x 0→ −tan x x 0
→
tan x → tan x → x → x
2
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Thầy Phạm Ngọc Lam Trường 2
α x ~ β x( )
( ) e
−cos2x
3) lim
tanx
= lim e
tanx
− + −1 1 cos2x = lim tanx +lim 2sin x
2
= lim x +lim 2x
2
= 1 x→0 arctan
sin2x( ) x→0 arctan sin2x( ) x→0 sin2x x→0 sin2x x→0 2x x→0 2x 2
α x ,β x( )( ) là các VCB cùng bậc
4) lim cosx 1− = lim 1−cosx = lim x / 2
2
1/3
=−
1limxx 0
5/3
=0 β x( )= o(α( )x ) x 0→ 3 x − x x 0→ x
− 3 x x 0→ −x 2 →
Bài 3: So sánh cặp VCB sau đây:
1) Khi x 0→
+
: α x( )= x− −
3
x
4
x , β x( )= − −1 x 1
2) Khi x→+ : α x( )= +1 12 , β x( )= ln x2 +2 1 x x x
3) Khi x 0→ : α x( )=
3
x x
2
+
3
, β x( )= e
sinx
−1
4) Khi x 0→ : α x( )= tanx, β x( )= e
sinx
− −x
2
1
5) Khi x 0→ : α x( )= 3x
3
−2x
2
, β x( )= ln cos2x( )
6) Khi x 0→
+
: α x( )= x x
3
+
2
, β x( )= e
sinx
−cos2x
Hướng dẫn giải
x( ( )) x~ x−
1/4
) 1) Đáp án: β x( )=o α
x
2
+1
2) L = lim= lim x→+ 1 x→+
x x2 x2
(G
ợ
i ý:
3
4
x
x
−−
2
1
ln1
x
1
1
x
+
+
2
ln
x
1
+
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Thầy Phạm Ngọc Lam Trường 3
Đặt:
1
= t L = lim
+
ln 1 t
(
+
2
2
) = lim
+
t
2
= limt
+
= 0 β x( )=o α( ( )x )
x t→0 t+t t→0 t t→0
3) Đáp án: β x( )=o α x( ( )) (Gợi ý: α x( )=
3
x x ~x
2
+
3 2/3
)
4) Đáp án: α x ~β x( ) ( )
5) Đáp án: α x ~β x( ) ( )
6) Đáp án: α x ~ β x( ) ( )
Bài 4: Với α x ,β x ,γ x( ) ( ) ( ) là các VCB trong cùng một quá
trình x → a, hãy chứng minh:
a) Nếu α x β x( ) ( ) và β x γ x( ) ( ) thì α x γ x( ) ( ).
b) Nếu α x( ) β x( ) và β x( )= o γ x
( )
thì α x( )=o γ x
( )
.
Hướng dẫn giải
α x
( )
α x
( )
a) limx a→ β x( ) =1 limx a γ x( ) =1 (Vì β x( ) γ x( )) α x( ) γ x( ) , đpcm.
→
b) β x( )= o γ x ( ) limx a→ γ xβ x( )( ) =0 limx a→ α xγ x( )( ) =0 (Vì α x β x( ) ( ) )
α x( )=o γ x ( ) , đpcm.
Bài 5: Tính các giới hạn sau:
1) lim x
2
+ + −3x 1 5x 4) limx→ ( x
2
+ 4x− −1 x
2
+ −x 5) x→−3 2x+1
1
2x+ −1 1
2) limx 0
→
3 x+ −1 1 5) limx→ 54xx22 + +− +xx 11 x
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Thầy Phạm Ngọc Lam Trường 4
3) xlim→− ( x2 −2x + x) 6) limx 3 1 5(− xsin x) (2 −1)
→
0 tan −x ln cos3x
Hướng dẫn giải
2 + + − (− + − + − −3)
2
3. 3 1 5. 3 −
1) lim x 3x 1 5x = = 16
x
→−
3
2x+1
2.− +3 1 5
2) lim 3 2x+ −11 = lim (2x+ −1 1)( x+ +1 1) = lim
2( x+ +1 1) = 4
x→0 x+ −1 1 x→0 (x+ −1 1) 3 (2x+1)2 + 3 2x+ +11 x→0 3 (2x+1)2 + 3 2x+ +11 3
3) Đáp án: xlim ( x
2
− + =2x x) − (Gợi ý: nhân liên hợp)
→−
4) lim( x
2
+ − −4x 1 x
2
+ − =x 5)
3
(Gợi ý: nhân
liên hợp) x→ 2
1 ln 5x + +x 1
5) lim 5x22 + +− +xx 11 x = exlim→ 4x22x− +x 1 = exlim→ ln 5
14 1+− /x 1/x/x 1/xx ++ 22 = e0 =1 (Chú ý: xlim A x→x0
( )B x( ) = ex xlim B x lnA x→ 0 ( ) ( ) )
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Thầy Phạm Ngọc Lam Trường 5
x→ 4x
= 5x
2
Bài 6: Tìm các giới hạn sau
1 2tanx(ex
+
−cosx ) 1) limx 0 sinx3x 7) lim
x 0
→
ln
13) lim
x 0
→
3xln 1 2xx2 −(4
+
sin x3)
→ e −1
2) limx 0 8) lim 1 2x( + )cotx 14) limx 0
→x→0 →
3) limx 0 arcsinx2 9) limx 0→ (arcsin x(
6
))ln 1( −2x
10
)
6)
(
)
(
)
)
(
(
)
)
(
2
3
2
x0
x0
2
3
2
2
2
3
5
1
1
xsinx
1
5
xsinx
1
lim
lim
tan
xlncos3x
3
x
x.ln1
2
sin
5
xsinx
1
5
xsinx
1
1
2
→
→
−
−
−
−
=
−
+
−
+
−
−
−
)
(
2
x0
2
3
2
2
2
3
5
sinx
lim
x
3
1
ln12sin
15
xsinx
15
xsinx
2
→
=
−
+
+
−
−
)
(
x0
2
3
2
2
2
3
lim
3
x
xsinx
15
xsinx
15
2
sin
1
2
→
−
+
−
−
+
(
)
2
2
x0
2
3
2
2
3
5
x
lim
x
3
1
5
xsinx
1
xsinx
1
5
.
2.
2
→
=
−
+
−
+
−
)
(
x0
2
3
2
2
3
1
10
7
lim
9
5
xsinx
1
1
1
5
xsi
0
2
nx
1
→
−
=
=
+
−
−
−
+
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Thầy Phạm Ngọc Lam Trường 6
→ x+ 2x
x
( ) 15) limx 0 16 e2x −1 e −cosx−ln 1 x
+
→ sin x
4) lim ( ) 1 4+ −x 1
x 0
→
ln
1−3x16) lim
x 0
→
ln 1( +3x)
0lim→
+
5) x
xe2x2+−x13 17) limx 0→ eαx −x eβx
6) limx 0 18) lim eαx −eβx
→
x 0
→
sinαx−sinβx Hướng dẫn giải sinx3x = limx 0 x
=
1
1) Cách 1: Thay VCB tương đương: lim
x 0
→
e −1
→
3x 3
Cách 2: Sử dụng quy tắc L’Hospital: lim sinx3x = limx 0
cosx
3x
=
1
x 0→ e −1
→
3e 3
2) lim tanx
=
1
(Gợi ý: tương tự ý 1)
x 0
→
ln 1( + 4x) 4
3) lim arcsinx = lim x = 1(Ở đây đã sử dụng quy tắc ngắt bỏ VCB bậc cao:
x 2x ~x
+
2
) x 0→ x+ 2x2 x 0
→
x
e
2x
−1 =−2
4) lim
x
0
→
ln(1 3− x) 3
5) lim+ e2x2 −13 =1 x→0 x +x
x
10)
2
x0
lim
x
→
11)
x
2
x0
e
1
lim
x
x
+
→
−
+
12)
(
)
2
2
x0
1
lnsinx
sinx
x
lim
tanx
xcosx
→
++
−
−
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Thầy Phạm Ngọc Lam Trường 7
6) limx 0 1(−cosx2) = limx 0
2
(
sin
2
22) = limx 0 x / 2
2
2 = 1
→ ln 1+x → ln 1+x → x 2
1
ln 1+ 2tanx
7) lx 0im
→
(
ex −cosx
)
= limx 0
→
ex2−tanxcosx
00 =
L
limx 0
→
e
2
x cos x+sinx2 = =12 2
8) lim 1( + 2x)
cotx
= e
2
(Gợi ý: Giải tương tự Bài 3, ý 5) x→0
9) limarctanx = 1 (Gợi ý: tương tự ý 3) x 0→ 2x+x2 2
x
0
→
x
x2 ( ) 00 =L limx 0 x
+
sinx
−
1+1x 0 =L l mx 0i→ e
x
+cosx2
+
(1+
1
x)
2
= 23 e
−cosx−ln 1+ x
e
10) lim → 2x 0
11) x 0lim
+
e
x
−
1
2
=
x 0lim
+
x
=1
→ x+ x → x
sinx−ln sinx( + +1 x
2
)
12) lim
x 0
2
→
tanx− xcos x
sinx = − +x
x
3
o x(
3
) 1+x
2
− =1
1
x
2
+o x(
3
) ln 1 u( + = − + +) u
u
2
u
3
o u(
3
)
6 2 2 3
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Thầy Phạm Ngọc Lam Trường 8
ln sinx( + 1+x
2
)=(sinx+ 1+x
2
− −1)
1
(sinx+ 1+x
2
−1)
2
+
1
(sinx+ 1+x
2
−1)
3
+o x(
3
)
23
= x−x3 +x2 +o x( 3) −1 x−x3 + x2 +o x( 3) 2 +1 x−x3 + x2 +o x( 3) 3 +o x( 3)= −x x3 +o
x( 3)
6 2 2 6 2 3 6 2 3
TuSo
= +
x
3
o
x(
3
) 6
tanx = + +x
x
3
o x(
3
) cosx = −1
x
2
+o x(
2
) xcos x
2
= x 1 −
x
2
+o x(
2
)
2
= −xx
3
+o x(
3
)
3 2 2
MauSo =
4x
3
+o x(
3
)
3
Từ đó lim sinx−ln(sinx
+ +
2 1 x
2
)
= limx 0 x /633 = 1 x
0
→
tanx− xcos x
→
4x / 3 8
13) limx xln 12 ( + 2x3
)
= limx 0 x.2x2 = 2
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Thầy Phạm Ngọc Lam Trường 9
→
0 3x −4sin x
→
3x 3
14) lim 1−cos2x = lim 2sin x2 = lim 2x2 = lim 2x2 = −2
x 0
→
ln 1( − 3xarctanx)
x 0
→
ln
1( −3xar tc anx)
x 0
→
−3xarctanx
x 0
→
−3x.x 3
(arcsin x(
6
))ln 1( − 2x
10
) x .
6
(−2x
10
)
15) lx 0im 16 = limx 0 16 =−2
→ sin x → x
16) lim 1 4+ −x 1 = lim 1 4+ −x 1 = lim 1 4+ −x 1 = lim 4 = 2 x 0
→
ln 1 3x(
+
) x 0
→
3x x 0
→
3x( 1 4
+ +
x 1)
x 0
→
3( 1
+ +
4x 1) 3
1) limx 2
7) lim → x
x→+ x3 − +4x
8
2) xlim sin x( + −1 sin x)
2
1
→+ limx 1+ sin
8)
3) xlim 3x−sin5x2x x 0→ x
→+ e sin2x+cos2x
9) lim
e
αx
−eβx
17) lim =α−β (Tương tự ý 1) x 0→
x
eαx −−esinβx 0βx
lim
0 = =L ... 1
18)
x 0
→
sinαx
Bài 7: Tìm các giới hạn sau
( 1)−
x
.cos2x
x x + −1 cosx
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Thầy Phạm Ngọc Lam Trường 10
4) lim x( −2)cos 2
x
x→ 2x−6 + 5
x 2→ x
− +
5x 6 sin(2x+7).cos(x2)+cos (42 −x )3
1 10)
lim
5) limx e2 sin x x→ x
x→0
2 3
x (sinx+cos x)
6) limx ( 1)−2
x
11) xlim→− (x2 +1)(x−3)
→ 2x +7
Hướng dẫn giải
lim x→
=0
x→+
Gợi ý: 2cos
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Thầy Phạm Ngọc Lam Trường 11
x→+
7) lim
= xlim→+ 1 4−
1
1
/ x+2 +
/ x
8/ x3 −xlim→+ x3
cosx
−4x+8 = − =1 0 1
Ở đây, xlim→+
x3
cosx
−4x+8
=0 theo nguyên lý kẹp, các bạn tự chứng minh.
Các ý 8), 9), 10), 11) giải tương tự.
8) Đáp án: 0
9) Đáp án: 5
10) Đáp án: 0
11) Đáp án: 0
Bài 8: Tính các giới hạn sau:
1) xlim→+ π 2ln 1− arctanx+ 1 8) limx 1
−
m
xm
−
1−
n
xn
(m,n N
*
)
→
1
x
9) lim(a R
*
)
2) lim
−
(π 2− x)cosx
x 0
→
1
) xlim π 2− arctanx+ x = xlim π−2arctanxx 0 =L xlim→+ 1+−x12x2
= xlim→+ 12+xx22 = xlim→+ 1/ x22 +1 = 2
→+ ln 1 1 →+ 1 0
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Thầy Phạm Ngọc Lam Trường 12
x
)
x→
10) lim lnx
3) lim tanx
−
( )
1
cosx x 0→ + ln sinx( )
x→x −e−
x
e
4) xlim→+
(π 2− arctanx lnx)
11)
limx 0
→
5)
lim
x 1
→
xx−1− lnx1 12) limx 0→ sinx x2
1
6) limx 0
→
13) lim ex 0→ (
x
+ x
x
7) x 1lim lnxln x 1→ + ( ( − )) 14) xlim→+ π2 arctanx x
Hướng dẫn giải
−2
+ x. 1
−
1
1
) lim
−
(π 2− x)
cosx
=1 (Gợi ý: Giải tương tự Bài 3, ý 5).
x→ 3) lim tanx
−
( )
2cosx
=1 (Gợi ý: Giải tương tự Bài
3, ý 5).
x→ x→+ (π− 2arctanx lnx) = xlim
→+
π
−
2arctanx
0
0
= =
L
... 0 4) lim
lnx
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Thầy Phạm Ngọc Lam Trường 13
5) limx 1
→
x− − 1 = limx 1
→
xlnx( − − +)x1 0 =L limx 1
→
lnx
x−
=
limx 1
→
lnx
0 =L limx 1
→
1/ x1+/ x1/ x2 = 12
x 1 lnx x 1 lnx
0 lnx+ x 1 lnx+
−1 1 0 x x
ln
sin2x
6) lim
x 0
ln sinx(( )) =
L
lim
x 0
→
= lim
x 0
→
2sin2x.cosxcos2x.sinx = lim
x 0
→
22cos2x.sinxsinxcos x2
= limx 0
→
cos2xcos x2 = 1
→
ln x( −1)
7) x 1lim lnxln x+ ( ( −1))
=
x 1lim+
= =
L
... 0
→ →
lnx
8) limx 1→ −m m − −n n = limx 1→ m(1−−mxxmn)(− +1n−xnxn)m = limx 1→
m1−−xmxn −nx− +mn+xnxm n+m 00
1 x 1 x
−
=L limx 1n 1− −mnxmxm 1n 1−− ++nmx(m+m 1n x−) m n 1+ −
00 = =L ... m2−n
→ −nx
9) limx 0
→
ln 1ax( + xax) = limx 0
→
eax −xe−ax 00
=L lim aex 0
→
( ax + ae−ax)= 2a e −e− lnx
L
10) x 0lim→
+
ln sinx( )
= =... 1
1
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Thầy Phạm Ngọc Lam Trường 14
11) limx 0
→
sinxcosxex −e−x = 2limx 0
→
esin2xx −e−x = limx 0
→
ex −xe−x 00 =L lim
ex 0→ ( x +e−x)= 2
Các ý 12), 13), 14) giải tương tự Bài 3, ý 5
1
12) l mx 0i
→
sinxx x2 = −61
1
13) l mi (e
x
+x)
x
= e
2
x 0→
14) xlim→+ π2 arctanx x = e−π2
Bài 9: Giải thích vì sao các giới hạn sau không dùng được quy tắc L’Hospital rồi tìm chúng bằng
cách khác
x−sinx x+cosx
1) lim 3) lim x→+ x+sinx x→+ x
2) limx 0→ x sin 1/ x
2
sinx(
)
4)
limx
0→ x+cotxsinx
Hướng dẫn giải
Không dùng được quy tắc L’Hospital vì khi x→+ các hàm sinx và cosx không tồn tại giới hạn.
x− 1− sinx −
sinx
=1
1) lim
sinx
= lim
x
, có lim
sinx
= 0 (Theo
nguyên lý kẹp) nên lim
x
x→+ x+sinx x→+ 1+ sinx x→+ x x→+ x+sinx
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Thầy Phạm Ngọc Lam Trường 15
x
2) −x2 x sin 1/ x2 ( ) x2 mà lim x2 = lim −x2
=0 lim x sin 1/ x2
( ) =0 (Nguyên lý kẹp) sinx
sinx sinx
x
→
0
sinx
x
→
0
sinx
x
→
0
sinx
x−1 x+cosx x+1 x+1 = x 1− = x+cosx =
3) mà lim lim 1 lim x x x x→+ x x→+ x x→+
1 (Theo nguyên lý kẹp)
x
4) lim x+sinx = lim(x+sinx sinx)
=0
x→0 cotx x→0 cosx
Bài 10: Tính các giới hạn sau
1) lim x
0→
2) limx 1 sin3 πx
→ x −1
5) lim x→
x
3
+8
6) xlim
2
2
→−
x −3x−10
8) lim x 0→
x
arcsin
9) lim 1−x2 x 0→ ln
1 x( − )
10) lim x 0→
11) lim
x→1/2
12) lim
x 0
+
2( 2(3)x)
ln 1
sin
→
tan 5x
18) lim n a+ − −x n a
x
x 0→ x
19) limx0 +
1+tanxsinx
→
1
22) limsin2x+2arctan
3x+3x
2
x 0
→
ln
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Thầy Phạm Ngọc Lam Trường 16
1( +3x+sin x
2
)+xe
x
23) xlim x→+
π
2 −arcsin
x2
x
+1
cosx−cos3x
3) limx 0→ x2
4) lim x 0→
7) xlim 2x→− ( +1) x43x++x1
5
(1 3+ x)
2
−1
13) lim
x 0
→
sinx+2sin x2 sin 2x(
2
−
−4x 6)
14) lim x 3→ ln x( 3 −26)
15) lim
3→
x
16) lim
2→ x
17) limx→0 x2 +sin x3
2x+ 2 3x
+
2
20) limx→ 2x−2
3
1+tan x
2
−
3
1−tan x
2
21) lim
x 0→ x+ 3 x2
24) lim x
4/3
(
3
x
2
+ −1
3
x
2
−1)
x→+
tanx−sinx
25) limx 0→ sin x3
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Thầy Phạm Ngọc Lam Trường 17
Hướng dẫn giải
1) limtan3x = lim 3x = 3 x 0→ tan2x x 0→ 2x 2
2) limx 1 sin3 πx 0 =
L
limx 1→ πcosπx3x2 = −3π
→ x −1 0
3) Cách 1: Sử dụng quy tắc L’Hospital:
cosx−x2cos3x 0
limx
0→
0 =
L
limx 0→ −sinx+2x3sin3x 00 =
L
limx 0
→
−cosx+29cos3x = 4
Cách 2: Thay VCB tương đương: limcosx−2cos3x = lim−2sin2xsin2(−x) = limx 0 4x22 = 4
x 0→ x x 0
→
x → x
4) limx 0→ sin5x−sin3x = limx 0→ sin5x−sin3x 0 = =L ... 2 (Có thể dùng thay
VCB như ý 3) sinx x 0
5) limx 3x
3
3 + +5x2 1 = limx 3+ 5 / x
2
+1/ x3
3
= 3
→
2x + −x 9
→
2 1+ / x−9/ x 2
6) xl→−im2 2 x
3
+8 0 = =
L
... −12 x −3x−10 0 7
7) xlim 2x( +1) 43x+1 = xlim 2( +1/ x) 4+1/ x2 = 4
→− x + x →− 1+1/ x
8) limx 12 −cos3x3 4 = limx 0 2sin2
2
(33x / 2)4 = limx 0 2. 3( x /2 2)
2
= 9
→0 2x + 3x − x → 2x + 3x − x → 2x 4 x
x arcsin
−
9) lim 1−x2 = lim 1−x2 = lim 1 =−1 x 0→ ln 1 x(
−
) x 0→ −x x 0→ 1−x2
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Thầy Phạm Ngọc Lam Trường 18
10) lim ln 1(( + 4x22 − 5x33) = lim 4x22 − 5x33 = limx 0 4x22 = 2 x 0→ ln 1+ 2x + 3x ) x 0→ 2x + 3x
→ 2x
11) xlim→1/2 arcsin 1(2 −− 2x) = xlim→1/2 41x−22−x 0 =L xlim→1/2
8−x2 = −21
4x 1 1 0
ln 1 sin
12) lim
x 0
+ 2( 2(3)x) = limx 0 sin22((3x)) = limx 0 ((3x))22 = 9
→
tan 5x
→
tan 5x
→
5x 25
13) limx 0
→
sinx5 (1++32xsi)2n−2 x1 = limx 0
→
5 (1+ 3xx)2 −1 00 = =L ... 65
14) lim sin 2x
(
2
− 4x
−
6
)
= lim sin 2x
(
2
−
4x−6
)
2x23− 4x−6 0
= =L ...
8
x 3
→
ln x( 3 − 26) x 3
→
ln 1( + x3 − 27) = limx 3
→
x − 27
0 27
15) lx 0im
→
ln 1ln cos3x
(
(+tan x2)
)
= limx 0
→
ln cos3xtan x( 2 ) = limx 0
→
ln cos3x( x2 )
0
0
= =
L
...
−
2
9
16) limx
→
2 sin 3xtan 4( (2 −−2xx2−) 8) = limx
→
2 3x24−−2xx2−8 0 0 = =L ... −25
17) limx 0→ ln 1 xtx(2 ++sin xanx3 ) = limx 0→ xtanxx2 =1
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Thầy Phạm Ngọc Lam Trường 19
18) lim
→
n a+ −x x n a− x 00 = =L ... n2 a1/n−1
x 0
19) limx 0 +
+1 tanx
1+tanxsinx = ex 0lim→ ln 1sinx+sinx
= =... 1
→ 1
1 1
2 2
= 2limx4/3 =0
3 x→0
22) Cách 1: Thay tương đương
Khi x → 0,sin2x + 2arctan3x + 3x ~ 8x
2
và ln 1 3x( + +sin x
2
)+xe ~ 4x
x
(Các bạn tự chứng
minh). sin2x+2arctan3x+3x
2
= lim8x = 2
Nên lim x 0→ ln 1 3x( + +sin x2 )+xex x 0→
4x
sin2x+2arctan3x+3x
2
0 Sử dụng quy tắc L’Hospital. Cách 2:
lim
x 0
→
ln 1 3x( + +sin x
2
)+xe
x
0
π−
x
20)
x
ln
l
2
6
im
x
3
1
3
2
x
x
x
2
2
...
e
lim
e
2
x
2
→
+
+
+
→
+
==
=
−
21)
(
)
3
3
2
2
3
3
2
2
2
/
3
3
2
0
0
x
x
1
1
1
1
tanx
tanx
1
tanx
1
tanx
lim
lim
x
x
x
→
→
+
−−
−
−
−−
+
=
+
3
3
2
2
2
2
2
2
/
2
/
3
2
2
3
3
/
2
2
3
2
/
3
3
/
/
x
0
x
x
0
x
0
x
0
x
0
0
x
tanx
1
tanx
1
1
tanx
1
tanx
tanx
lim
lim
lim
lim
lim
lim
3
3
3
3
x
x
x
x
x
x
→
→
→
→
→
→
−
−
−
+
=
=
+
=
=
−

Tài liệu khác của Đại học Giao thông vận tải

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lOMoAR cPSD| 58833082
BÀI TẬP: GIẢI TÍCH I
CHƯƠNG III: GIỚI HẠN VÀ SỰ LIÊN TỤC CỦA HÀM SỐ
GIẢI BÀI TOÁN TÌM GIỚI HẠN
Bài 1: CMR khi x 0→ 1) 1−cosax a x ( 2 2 a 0) 2
2) sinax+ sin bx2 ax (a 0) 3) ax −1 xlna (1 a 0)
4) ln 1( + ax) ax a(  0) 5) (1+kx) − α 1 kαx (kα 0 ) 6) ax (
α + a x1 α 1+ + +... a xn α n+ axα
α 0,a 0)
Hướng dẫn giải
1) Ta có: 1−cosax = 2sin2 ax 2.  ax   2 = a x2 2 khi x 0→ , đpcm. 2 2 2  
2) limx 0 sinax+sin bx2 = limx 0 sinax + sin bx2
 = limx 0→ sinaxax +limx 0→ sin bxax2 = +1 lim (
x 0→ bxax)2 = + =1 0 1 → ax →  ax ax
sinax+sin bx2 ax, đpcm.   3) lim → = → → x 0
axlna 0x −1 0 
L limx 0 a lnaxlna = limax 0 x =1 đpcm.   4) lim ( ) ( x 0→   =L limx 0→ a
) = limx 0→ 1+1ax = 1 đpcm. ln 1+ ax  0 a / 1+ ax ax   0
1+kx −1    5) lim ( ) ( + x 0 α 0 =L limx 0→
α 1( +kakx)α 1− k = lim 1 kxx 0→
)α 1− =1 đpcm. → kαx   0 lOMoAR cPSD| 58833082
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ 0  6) lim ( ) ( ) = x 0 1 n 1 lim =   n
lim 1( + o x( ))= 1 α → α1 − x0 + ax0a αx + ax + α
a xα 1+ + +...a xα n+ L
aαxα 1− + a α+1 xα
... a α+n xα n 1+ − → α α1 + αn + αx 0→ ax + ++ 1 ax ... n ax ax , đpcm .
Bài 2: Khi x 0→ , cặp VCB sau có tương đương không?
1) α x( )= +x x , β x( )= esinx −cosx
2) α x( )= ln cosx( ), β x( )=−
3) α x( )= arctan sin2x(
) , β x( )= etanx −cos2x
4) α x( )= 3 x− x , β x( )= cosx−1
Hướng dẫn giải
1) limx→0 esinx −cos =
x l im esinx − + −1 1 cosx = limx→0    esinx1/4−1+1−cosx1/4   = limx→0 x → 0 esinxx1/4x +
−1+lxi mx→0 1−xcosx1/4 x 2 7 / 4 sinx x/2 x x 1 /4 x x = lim x 1 + + /4 lx 0im = limx 0 1/4 lx 0im
= limxx 0 3/4 +limx 0
7/4 = 0 β
x( )= o(α( )x ) x 0→ x → x → x → 2 → → 2
ln 1 2sin  2 x2 x ln cosx ln cosx −  −2sin x/ 2 2) lim ( ) ) ( ) 2 =−2lim ( → 2
=−2limx 0 2 2 =−2limx 0 2 2 = 4limx 0 2 2 =1 x 0→ −tan x x 0
tan x → tan x → x → x 2
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Thầy Phạm Ngọc Lam Trường 1 lOMoAR cPSD| 58833082
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ α x ~ β x( ) ( ) e −cos2x
3) lim tanx = lim etanx − + −1 1 cos2x = lim tanx +lim 2sin x2 = lim x +lim 2x2 = 1 x→0 arctan
sin2x( ) x→0 arctan sin2x( ) x→0 sin2x x→0 sin2x x→0 2x x→0 2x 2
α x ,β x( )( ) là các VCB cùng bậc =−
4) lim cosx 1− = lim 1−cosx = lim x / 22 1/3
1limxx 0 5/3 =0 β x( )= o(α( )x ) x 0→ 3 x − x x 0→ x
− 3 x x 0→ −x 2 →
Bài 3: So sánh cặp VCB sau đây:
1) Khi x 0→ +: α x( )= x− −3 x 4 x , β x( )= − −1 x 1
2) Khi x→+ : α x( )= +1 12 , β x( )= ln x2 +2 1 x x x
3) Khi x 0→ : α x( )= 3 x x2 + 3 , β x( )= esinx −1
4) Khi x 0→ : α x( )= tanx, β x( )= esinx − −x2 1
5) Khi x 0→ : α x( )= 3x3 −2x2 , β x( )= ln cos2x( )
6) Khi x 0→ +: α x( )= x x3 + 2 , β x( )= esinx −cos2x
Hướng dẫn giải (G ợ i ý: 3 4 x −− x
1) Đáp án: β x( )=o α (
x ( )) x~ x− 1/4 )  1ln1 +   ln 2 2xx 1 1 1 + + x2 +1 x
2) L = lim= lim x→+ 1 x→+ x x2 x2
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Đặt: 1 = t L = lim ln 1 t( ) = t2 = = + +2 2 lim+
limt+ 0 β x( )=o α( ( )x ) x t→0 t+t t→0 t t→0
3) Đáp án: β x( )=o α x( ( )) (Gợi ý: α x( )= 3 x x ~x2 + 3 2/3 )
4) Đáp án: α x ~β x( ) ( )
5) Đáp án: α x ~β x( ) ( )
6) Đáp án: α x ~ β x( ) ( )
Bài 4: Với α x ,β x ,γ x( )
( ) ( ) là các VCB trong cùng một quá
trình x → a, hãy chứng minh: a) Nếu α x β x( ) ( ) và β x γ x( ) ( ) thì α x γ x( ) ( ).
b) Nếu α x( ) β x( ) và β x( )= o γ x  ( )  thì α x( )=o γ x  ( )  .
Hướng dẫn giải α x( ) α x( )
a) limx a→ β x( ) =1limx a γ x( ) =1 (Vì β x( ) γ x( ))  α x( ) γ x( ) , đpcm. →
b) β x( )= o γ x  ( )   limx a→ γ xβ x( )( ) =0 limx a→ α xγ x( )( ) =0 (Vì α x β x( ) ( ) )
α x( )=o γ x  ( )  , đpcm.
Bài 5: Tính các giới hạn sau: ( )
1) lim x2 + + −3x 1 5x 4) limx→
x2 + 4x− −1
x2 + −x 5 x→−3 2x+1 1 2x+ −1 1  2) lim → x 0 3 x+ −1 1
5) limx→  54xx22 + +− +xx 11   x
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Thầy Phạm Ngọc Lam Trường 3 lOMoAR cPSD| 58833082
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___________________________________________________________________________________________________________________________________________________________________________________________________ ______________ ( ) 3) xlim→−
x2 −2x + x
6) limx 3 1 5(− xsin x) (2 −1)
→0 tan −x ln cos3x
Hướng dẫn giải 2 + + −
(− + − + − −3)2 3. 3 1 5. 3 − 1) lim x 3x 1 5x =
= 16 x→−3 2x+1 2.− +3 1 5 )
2) lim 3 2x+ −11 = lim (2x+ −1 1)( x+ +1 1 = lim ( ) 2 x+ +1 1 = 4 ( ( ( x→0
x+ −1 1 x→0 x+ −1 1)  3 2x+1)2 + 3 2x+ +11  x→0   3 2x+1)2 + 3 2x+ +11  3     ( )
3) Đáp án: xlim x2 − + =2x x
− (Gợi ý: nhân liên hợp) →− ( ) 3 4) lim
x2 + − −4x 1 x2 + − =x 5 (Gợi ý: nhân
liên hợp) x→ 21
ln5x + +x 1  
5) lim5x22 + +− +xx 11   x = exlim→  4x22x− +x 1 = exlim→ln 5
14 1+− /x 1/x/x 1/xx ++
22     = e0 =1 (Chú ý: xlim A x→x0 ( ) ( ) ( )
B x( ) = ex xlim B x lnA x→ 0 )
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ x→  4x − − ( 2 3 2 1 −5 xsinx 1 − )
1 5 xsinx 1 6) lim =l im x0 → tan( − ) xl nc ( os3x ) x0 → (    2  − ) 2 3 x − 3
x.ln12 sin   ( 2 1 −5 xsinx ) 3 2
+ 1 −5 xsinx 1 +   2    2 5 sinx =lim x0 →     2 x 3 − 3 ln12sin   ( 2 2 15 − xsinx ) 3 2
+ 15 − xsinx 1 +   2    = 5x2 lim x0 →   2 3 x − 3 2 sin ( 2 2 15 − xsinx ) 3 2
+ 15 − xsinx 1 +   2   2 5 x =lim → 2 x0x 3    − 3
2.   .  ( 2 2 1 −5 xs inx ) 3 2
+ 1 −5 xsinx 1 +   2    10 − 1 1 − 0 = lim = x0 9 →   27 3  ( 2 2 1 −5 xsinx ) 3 2
+ 1 −5 xsi nx +1   
Bài 6: Tìm các giới hạn sau + 1) limx 0 sinx3x 7) limx 0→ ln 1 2tanx(e − x cosx ) + 13) lim )
x 0→ 3xln 1 2xx −( 2 4 sin x3 → e −1
2) limx 0 8) lim 1 2x( + )cotx 14) limx 0 →x→0 → ( 3) lim )) ) x 0 arcsinx2 9) limx 0→
arcsin x( 6 ln 1( −2x10
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ ( ) + → x+ 2x x 15) lim − x 0 16 e2x 1
e −cosx−ln 1 x → sin x 10) → 2 x0 lim x
4) lim ( ) 1 4+ −x 1 x 0→ ln
1−3x16) lim x e 1 − 11) lim + x0 → 2 x 0→ + ln 1( +3x) x x sinx −l ( 2 nsinx ++1 x ) 5) x 0lim→ + 12) lim → 2 x0 tanx −xc osx xe +− 2x2 x13 17) limx 0→ eαx −x eβx 6) limx 0
18) lim eαx −eβx =
→ x 0→ sinαx−sinβx Hướng dẫn giải sinx = 3x limx 0 x 1
1) Cách 1: Thay VCB tương đương: lim
x 0→ e −1 → 3x 3 =
Cách 2: Sử dụng quy tắc L’Hospital: lim sinx = cosx 1 3x limx 0 3x
x 0→ e −1 → 3e 3 =
2) lim tanx 1 (Gợi ý: tương tự ý 1) x 0→ ln 1( + 4x) 4
3) lim arcsinx = lim x = 1(Ở đây đã sử dụng quy tắc ngắt bỏ VCB bậc cao: x 2x ~x+ 2 ) x 0→ x+ 2x2 x 0→ x
e2x −1 =−2 4) lim
x 0→ ln(1 3− x) 3 5) lim − = +
e2x2 13 1 x→0 x +x x
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ 6) lim ) = 2(sin ) = = x 0 1(−cosx2 limx 0 2 22 limx 0 x / 22 2 1 → ln 1+x → ln 1+x → x 2 1 ln 1+ 2tanx ( )   7) l → →
x 0im→ ex −cosx
= limx 0 ex2−tanxcosx 
  00 =L limx 0 e2x cos x+sinx2 = =12 2
8) lim 1( + 2x)cotx = e2(Gợi ý: Giải tương tự Bài 3, ý 5) x→0
9) limarctanx = 1 (Gợi ý: tương tự ý 3) x 0→ 2x+x2 2 ( ) + +( x → sinx− 0 x x2
  00 =L limx 0 x
1+1x     0 =L l mx 0i→ ex +cosx2
1+1x)2 = 23 e
−cosx−ln 1+ x   e
10) lim   → 2x   0 = 11) e x x 0lim x −1 = + 2 x 0lim+ 1 → x+ x → x ( ) sinx−ln sinx + +1 x2 12) limx 0 2 → tanx− xcos x sinx = − + ) ) x x3
o x( 3 1+x2 − =1
1x2 +o x( 3 ln 1 u( + = − + +) u u2 ) u3o u( 3 6 2 2 3
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ ( ) ) ) + )  ln sinx + 1+x 1( 1( ) 2
=(sinx+ 1+x2 − −1
sinx+ 1+x2 −1 2
sinx+ 1+x2 −1 3 +o x( 3 23  = x−x ) ) ) )
3 +x2 +o x( 3   −1  x−x3 + x2 +o x( 3   2 +1  x−x3 + x2 +o x( 3   3 +o x( 3 = −x x3 +o x( ) 3 6 2 26 2 36 2 3 = +  TuSo x3 o x( ) 3 6 tanx = + + ) ) x x3 o x( 3 cosx = −1
x2 +o x( 2 xcos x2 = x 1 − x ) )
2 +o x( 2   2 = −xx3 +o x( 3 3 2 2   MauSo = 4x )
3 +o x( 3 3 ( ) + +
Từ đó lim sinx−ln sinx 2 1 x2
= limx 0 x /633 = 1 x → 0
tanx− xcos x → 4x / 3 8 ) 13) lim ( + = = x xln 12 2x3 limx 0 x.2x2 2
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→0 3x −4sin x → 3x 3
14) lim 1−cos2x = lim 2sin x = = = − 2 lim 2x2 lim 2x2
2 x 0→ ln 1( − 3xarctanx) x 0→ ln
1( −3xar tc anx) x 0→ −3xarctanx x 0→ −3x.x 3 ( arcsin x( )) ) ( ) 6 ln 1( − 2x10 x .6 −2x10 15) lx 0im = 16 limx 0 16 =−2 → sin x → x ( + + )
16) lim 1 4+ −x 1 = lim 1 4+ −x 1 = lim 1 4+ −x 1 = lim 4 = 2 → → → x 0
ln 1 3x( + ) x 0 3x x 0 3x 1 4 x 1 + + → ( ) x 0 3 1 4x 1 3 eαx −eβx 17) lim
=α−β (Tương tự ý 1) x 0→ x lim
eαx −−esinβx 0βx
    0 = =L ... 1 18) x 0→ sinαx   1) limx 2 7) lim → x
Bài 7: Tìm các giới hạn sau x→+ x3 − +4x 8 ( 1)− x.cos2x x x + −1 cosx ( ) 2) + − xlim sin x 1 sin x 2   1
→+ limx 1+ sin   8)
3) xlim 3x−sin5x2x x 0→    x  →+ e sin2x+cos2x  9) lim
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4) lim x( −2)cos x 2
x→   2x−6 + 5  x 2→ x − +5x 6
sin(2x+7).cos(x2)+cos (42 −x )3   1 10) lim  
5) limx e2 sin x x→ x x→0 2 3 x (sinx+cos x) 6) lim + x
( 1)−2 x 11) xlim→− (x2 1)(x−3) → 2x +7
Hướng dẫn giải lim x→ =0 x→+ Gợi ý: 2cos
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ x→+ 7) lim = cosx
xlim→+ 1 4− 1 1/ x+2 +/ x8/ x3 −xlim→+     x3
−4x+8     = − =1 0 1   Ở đây, cosx xlim→+    x3
−4x+8    =0 theo nguyên lý kẹp, các bạn tự chứng minh.
Các ý 8), 9), 10), 11) giải tương tự. 8) Đáp án: 0 9) Đáp án: 5 10) Đáp án: 0 11) Đáp án: 0
Bài 8: Tính các giới hạn sau:  −  1) ( )
xlim→+ π 2ln 1−   arctanx+ 1  8) limx 1
−mxm 1−nxn   m,nN* →  1x
9) lim(aR ) * 2) lim (
− π 2− x)cosx x 0→ 1 ) xlim
π 2−  arctanx+ x   = xlim
π−2arctanxx     0 =L xlim→+ 1+−x12x2 = xlim→+
12+xx22 = xlim→+
1/ x22 +1 = 2
→+ ln 1  1 →+ 1   0
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ x→ 10) lim lnx 3) lim tanx ( )1 − cosx
x 0→ + ln sinx( ) x→ − x e−x e  4) xlim→+
 (π 2− arctanx lnx)   11) limx 0→  5) lim →  x 1
xx−1− lnx1   
12) limx 0→    sinx x21 x  ) 6) lim → (
x 0 13) lim ex 0→ x + x x 7) ( ( − )) x 1lim lnxln x 1→ +
14) xlim→+   π2 arctanx   x
Hướng dẫn giải −2 + x. 1 −1 1 ) lim (
− π 2− x)cosx =1 (Gợi ý: Giải tương tự Bài 3, ý 5). ( ) x→ 3) lim tanx 2cosx = −
1 (Gợi ý: Giải tương tự Bài 3, ý 5). π
x→ x→+   (π− 2arctanx lnx)   = xlim→+ −  2arctanx = =
      00 L ... 0 4) lim lnx
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ 5) lim lnx
x 1→   x− − 1   = limx 1→ xlnx( − − +)x1 0    =L limx 1→ x− = limx 1→ lnx
    0 =L limx 1→ 1/ x1+/ x1/ x2 = 12x 1 lnx x 1 lnx
  0 lnx+ x 1 lnx+
−1 1   0 x x ln sin2x  6) limx 0 ln sinx((
))   =L limx 0→= limx 0→ 2sin2x.cosxcos2x.sinx = limx 0→ 22cos2x.sinxsinxcos x2 →
= limx 0 cos2xcos x2 = 1 →
ln x( −1)   7) ( 1 ( x 1lim lnxln x+
−1))= x 1lim+   = =L ... 0 → → lnx  8) lim )( )
x 1→   −m m − −n n    = limx 1→ m(1−−mxxmn
− +1n−xnxn m = limx 1→
m1−−xmxn −nx− +mn+xnxm n+m       00 1 x 1 x − = )
L limx 1n 1− −mnxmxm 1n 1−− ++nmx(m+m 1n x− m n 1+ −
      00 = =L ... m2−n → −nx 9) lim → ( + ) = → − x 0 ln 1ax xax
limx 0 eax xe−ax       00 = → ( + )= L lim aex 0 ax ae−ax
2a e −e− lnx    L
10) x 0lim→ + ln sinx( )     = =... 1
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ 11) lim → → → x 0
sinxcosxex −e−x = 2limx 0 esin2xx −e−x = limx 0 ex −xe−x       00 =L lim e ( ) x 0→
x +e−x = 2
Các ý 12), 13), 14) giải tương tự Bài 3, ý 5 1
12) l mx 0i→    sinxx    x = − 2 61 1
13) l mi (ex +x)x = e2 x 0→ 14)
xlim→+    π2 arctanx   x = e−π2
Bài 9: Giải thích vì sao các giới hạn sau không dùng được quy tắc L’Hospital rồi tìm chúng bằng cách khác x−sinx x+cosx 1) lim
3) lim x→+ x+sinx x→+ x
2) limx 0→ x sin 1/ x2 sinx( ) 4) limx
0→ x+cotxsinx
Hướng dẫn giải
Không dùng được quy tắc L’Hospital vì khi x→+ các hàm sinx và cosx không tồn tại giới hạn. x− 1− sinx −sinx =1 1) lim sinx = lim x
, có lim sinx = 0 (Theo
nguyên lý kẹp) nên lim x
x→+ x+sinx
x→+ 1+ sinx x→+ x
x→+ x+sinx
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_________________________________________________________________________________________________________________________________________________________________________________________________________________ x 2) −x ( ) = =
2 x sin 1/ x2
x2 mà lim x2 lim −x2
0 lim x sin 1/ x2 ( ) =
0 (Nguyên lý kẹp) sinx sinx sinx x→0 sinx
x→0 sinx x→0
sinx x−1 x+cosx x+1
x+1 = x 1− =  x+cosx =
3) mà lim lim 1 lim x x x x→+ x x→+ x x→+
1 (Theo nguyên lý kẹp) x
4) lim x+sinx = lim(x+sinx sinx) =0 10) lim x 0→ x→0 cotx x→0 cosx
Bài 10: Tính các giới hạn sau 1) lim 11) lim x x→1/2 0→ 12) limx 0  + ( (3)x) 2) lim 2 2 x 1 sin3 πx   ln 1 → x −1 sin 5) lim x→ → tan 5x x3 +8
18) lim n a+ − −x n a 6) xlim2 2 x
→− x −3x−10 x 0→ x 8) lim x 0→ 19) limx0   +  x
1+tanxsinx   arcsin →  1 9) lim
1−x2 x 0→ ln 1 x( − )
22) limsin2x+2arctan
3x+3x2 x 0→ ln
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1( +3x+sin x ) 2 +xex
20) limx→  2x−2 
23) xlim x→+    π2 −arcsin
3 1+tan x2 − 3 1−tan x2 21) lim x x
2 +1     x 0→ x+ 3 x2 cosx−cos3x 3) limx 0→ x2 ( ) 24) lim x4/3
3 x2 + −1 3 x2 −1 4) lim x 0→ x→+ tanx−sinx 7) ( xlim 2x→−
+1) x43x++x1 25) limx 0→ sin x3
(5 1 3+ x)2 −1
13) limx 0→ sinx+2sin x2 sin 2x( 2 − −4x 6)
14) lim x 3→ ln x( 3 −26) 15) lim x 3→ 16) lim x 2→ 17) lim + x→0 x2 sin x32x+ 2 + 3x 2
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Hướng dẫn giải
1) limtan3x = lim 3x = 3 x 0→ tan2x x 0→ 2x 2 2) limx 1
sin3 πx 0    =L limx 1→ πcosπx3x2 = −3π
→ x −1   0
3) Cách 1: Sử dụng quy tắc L’Hospital: limx 0→
cosx−x2cos3x 0  
0 =L limx 0→ −sinx+2x3sin3x       00 =L limx 0→ −cosx+29cos3x = 4  
Cách 2: Thay VCB tương đương: limcosx− (− =
2cos3x = lim−2sin2xsin2
x) = limx 0 4x22 4 x 0→ x x 0→ x → x
4) limx 0→ sin5x−sin3x = limx 0→ sin5x−sin3x 0      = =L ...
2 (Có thể dùng thay VCB như ý 3) sinx x 0 5) limx 3x33 + +5x2 1 = limx
3+ 5 / x2 +1/ x33 = 3
→ 2x + −x 9
→ 2 1+ / x−9/ x 2
6) xl→−im2 2 x3 +8      0  = =L ... −12 x −3x−10 0 7 7) xlim 2x(
+1) 43x+1 = xlim 2( +1/ x) 4+1/ x2 = 4 →− x + x →− 1+1/ x 8) lim − = ( = x 12 cos3x3 4
limx 0 2sin2 2 33x / 2)4 limx
0 2. 3( x /2 2)2 = 9
→0 2x + 3x − x
→ 2x + 3x − x → 2x 4 x x arcsin −
9) lim 1−x2 = lim 1−x2 = lim 1 =−1 x 0→ ln 1 x( − ) x 0→ −x x 0→ 1−x2
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10) lim ln 1(( + 4x ) 22 − 5x33
= lim 4x22 − 5x33 = limx 0 4x22 = 2 x 0→ ln 1+ 2x + 3x ) x 0→ 2x + 3x → 2x 11) −− xlim→1/2 arcsin 1(2
2x) = xlim→1/2
41x−22−x    0  =L xlim→1/2
8−x2 = −21 4x 1 10 ln 1 sin 12) lim ( ( (( (( x 0
 + 2 2 3)x)  = limx 0 sin22 3x)) = limx 0 3x))22 = 9 → tan 5x → tan 5x → 5x 25 13) lim → ( → ( x 0
sinx5 1++32xsi)2n−2 x1 = limx 0 5 1+ 3xx)2 −1      00 = =L ... 65 ( ) ( ) − −  
14) lim sin 2x 2 − 4x 6 = lim sin 2x 2 4x−6
2x23− 4x−6 0 = =L ... 8 → → → x 3
ln x( 3 − 26) x 3
ln 1( + x3 − 27) = limx 3
x − 27     0 27 15) l ) ) → ) → )
x 0im→ ln 1ln cos3x( (+tan x = = 2
limx 0 ln cos3xtan x( 2
limx 0 ln cos3x( x2     = = −  00 L ... 29 16) lim → →
x 2 sin 3xtan 4( (2 −−2xx2−) 8) = limx 2 3x24−−2xx2−8 0      0 = =L ... −25 17) lim ++ ) = =
x 0→ ln 1 xtx(2 sin xanx3
limx 0→ xtanxx2 1
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18) lim→ n a+ −x x n a− x       00 = =L ... n2 a1/n−1 x 0 19) limx 0  +  +1 tanx
1+tanxsinx  = ex 0lim→ ln1sinx+sinx   = =... 1 →  1   + ln     + l 3 2 im → x x1
2 x +2  + 20) 6 = 3 x 2 lim   e ... == e
x →  2 x −23 2 + −− − 3 2 3 2 + −− 1 t anx 1 − 1 tanx 1 tanx (3 2 1 tanx ) 1 21) lim =l im → → 2 x 0 3 /3 2 x 0 + x x x 3 2 3 2 2 2 2 2 1 +t anx 1 −
1 −t anx 1 − tanx tanx tanx x =l im −l im = lim + lim = lim = lim → 2 /3 → 2 /3 → 2 /3 → 2/ 3 → 2/ 3 → 2 /3 x 0 x 0 x 0 x 0 x 0 x 0 x x 3 x 3 x 3 x 3 x 1 1 2 2 = 2limx = 4/3 0 3 x→0
22) Cách 1: Thay tương đương
Khi x → 0,sin2x + 2arctan3x + 3x ~ 8x )
2 và ln 1 3x( + +sin x2 +xe ~ 4xx (Các bạn tự chứng
minh). sin2x+2arctan3x+3x2 = lim8x = 2 Nên lim )
x 0→ ln 1 3x( + +sin x2 +xex x 0→ 4x
sin2x+2arctan3x+3x2   0  Sử dụng quy tắc L’Hospital. Cách 2: lim )
x 0→ ln 1 3x( + +sin x2 +xex     0 π− x
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Thầy Phạm Ngọc Lam Trường 19

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