高等数学 求极限方法小结及举例

lim( 1 + f ( x ))
g( x )
=e
c
3
c = lim[ f ( x ) g( x )]
( x + 1)( x − 1) x +1 例 1 . lim 2 = lim = lim = −2 . x →1 x − 3 x + 2 x →1 ( x − 2)( x − 1) x →1 x − 2
( (幂指型)
lim u( x )
[
]= e
=e
5 . " (1 ± 0 )∞ " 型 ,
(1 +
f ( x ))
g( x )
g( x )
g ( x )⋅ln (1+ f ( x ) )
,
lim( 1 + f ( x ))
=e
lim[ g ( x )⋅ln (1+ f ( x ) )]
,
(经验公式)
10
例 11 .
f ( x ) = ( x − a )n ϕ ( x ) , ϕ ( x ) 在 U (a ) 内 n − 1 次可导 , 求 f ( n ) (a ) . f ( n −1) ( x ) = n !( x − a ) ⋅ ϕ ( x ) + ( n − 1) ⋅
解.
n! ( x − a ) 2 ⋅ ϕ ′( x ) 2!
=0.
x → −0 x → −0
如果 n > 2 如果 n > 1
lim f ′( x ) = lim n x n −1 = 0 .
∴ 当 n > 1 时 , f ′( x )存在 , 当 n > 2 时 , f ′( x )连续 .
9
例9 . y = f x 2 , f 二次可导 , 求 y′′ . 解.
2 2 2 2 1 y′′ = 3 ⋅ y′ = 3 − 2 − 1 = − 5 − 3 . y y y y y
12
[
v( x )
] = [ lim u( x ) ]
lim v ( x )
.
当 x → 0 时, x ~ sin x ~ tan x ~ arcsin x ~ arctan x x ~ ln(1+ x) ~ e x −1 , a x −1 ~ x ln a .
1
当 x → 0 时, x2 , (1+ x)α −1 ~ α x , 1+ x −1 ~ x . 1− cos x ~ 2 2
+ ⋯⋯ + ( x − a )n −1ϕ ( n −1) ( x ) = n ! ϕ (a ) .
11
x = f ′( t ) d2y 例 12 . f ′′( t ) ≠ 0 求 . 2 dx y = t f ′( t ) − f ( t ) d y y′( t ) f ′( t ) + t f ′′( t ) − f ′( t ) 解. =Βιβλιοθήκη Baidu= =t d x x′( t ) f ′′( t )
x 2 −1
e x −1 e x −1 ~ x x 例 2 . lim =========== lim =1. x →0 x x →1 x t = e −1 ln(1 + t ) lim x ========== lim = lim ln(1 + t ) x →0 e − 1 t →0 t t →0 x
求极限方法小结及举例
一. 常用办法 : 1 . 利用初等函数的连续性 : "能 则 " 代 代 lim f ( x ) = f ( x 0 )
x → x0
lim f [u( x)] ====== f [limu( x)]
f 连续
2 . 幂指函数取极限 :
lim u( x )
∞ 3 . 洛必达法则 : " 0 型和 ∞ 型 " 0 4 . 乘除法运算中的等价无穷小代换 .
−1
( x ) 的二阶导数 .
( x ) 的直接函数是 x = f ( y ) .
dy 1 = , d x f ′( y )
d2y d 1 − f ′′( y ) d y − f ′′( y ) = ⋅ = . = 2 2 dx 3 d x f ′( y ) [ f ′( y ) ] dx [ f ′( y )]

2( x +1) 2 x +1 x →∞ 2 x +1 2 lim

=e =e.
1
可用经验公 式验证
例 5 . lim x +x
2
x →∞
x 2 − 10000
= lim
1+ 1 x 1 − 10000 x2
x →∞
=1.
"∞" ∞
6
1 x −π 2 例 6 . lim x →π tan x
2
t =π − x −1 2 t ========= lim t →0 cot t
tan t = − lim = −1 . t →0 t
"∞" ∞
例 7 . lim ( x ⋅ cot x )
x →0
x = lim =1. x →0 tan x
( 有界量乘无穷小 )
"0⋅ ∞"
lim x cos 1 = 0 . x x →0
= lim
2x x2 + x + x2 − x
2 1+ 1 + 1− 1 x x
x +1
x → +∞
x → +∞
2x + 3 例 4 . lim x → ∞ 2 x + 1
2 = lim 1 + x → ∞ 2 x + 1
2( x +1) 2 x +1 2 x +1 2
= lim x
x → +0 n −1
sin 1 = 0 . x
如果 n > 1
f ( x ) − f ( 0) xn ′ f − (0) = lim = lim = 0 . 如果 n > 1 x →−0 x −0 x → −0 x
如果 n > 1 ,
则 f ′(0) = 0 .
8
∴ 当 n > 1时 ,
5 . 约去公因子 .
6 . 代数变换 , 如取对数等 . f ′( x) = f ( x) [ln f ( x)] ′
二 . 常见未定式及对策 :
1 . " 0 "型 , 常见 . 0 2 . " ∞ " 型 , 常见 . ∞ 3. "0 ⋅ ∞" 型 , g( x ) f ( x) f ( x ) ⋅ g( x ) = 或 f ( x ) ⋅ g( x ) = . 1 1 f ( x) g( x ) 2
( ) y′ = f ′(x 2 )⋅ 2 x = 2 x f ′(x 2 ) . y′′ = 2 f ′(x 2 ) + 2 x f ′′(x 2 )⋅ 2 x = 2 f ′(x 2 ) + 4 x 2 f ′′(x 2 ) .
−1
例 10 . 设 f 二次可导 , 求 y = f 解. y= f
+ ⋯ + ( x − a )n ⋅ ϕ ( n −1) ( x ) ,
f ( n −1) (a ) = 0 .
f ( n −1) ( x ) − f ( n −1) (a ) f ( n ) (a ) = lim x →a x−a
n ! ϕ ( x ) + ( n − 1) ⋅ n ! ( x − a )ϕ ′( x ) = lim 2 x →a
n x n −1 sin 1 − x n − 2 cos 1 x>0 x x f ′( x ) = 0 x=0 n x n −1 x<0 ′( x ) = lim n x n −1 sin 1 − x n − 2 cos 1 lim f x x x → +0 x →+0
dt 1 d2y d 1 ′⋅ (t ) = 1( t )? = = . = 1⋅ 2 dx dx x′(t ) f ′′( t ) dx
例13 . y = tan( x + y ) , 求 y′′ .
解 . y′ = sec2 ( x + y ) ⋅ (1 + y′ ) ,
sec2 ( x + y ) 1 + tan 2 ( x + y ) 1 + y 2 1 ′= y = = 2 2 2 = − 2 −1 . 1 − sec ( x + y ) − tan ( x + y ) −y y
x
1 t
= ln lim (1 + t )
t →0
1 t
= ln e = 1 .
∴
e −1 lim =1. x →0 x
x
当 x → 0 时, 证明e x −1 ~ x ?
4
例3 .
x → +∞
lim
(
x +x− x −x
2 2
)
=1.
"∞ − ∞"
" ∞", 用 必 法 ? 洛 达 则 ∞
= lim
lim x cos 1 = lim x ⋅ lim cos 1 = 0 . x x→0 x→0 x x→0
错!
7
x n sin 1 x > 0 x 例 8 . f ( x ) = 0 x=0 n − 正整数 . n x<0 x 讨论 f ′( x ) 的连续性及 n 的取值范围 . x n sin 1 f ( x ) − f ( 0) x ′ 解 . f + (0) = lim = lim x → +0 x → +0 x −0 x
x +1
"(1+0 )∞"
2 = lim 1 + x →∞ 2 x + 1

"幂指函数取极限 "
5
2 = lim 1+ x→∞ 2x +1
lim 1 + 2 = x →∞ 2 x + 1
2( x+1) 2x+1 2x+1 2
4 . "∞ ± ∞" 型 ,
1 ± 1 = f ( x ) ± g( x ) . f ( x ) g( x ) f ( x ) ⋅ g( x )
5 . " ( 1 ± 0 ) ∞ " 型 , 0 " "0 型, u( x ) v ( x ) = e v ( x )⋅ln u( x ) 6. (指数型) " ∞0 " 型 , 7. lim [v ( x )⋅ln u( x ) ] v( x )