f:RRx0R
lim
xx0
f(x) = −∞
AR,α > 0,xR\ {x0},|xx0|< α f(x)6A
A x R\ {x0}
0<|xx0|< α AR,α > 0,xR,0<|xx0|<
α f(x)< A
f x0
 > 0,α > 0,xR,|xx0|< α |f(x)f(x0)|< 
x x0
f: [a, b]Rf
f f(a)f(b)f(a).f(b)60f a b
c[a, b]f(c)=0 f(a)f(b)f(a).f(b)<0c
a b c ]a, b[
f y f(a)f(b)
y[f(a), f(b)] [f(b), f(a)] f y a b x [a, b]y=f(x)
y f(a)f(b)x a b
f
x a b y f(a)f(b)y=f(x)
f(x)f(a)f(b)
f:R+Rflim
x+f(x)=0
fR+
lim
x+f(x) = 0
 > 0,AR,xR+, x >A|f(x)|< 
BR+x>B⇒ |f(x)|61
f+= 1
B x R+x>B|f(x)|<1|f(x)|61x
xR+B B B 0
g: [0, B]Rg(x) = f(x)x[0, B]g
g f [0, B]f[0,B]
g(x) = f(x)x[0, B]g[0, B]f
[0, B]g[0, B]
m M x [0, B]m6g(x)6M g
CR|f(x)|6C x [0, B]
g(x) = f(x) [0, B]m6f(x)6M x [0, B]|f(x)|6
max(|m|,|M|)C= max(|m|,|M|)
a6c6b|c|6max(|a|,|b|)
a6c6b06a a b c |a|=a|b|=b|c|=c
|a|6|c|6|b|b60a b c |a|=a|b|=b|c|=c|b|6|c|6|a|
a60b>0|a|6|b| −|b|6−|a|=a6c6b=|b| |c|6|b|
|b|6|a| −|a|=a6c6b=|b|6|a| |c|6|a|
|c|6|a| |c|6|b| |c|6max(|a|,|b|)
fR+
x[0, B]|f(x)|6C x >B|f(x)|61
xR+|x|6max(C, 1) fR+
f:R
+R
f(x)=1xE1
x,
tRE(t)E(t)6t <
E(t)+1
f0
tRt1< E(t)6t
E(t)6t < E(t)+1 t1< E(t)
t1< E(t)6t
x > 0x>f(x)>0
t= 1/x
1/x 1< E(1/x)61/x
1x < xE(1/x)61x
x1>xE(1/x)>11
x > 1xE(1/x) = f(x)>1
f0
lim
x0f(x) = 0  > 0α=
0< x < α |f(x)|=f(x)f(x)>0|f(x)|< x < α =
f0 0
g:R+Rg(x) = f(x)x > 0g(0) = lim
x0f(x) = 0
x > 1f(x)=1
x > 1 0 61/x < 1E(1/x) = 0
f(x) = 1 xE(1/x) = 1
f(1) f1
f(1) = 1 1×E(1/1) = 0 E(1/1) = E(1) = 1 f(x) = 1 x > 1 lim x1+f(x) =
16=f(1) f1
lim
x+
x2+ 5 + exp(1/x)
(x+ 1)3
lim
x+
x2+ 5 + exp(1/x)
(x+ 1)3= lim
x+
x2
(x+ 1)3+ lim
x+
5
(x+ 1)3+ lim
x+
exp(1/x)
(x+ 1)3
x+
(x+ 1)3+5/(x+ 1)30
1/x 0e1/x 1e1/x/(x+ 1)30
x2
(x+ 1)30
x2
(x+ 1)3=x2
x3+ 3x2+ 3x+ 1 =x2
x2(x+ 3 + 3/x + 1/x2)=
1
x+ 3 + 3/x + 1/x21/x20 3/x 0 3 3x+x+3+3/x + 1/x2+
1/(x+ 3 + 3/x + 1/x2)0
lim
x+
x2+ 5 + exp(1/x)
(x+ 1)3= 0
lim
x+(px2+x+ 1 x)
+xx2=|x|=x(x2+x+ 1)2=
|x2+x+ 1|=x2+x+ 1
px2+x+ 1 x=(x2+x+ 1 x)(x2+x+ 1 + x)
x2+x+ 1 + x
=x2+x+ 12x2
px2(1 + 1/x + 1/x2) + x
=x+ 1
xp1 + 1/x + 1/x2+x
=x(1 + 1/x)
x(p1 + 1/x + 1/x2+ 1)
=1 + 1/x
p1 + 1/x + 1/x2+ 1
x→ ∞ lim
x+
(px2+x+ 1x) = 1
2
lim
x0
(tan x)2
cos(2x)1
cos(2x) = 2 cos2x1 cos 2x1 = 2 cos2x2 = 2(1
cos2x) = 2 sin2x x 6= 0 0 x[π/4, π/4]
tan2x
cos 2x1=
sin2x
cos2x
2 sin2x
=sin2x
2 cos2xsin2x
=1
2 cos2x
x0 cos x1 cos2x1 lim
x0
(tan x)2
cos(2x)1=1
2
lim
x1+(x31) ln(2x2+x3)
2x2+x3x= 1
2x2+x3x1 2x2+x3 = (x1)(x+ 3) x31
x1
x31 = (x1)(x2+x+ 1)
(x31) ln(2x2+x3) = (x1)(x2+x+ 1) ln((x1)(x+ 3))
= (x2+x+ 1)(x1) ln(x1) + (x2+x+ 1)(x1) ln(x+ 3)
x1+x x2+x+ 1 3x10x > 1
ln(x1) → −∞ (x1) ln(x1) 0 (x2+x+ 1)(x
1) ln(x1) 0x+ 3 4 (x1) ln(x+ 3) 0 (x2+x+ 1)(x1) ln(x+ 3) 0
lim
x1+(x31) ln(2x2+x3) = 0
nNxR+
fn(x) = 1 + x+x2+· · · +xn1+xn.
nNfn
fnx7→ −1x7→ xk
k= 1,...,n R+RfnR+
R+x7→ xk
fn
fn(0) fn(1) xn[0,1] fn(xn) = 0
(xn)nN
fn(0) = 1 + 0 + 02+··· + 0n=1fn(1) = 1+1+12+··· + 1n=1 + n=n1
fn(0) fn(1) fnR+[0,1]
xn[0,1] fn(xn) = 0 n > 1
fn(1) = n1>0fn(0) = 1<0xnxn]0,1[
nNxR+fn+1(x)>fn(x)
fn+1(x) = 1 + x+x2+··· +xn+xn+1 =fn(x) + xn+1 xR+xn+1 >0
fn+1(x)>fn(x)
fn+1(xn)>0
x=xnfn+1(xn)>fn(xn)xn
fn(xn) = 0 fn+1(xn)>0
fn+1(xn)>fn+1(xn+1)xn>xn+1
fn+1(xn)>0xn+1 fn+1(xn+1) = 0 fn+1(xn)>fn+1(xn+1)
xn< xn+1 fn+1 fn+1(xn)< fn+1(xn+1)xn>xn+1
(xn)
(xn)xn
[0,1] n
1 / 4 100%
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans linterface ou les textes ? Ou savez-vous comment améliorer linterface utilisateur de StudyLib ? Nhésitez pas à envoyer vos suggestions. Cest très important pour nous !