第十次（打印版）

Tiao Lu (lutiaopku@sina.com)
2006-11-2
x → 0,
1. sin x ∽ x; 2. tanx ∽ x; 3. ln(1 + x) ∽ x; 4.
(1 + x) 1
n − 1 ∽ 1
1
x 5. 1 − cosx ∽ n 2x2 ; 6. ex − 1 ∽ x;
7. arcsinx ∽ o(x) 8. arctanx ∽ o(x)
4225
tanx
lim
x→0 x
ln(1 + x)
lim
x→0 x
lim
x→0
= lim
x→0
t = (1 + x) 1
n
(1 + x) 1
n − 1
1
n x
sin x
x
1
cosx
sin x
= lim
x→0 x lim
x→0
1
cosx
= 1
= lim ln(1+x)
x→0 1
x = ln lim(1+x)
x→0 1
x = ln e = 1
= lim
t→1
= lim
t→1
t − 1
1
n (tn − 1)
2 x
1 − cosx = 2 sin
e
lim
x→0
x − 1
x (t = ex − 1) = lim
t→0
t − 1
1
n (t − 1)(tn−1 + t n−2 + · · · + 1)
1
2
∽ 2(x
2 )2 = 1
2 x2
t
ln(1 + t)
= 1
= 1

:
:
:
:
lim
x→0
tanx − sin x
lim
x→0 x3 lim
x→0
(1 + x 2 ) 1
3 − 1
1 − cosx
(1 + x 2 ) 1
3 − 1
1 − cosx
= lim
x→0
tanx − sinx
lim
x→0 x3 = lim
x→0
1
3 x2
1
2
sinx(1 − cosx)
cosx
2
=
x2 3
1
= lim
x3 x→0
x · 1
2 x2
cosx
,
,
tanx − sinx
lim
x→0 x3 = lim
x→0
x − x
= lim
x3 x→0
0
= 0
x3 1 1
=
x3 2
43 .2.1
f(x)x0 f(x)
x0
2

1
y
−1
f(x)x0 f(x)
x0
f(x)x0 f(x)
x0
II
,,
,
,
76
3
x

1
1
1
y
y
y
−1
−1
−1
4
x
x
x

y
1
0.1
c.
d
d
dx (c) = 0 dx (cx) = c
d
dx (cxn ) = cnxn−1
d c , dx xn
1 = c(−n) xn+1 :
−1
d
dx (xa ) = ax a−1 , a ∈ R
d 1
dx xa
= −a 1
xa+1, a ∈ R
:(1) f(x) = 3; (2) g(x) = 4x 5 ; (3)
h(x) = 1
x 4/5, x > 0.
: (1) f ′ (x) = 0; (2) g ′ (x) = 20x4 ; (3) h(x) =
−4 1
5 x9/5, x > 0.
5
x

d
dx
d
dx
d
dx
(sinx) = cosx
d
dx
(tanx) = 1
cos 2 x = sec2 x d
d
(sec x) = sec x tanx
d −1 1
dx sin x = √
1−x2
d −1 1 tan x = dx
1+x2
−1 1 sec x =
d
dx
|x| √ x 2 −1
d
dx
d
dx
d
dx
dx
dx
/a
d
dx (ax ) = ax ln(a)
d
1 (ln(x)) = dx x
d
dx (loga x) = 1
xln a
0.2
d , x > 0
, x > 0
(cosx) = − sin x
1 (cotx) = −
(csc x) = − csc x cotx
−1 1
cos x = −
−1 1
cot x = −1+x2
−1 1
csc x = −
d
dx (ex ) = ex 1 (ln |x|) = dx x
sin 2 x = − csc2 x
√ 1−x 2
|x| √ x 2 −1
, x = 0
u, vx, C.
(1) (Cu) ′ = Cu ′ ; (2) (u ± v) ′ = u ′ ± v ′ ; (3)(uv) ′ =
u ′ v + uv ′ ;
1
(4)
u ′ u
v = ′ v−uv ′
v2 , (v = 0),
1 ′ v
v = − ′
v2; (5) y = y(u), u = u(x), yx = yu · ux;
(6) y = f(x)x = f −1 (y),x ′ y = 1
(7) (u v ) ′ = u v v ′ lnu + v u′
u
.
6
y ′ x
(y ′ x = 0);

1
lim(1
+ 2x) x
x→0
= lim(1
+ u)
u→0 2
u
= lim
u→0
1
= lim(1
+ u) u
u→0
= e 2
(1 + u) 1
u
1
(20)
√ ×
1.f(x) = x2 + x, x ∈ (−∞, +∞)
2.
2
(45).1.; 2.,
; 3.,; 4.,
.
{xn} =
2n+5
n+3 ( )
(A) 0; (B) 2; (C) 1; (D).
3
(15)
y = 1 + (sin x)ex
arctanx
4
(10)
limx→1
lnx
(x−1)arctanx
5
(10)x · 3 x = 0(0, 1)(:
)
7
2
2