Devoir 1 exercice 2 : Pour tout entier n ∈ N Feuille 1 exercice 3

n∈N?En:= {k+n
k, k ∈N?}
•Eninf En:= inf{k+n
k,1≤k≤n}
•n∈N?inf En≥√4n
•En⊂R+0 limk(k+n
k) = +∞En
k=n k +n
k=n+ 1 k≥n+ 1 k+n k≥n+1+n k> n + 1
inf En:= inf{k+n
k,1≤k≤n}
•inf En≥√4n k +n
k≥√4n
(k+n
k)2≥4n(k−n k)2≥0 inf En≥√4n
(k−n
k)2= 0 n=k2n
f:R→Rf
f+∞
f:R→Rl+∞
limnf(xn) = l(xn)n+∞f
a6=b f(a)6=f(b)f T f(a+nT ) = f(a)6=f(b) = f(b+nT )
n∈Zlimnf(a+nT ) = f(a)6=f(b) = limnf(b+nT )
limna+nT = limnb+nT = +∞f
∀n∈N?:√n+ 1 −√n < 1/2√n < √n−√n−1
1
21
√2+1 √3+··· +1 √10000 
√n+ 1 −√n=1
√n+1+√n<2 √n√n−
√n−1 = 1
√n+√n−1>2
√n√n+ 1 −√n <
1/2√n < √n−√n−1n∈ {2,3,...,10000}100 −√2 = √10000 −√2<
1
21
√2+1 √3+··· +1 √10000 <√10000 −1 = 99 98
f(x) = sin(x2)
•f(x)=0 x=±√kπ, k ∈Zf T > 0T
f T = 0 f≡0f(T) = f(T+ 0) = f(0) = 0 k0∈N?
T=√k0π√π√2π T f(√π+√2π) = −sin(2√2π)6= 0
f(T+√π) = f(√π)=0 √k0π+√π=√kπ k ∈Zk0+1+2√k0=k
k0=q2q∈Qf(T+√2π) = f(√2π)=0
1+2√2k0=k02ko=q02, q ∈Q2ko=q02k0=q2
√2 = p0/p ∈Q
•f xk:= √kπ, k ∈N
xk+1 −xk=√π
√k+1+√kk√π
2√k
f T f T [√kπ, √kπ +T]
k2T√k
√π
√mπ < √kπ+1 f T [√kπ, √kπ+T]
[0, T ]f
•R R f(x) =
sin(x2)R|f(xn)−f(xn+1)|xn=√2nπ

•f T f0T
f(x) = sin(x2)T f0(x) = 2xcos(x2) + sin(x2)
f0(√2nπ)=2√2nπ f0R
R→Rf
x1< x2< x3f(x1)> f(x2)f(x2)< f(x3)f
u f(x2)<u<inf{f(x1), f(x3)}
x1< s < x2x2< t < x3f(s) = u=f(t)s < t f
R→Rf◦f(x) = −x x f
f
f◦f f ◦f(x) = −x x
f∈C0(R)f◦f◦f(x) = x, ∀x∈R
f◦f◦f(x) = f◦(f◦f)(x) = x f f(a) = f(b)
a=f◦f◦f(a) = f◦f◦f(b) = b
f f ◦f f ◦f◦f
f x ∈Rx f(x)
f(x)< x f(x)> x f(x) = x f(x)> x f
x=f◦f◦f(x)> f(f(x)) > f(x)> x f(x)< x f(x) = x x
f(x) = x, x ∈R
f∈C0(R)f◦f(x) = f(x),∀x∈R
Ef:= {x∈R:f(x) = x}f
f(R)x∈Rf(x) = f(f(x)) f(x)∈Ef
Ef⊂f(R)Ef=f(R)f(x) = x
x∈f(R)f◦f=f f(R)
A f f(x) = x A A
A A = [−1,1] f
f(x) = x x ∈[−1,1] f(x) = 2 −x x ∈[1,2] [2,+∞[f(R) = [−1,1]
f(x) = x f(R)=[−1,1]
m > n > 0 lim
x→0
(cos x)1/m −(cos x)1/n
x2.
•(cos x)1/m −(cos x)1/n
x2=(cosnx)1/nm −(cosmx)1/nm x2
amn −bmn = (a−b)(amn−1+bamn−2+··· +bmn−1)a= (cosnx)1/nm, b =
(cosmx)1/nm
(cosnx)1/nm −(cosmx)1/nm
x2=cosnx−cosmx
x2(mn
p(cosnx)nm−1+··· +mn p(cosmx)nm−1)

limx→0mn
p(cosnx)nm−1+··· +mn p(cosmx)nm−1=mn
lim
x→0
(cos x)1/m −(cos x)1/n
x2= lim
x→0
cosnx(1 −cosm−nx)
mnx2
= lim
x→0
(1 −cosm−nx)
mnx2
= lim
x→0
(1 −cos x)(1 + cos x+··· + cosm−n−1x)
mnx2
=m−n
mn lim
x→0
(1 −cos x)
x2=m−n 2mn .
•
f:R→R0<a<1
lim
x→0f(x) = 0 = lim x→0
f(x)−f(ax)
x.
lim
x→0
f(x)
x= 0.
limx→0f(x)−f(ax)
x= 0 ε > 0δ > 0x∈]−δ, δ[
|f(x)−f(ax)|< ε|x|n∈N
|f(x)−f(anx)|≤|f(x)−f(ax)|+|f(ax)−f(a2x)|+··· +|f(an−1x)−f(anx)|
≤ε|x|(1 + a+a2+··· +an−1) = ε|x|1−an−1
1−a<ε|x| 1−a.
x∈]−δ, δ[ 0 < a < 1
akx∈]−δ, δ[k∈Nnlimx→0f(x)=0
limn→∞ f(anx)=0
∀ε > 0,∃δ > 0 : x∈]−δ, δ[ =⇒ |f(x)| ≤ ε|x|
1−a
∀ε > 0,∃δ > 0 : x∈]−δ, δ[ =⇒
f(x)
x
≤ε
1−a
lim
x→0
f(x)
x= 0