微积分两个重要极限
1.4
两重要极限个s
n xi 证 l先im= 1 . x→0 x
+sin
x 1. iml =1x → 0
x
单位设圆O , 圆角 心∠OB A =x, 不妨 x 设 ∈0,
(
π2
),
过
A 单位圆的切作 线得, ΔO AC.
形 扇AOB的心圆角 x ,为ΔO BA高的 为D ,
B是有于 in xs BD, =x =弧 AB, ta xn= A C
,
oB
Cx
Δ
AB面积O
1 1 积1 而从sin x
D
A
∴si
nx
x
1 co1 sx ⇒> > ,is n xx sni
x
insx 同 乘- sin 得x:- 1
s2i x n xx x 故0
2 x lQmi 0,= x+→0 2si
nx 由 夹逼定 理il (1m) = ,0+ x→ 0
xsi nx∴ li m= 1. x→0 x+
ins xπ 证再 lm i=1. 妨不设 x∈ - ,( ),0令 =u − x ,-x → 0x 2s
i(nu) sin x- ins u=1得 l m =ilim =l m -i + x→0+ u→ u→0 x0 u-u
s n xi ∴ lmi= .1 x→0
xisΔnl mi = 1→0 Δ
⎛ ΔΔ可以是 意的函数, ⎞ ⎜ 任 ⎟0⎜ 注意必但须 型是 . ⎟0 ⎠⎝
上
述证中明我们,: 当有x ∈ (0 ,
π
故
当 x ∈(
-
π
2
2时 ) ,sni x
(-, x)
x
从而
.π
2
时
)1
− c osx . 例1(书例3) 中 求lmi 2x→ 0xx
2 x2 in ssi n1 2 = im l2解 原 式= im l→x x202 x→0 x (2 )2
2⎞x⎛ sn ⎜ ⎟ i 2 111 2 ⎟= ⋅1 =. =lim⎜ 22 x → ⎜0x ⎟ 2 ⎟ ⎝⎜ 2
⎠
2
sni x ( −例2:求 lim
→π / x
3π
3 1− c2o sx
)
s
n(i x− 解: il
m→x /π3
π
31 − 2 cosx
)令 = xt− π 3/
si t n=l imt 0→1 − c2s(o t + π/ 3
)isn t= l m i t → 01 −2 co(st c o sπ/ 3 − si tn sn πi/ 3 s)nit = lm t → 0 i −1cos t + 3 sin t
isnt si nt= l i m =ilmt → 0 1 − cs t o t+0 3→ is nt2 t 2 sin+ 3sin t
2= il
m
→ t
0⎛ t ⎜
⋅ ⎜2⎜ ⎜ ⎝
si
nt t 2 t ⎞si n ⎟⎟2 + t⎟ 2 ⎟
⎠is n 3⋅ tt
=
1
= 3
3
3lim 例
3:
xπ
→
1− t a xn 1−+ t a xnsin 2x
−1t anx − 1 tan x +解1:l i xm→ π sin 2 x
−
2at nx = 1lm i⋅x→ π ins2 x 1− at xn + +1 ta n
x−tan −x tn x a=li m =im x l →π2 sin xosc x x → π sni2 x
− 11= − = l im 2 2x→ π 2 cs x
o1− tnax −1 + ta nx 解 2 :ilm →π sin 2xx
− 2 tn x 1 a=li m x⋅π→ isn x 1 −2 ta x +n 1 + at xn
−at x n =im xl→ sπn 2i x
ta(nπ − t )tant = li m令 t =π − x ilmt → 0 s−in2 t t → 0s ni2 (π− t
) 1si n t t 1= −lim =⋅−t →0c so t 2 sn i2 t 2 2t
1x 2 l.i(1 m +)= e x∞→ x 1 (n1 +)= e 已:证ilm n →∞ n
1
x 证 l先i m1( +) =e . x→∞ x
+ [x ]设 n ,=
则有 n x≤ n
1, n1 1 n+x1(1+ )
n当x→ ∞时,+n → +∞ 1
+1n 1n 1l m i(1+ ) =lm i(1+ ⋅ l)im(1 +) n → + n ∞ +∞ →n→ + ∞ nn
=n ,
e
n1 1n 1 1+− 1il m( 1 ) += il m(1 ) ⋅ +ilm( 1+ ) = e n,→ ∞+ →n+∞
n→ +∞ +1 nn+1n +1
1
x ∴il m1(+ = ) e x . +→ ∞x1
x 证 再ilm(1 + ) e=. x →∞ -x令
u=− x,
1
x1 −u = li (1 m +1) u l∴i (1 +m) = ilm 1(− ) u →∞ +−1ux → −∞u →+∞ x u
1 u
1 1− =li (1 m+) ( 1 +)= →u∞+ u− 1u1−
e
.
x 1∴lim ( 1 + = e x→)∞ x1 令t
, =
1xx l i (1m + ) =l im(1 + t) →x t →∞ x
10
x
1
t=
e .
iml( +1x ) =
ex →
(0注 意到是 1 ∞型
)1 Δ lim(
1 +) = e Δ∞ Δ
→
il(1m Δ ) = e+
Δ0
→
1
Δ⎛
Δ可以是 意的函数, 任⎞⎜ ⎟ 但⎜注意必须是 1 型 ∞ ⎟ ⎠ .⎝
总
:结
ins xl m =i1 1 . x 0 →x
1x l m(1 i + ) e =. 2x → ∞
xin ◊s ilm =1 般一形: ◊ 式 0 →
1◊ l◊i(1 +m = ) e般形一式 ◊ →: ◊∞
ilm(1 + )x= e
→x0
1 x
l
im( 1 + ◊ ) e =般一形式:◊
→0
1 ◊
例4 解
1
x l求im 1(− x)∞→x
1 1−x −1 式 =原l im(1 [ + ) =]l im x→ ∞x→∞1 x −−x( +1 )x−1 .= e
例 解5
3+ 2x 求 lxim )( →∞ 2x +x
1+x2 2 −41 式原 l=im([1+ )] 1 (+ )= e . 2x→∞ x+ x22+
1
例
(书6例11) 中 求lm ico(s
→ x0
21
x
)
s
n
i2
x
: 解ilm (osc
x →0
2
x
)
isn
2
= xlim 1(− s in
→ x0
1 x22 sinx) 1 u
令u = −
in s 2
则当 x x →时,u 0 0
→
l=im( + 1 )
uu 0→
−
⎡ ⎤
= ilm ( 1 + u⎢ ) ⎥→ 0u ⎣
⎦ u1
−
1
=
e−
1形如[ f( x]
)g
( x)
(
f ( x) 1) ≠函数称的为指幂数.
函
定 设 l理i m (fx ) = A >0, iml g x( )= B ,则 g (x B li)m[ f( )x ] =A
极从定义限出发以可明以下两证个论:结证:
1)(如 li果 mf( x )= A 0>, 则illn mf x( = ln )A f ()x Al i mf ( x) A , == (e)如2果则 lie
由m论(结1)得
又由结论2)得
li([mg( x ⋅ )ln f( x) =]B ⋅ l n
lAi[ fm x )](
( gx)
= l
mie
g x ()ln⋅ f( x )
e=
B ⋅ n l
A
=A
B
论
若 l推mi f( x ) = ,0lim g x( =)∞ .
且l i m f x( )⋅ g ( )x= A ,
g x( )im[ l1+ (f x )]则
⎧
⎪ l=mi⎨[ 1+f (x )] ⎪⎩
f1( )x
⎫ ⎬⎪⎪ ⎭
f
( x ⋅ g ( x)
=)
e
A
该∞论对推1 极型可简化步限。骤
l
i m(1 + s ni x 例 ) (书6中例12) 求 x →
解0 1: lm(1i+ is nx )
→x 0 2x1
1 2
= lxm[(i1 +si xn) x
→ 0
1 si
nx s n x i2x
]
e=
1
2
l
i (m1 +insx ) 例 (6中书 1例2 求)x →0
2解属 1:型,
f ( x ) =s in x ,
g(1)x =2
x
∞12 x
1
s1n ix= Q l mi si x n ⋅= il mx→ 02x x→2 0x
21
∴
lmi( 1 + s inx
)x 0
→1
2x
=e
1 2
最
后一最 容步丢掉!易
1
li
m(cos x ) 例6 (书中 1例1) x求→ 0 ∞
1 解2:属 型
1
2sni 2
x1
il mc(osx )x
→0
2
si
n2
=x li [1m+ cos x −( 1)] 2
x→ 0
c1s o 2x− 1
s
n i x
2
⎧ ⎪= li m[⎨1+ ( cs ox 0→ ⎩⎪
2
− 1x ])
⎪ ⎫ ⎬ ⎭
cos⎪2 −x1 isn2 x
1
⎧ ⎫ ⎪ 2 2os xc − 1⎪= im ⎨[1l+ ( co xs− 1 )]⎬ x→ ⎪0 ⎩ ⎪
⎭-
sin 2 x sin 2 x
=e
−1
3
.结小两
重个要极 限sniα 0 1 im l= ; 1
α→0 α1 α 0 2 iml( 1+ )= e.
α →
∞α
ilm1 (+α )α = e
α.→0
1推论
若l imf ( x ) = 0 ,imlg ( x )= ∞. 且imlf x( ⋅ g)( x ) =A ,则
lim[
+1 f( x]g) (x) =e A
思考题
x ( im l 3+ 9求极 x限→ ∞
+
1 x
1 xx
)
错解: ilm3 +
9xx → +∞
(
x
)
1
x
1 ⎞ ⎛ =9l m ⎜ i1 +x ⎟ x → ∞+3 ⎠
⎝
1 1又 im lx = 0 x ⋅→ +∞3 x
1
出错原x:因极限非 此 1 ∞型
1 ⎞ 所⎛ 以lm ⎜ 1i+ x ⎟ = e 0= 1 x → +∞ 3⎠ ⎝ 故
l im 3x+ 9 x
x →+∞
(
) 1x
=
9
思考解答
x 题 →+
l∞i m( 3 + 9x
1 x
x1 x
)
=l im9
x→ ∞+
(
)x
x
11
⎞⎛ 1⎜ +x ⎟3 ⎠⎝
x
1
1 ⎛⎞ = 9 lim⋅ ⎜1 + x ⎟ x → +∞ 3 ⎠ ⎝ 1 ⎞ =⎛ 9⋅ ilm⎜ 1 +x ⎟ →x +∞3 ⎠⎝
1 x→ +∞ xlim
=
9×1 = 0
9业作:
P5: 1.0偶 . 2. 4.3
习预:
.15无穷 与无小穷大