un=Rπ/4
0tannxdxvn=R+∞
0
dx
xn+ex
n+∞
In=Z+∞
0
xn
1 + xn+2 dx
un=R+∞
0
sinnx
x2dx un=R+∞
0
xndx
xn+2+1 un=R+∞
0
xndx
x2n+1
un=Z+∞
0
sin(nt)
nt +t2dt
lim
n→+∞Z+∞
0
e−tsinn(t) dt
Z+∞
−∞ 1 + t2
n−n
dt−−−−−→
n→+∞Z+∞
−∞
e−t2dt
un= (−1)nZ+∞
0
dt
(1 + t3)n
n≥1
lim
n→+∞Z+∞
0
dt
(1 + t3)n
un
fn
fn(x) = ln(1 + x/n)
x(1 + x2)
R∗
+
un=nR+∞
0fn(x) dx
f: [0 ; +∞[→C
f+∞`
n→+∞
µn=1
nZn
0
f(t) dt
f∈ C0(R+,R+)n∈N
In=Z+∞
0
nf(t)e−nt dt
Inn→+∞
f:R+→R
n→+∞
Z+∞
0
nf(x)
1 + n2x2dx
f:R+→RC1
x > 0
lim
n→+∞Z+∞
0
ncos t(sin t)nf(xt) dt
lim
n→+∞Z√n
01−t2
nn
dt
lim
n→+∞Zn
01 + x
nne−2xdx
Zn
0r1 + 1−x
nndx
lim
n→+∞Zn
0cos x
nn2
dx
lim
n→+∞Z+∞
0
n!
Qn
k=1 (k+x)dx
FR R −∞
0 +∞h δ 0< h < δ
In=Z1
0
F√n(δt −h)dt
Sn=
n−1
X
k=0
F√nδk+ 1
n−h
Snn+∞
fC1[a;b] 0 <a<1< b f(1) 6= 0
(fn)
fn(x) = f(x)
1 + xn
(fn)
lim
n→+∞Zb
a
fn(t) dt=Z1
a
f(t) dt
Z1
a
tn−1fn(t) dt∼ln 2
nf(1)
n∈N∗x∈R
fn(x) = n
√π1−x2
2n22n4
gR[a;b]
lim
n→+∞ZR
fn(x)g(x) dx=g(0)
f: [0 ; 1] →Rf(1) 6= 0
n
In=Z1
0
tnf(t) dt
f: [a;b]→R+
n∈N∗
(x0, . . . , xn) [a;b]
∀i∈J1 ; nK,Zxi
xi−1
f(x) dx=1
nZb
a
f(x) dx
g: [a;b]→R+
lim
n→+∞
1
n
n
X
i=1
g(xi)
a b (an) (bn)
a0=a, b0=b∀n∈N, an+1 =an+bn
2, bn+1 =panbn
(an) (bn)
M(a, b)
T(a, b) = Z+∞
−∞
du
p(a2+u2)(b2+u2)
Ta+b
2,√ab=T(a, b)
u=1
2t−ab
t
T(a, b) = π
M(a, b)
In=Z+∞
0
dx
(1 + x3)nn∈N∗
(In)n≥1
P(−1)n−1In
[0 ; π/4[ tannxCV S
−−−→ 0|tannx| ≤ 1 = ϕ(x) [0 ; π/4[
un→Zπ/4
0
0 dx= 0
[0 ; +∞[1
xn+ex
CV S
−−−→ f(x)f(x)=e−x[0 ; 1[ f(x)=0
]1 ; +∞[
1
xn+ex≤e−x=ϕ(x)ϕ[0 ; +∞[
vn→Z1
0
e−xdx=e−1
e
In=Z1
0
xn
1 + xn+2 dx+Z+∞
1
xn
1 + xn+2 dx
In→Z+∞
1
dx
x2= 1
[0 ; +∞[
|sin x|ϕ(x)=1/x2
]0 ; +∞[
un=Z+∞
0
sinnx
x2dx=Z1
0
sinnx
x2dx+Z+∞
1
sinnx
x2dx
Z1
0
sinnx
x2dx≤Z1
0sinn−2(x)dx|sin x|≤|x|
Z1
0sinn−2(x)dx→0
Z1
0
sinnx
x2dx→0
Z+∞
1
sinnx
x2dx≤Z+∞
1
|sin x|n
x2dx
|sin x|n
x2
CS
−−→ f(x)f(x)=0 x6=π/2 [π]
|sin x|n
x2≤1
x2=ϕ(x)ϕ[1 ; +∞[
Z+∞
1
|sin x|n
x2dx→Z+∞
1
f(x) dx= 0
un→0
un=Z1
0
xndx
xn+2 + 1 +Z+∞
1
xndx
xn+2 + 1
Z1
0
xndx
xn+2 + 1≤Z1
0
xndx=1
n+ 1
Z+∞
1
xndx
xn+2 + 1 −−−−−→
n→+∞Z+∞
1
dx
x2= 1
xn
xn+2+1 ≤1
x2[1 ; +∞[
un→1
un=Z1
0
xndx
x2n+ 1 +Z+∞
1
xndx
x2n+ 1
Z1
0
xndx
x2n+ 1≤Z1
0
xndx=1
n+ 1
Z+∞
1
xndx
x2n+ 1≤Z+∞
1
dx
xn=1
n−1
un→0
fn:t7→ sin(nt)
nt +t2
fn]0 ; +∞[
t→0+fn(t)∼nt
nt+t2→1
t→+∞fn(t)=O1
t2
fn]0 ; +∞[
t∈]0 ; +∞[
n→+∞fn(t)=O1
n(fn)
t≤π/2|sin u|≤|u|
|fn(t)| ≤ nt
nt +t2≤1
t≥π/2
|fn(t)| ≤ 1
nt +t2≤1
t2
|fn| ≤ ϕ
ϕ:t7→ (1t∈[0 ; π/2]
1/t2t∈]π/2 ; +∞[
ϕ]0 ; +∞[
un→Z+∞
0
0 dt= 0
t=π/2 + πmod 2π
Z+∞
0
e−tsinn(t) dt≤Z+∞
0
e−t|sinnt|dt
fn(t) = e−tsinn(t)
CS
−−→ f(t)
f(t) = (0t6=π/2 mod π
e−t
fnf
|fn(t)| ≤ e−t=ϕ(t)
ϕ[0 ; +∞[
lim
n→∞ Z+∞
0
e−tsinn(t) dt=Z+∞
0
f(t) dt= 0
fn(t) = 1 + t2/n−n
R
(fn)f f(t)=e−t2
R
t∈R
ϕ:x7→ −xln(1 + t2/x)
[1 ; +∞[
ϕ00 ϕ0lim+∞ϕ0= 0
ϕ0
ϕ n ∈N∗
|fn(t)| ≤ 1 + t2
n−n
= exp(ϕ(n)) ≤exp(ϕ(1)) = 1
1 + t2
t7→ 1/(1 + t2)R
Z+∞
−∞ 1 + t2
n−n
dt−−−−−→
n→+∞Z+∞
−∞
e−t2dt
1 / 12 100%
La catégorie de ce document est-elle correcte?
Merci pour votre participation!

Faire une suggestion

Avez-vous trouvé des erreurs dans l'interface ou les textes ? Ou savez-vous comment améliorer l'interface utilisateur de StudyLib ? N'hésitez pas à envoyer vos suggestions. C'est très important pour nous!