exos continuite

•lim
x→2
2x3−4x2+x−2
x2−4•lim
x→+∞exsin(e−x)•lim x→+∞
x2
ln(ex+ 1)
•lim
x→+∞
√x2+x−1−x√x•lim
x→+∞
xln(x)
(ln(x))x•lim
x→+∞
cos(x+x2−1)
x
•lim
x→0
x
arccos(x)−π
2•lim
x→0xEnt 1
x•lim
x→+∞x1
x
•lim
x→0
x−Ent(x)
p|x|•lim
x→−1
x3+x2−x−1
x3−3x−2•lim
x→+∞ln(x)
x1
x
•lim
x→+∞
sh(x)
ex•lim
x→0
2
sin2(x)−1
1−cos(x)•lim
x→1+ln(x) ln(ln(x))
•lim
x→+∞qx+px+√x−√x•lim x→+∞xx+1 −(x+ 1)x•lim
x→1
x2−1
(√x−1) ln(x)
f1(x) = ex−1
x
f2(x) = 1−x
1−x2
f3(x) = x2ln(x)
sin(x)
f4(x) = Ent(x) + px−Ent(x)
f5(x) = 1
x−1−3 (x−1)2
f6(x) = e1
1−x+ 2x−3
f7(x) = xln x
x+ 1
f8(x) = xEnt 1
x
f9(x) = sup
n∈N
xn
n!
f f(x) = e−1
x2x > 0f(x)=0 x60f
R
n∈Nn f

f0 1 ∀x∈Rf(x) = f(x2)
fR∀x∈Rfx+ 1
2=f(x)
fRf(0) = 1 ∀x∈Rf(2x) = f(x) cos(x)
fR∀(x, y)∈R2fx+y
2=1 2(f(x) + f(y))
f(0) = f(1) = 0 f
f0∀(x, y)∈R2fx+y
3=1 2(f(x) + f(y))
fR∀(x, y)∈R2f(x+y) = f(x)+f(y)
f
R Z
k[0,+∞[ 0 6x<y⇒
√x+√y>1
k
[1,+∞[
fRf◦f
kRk < 1
f[0,1] f(0) = f(1) x
fx+1
2=f(x)x f x+1 n=f(x)
n>2
fC1R∀x∈Rf(f(x)) = ax +b
a∈]0,1[ b
∀x∈Rf(ax +b) = af(x) + b f0(ax +b) = f0(x)
un+1 =aun+b
f0f
a > 1

I
x2022 −x2021 =−1I= [−1,1]
ln x=x2−5
x+ 2 I= [1,10]
3x= 1 + ln(2 + x2)I= [0,1]
ex= 2 + x[ln 2,2 ln 2]
x3−3x2=−1I= [−1,1]
0.01
n∈Nfnfn(x) = xn+ 9x2−4
fn(x)=0
un
u1u2∀n∈N∗un∈0; 2
3
∀x∈]0,1[ fn+1(x)< fn(x)
(un)
(un)l
un
nl
fRf(x) = ex+x
fR
n f(x) = n
xn
(xn)
∀n>1 ln(n−ln n)6xn6ln n
(xn)xn
ln(n)
n fnfn(x) = x5+nx −1
fn
∀n>1unfn(un)=0
un61
n(un)
(nun)
n>1fnR+∗fn(x) = 1 + x+x2+··· +xn
fn(x)=2 un
∀n>2un∈]0; 1[
(un)
lim
n→+∞un
n= 0
vn=un−1
21 2+vnn+1
= 2vn

n>1gngn(x) = ex−1
nx
gn]0; +∞[gn(x)=0
un
0< un<1
n(un)
gn+1(x)−gn(x)gn(un+1)
(un)
lim
n→+∞nun
(un)
unnxn+1 −(n+ 1)xn= 1
n∈N∗fn(x) = nxn+1 −(n+ 1)xn
u1
fn]0,+∞[
fn(x) = 1
1 + 1
n6un61 + 2 n(un)
β∈Rvn=1 + β
nn
fn1 + β
n
(x−1)ex= 1 α
α∈]1,2[
ε0< ε < α −1
lim
n→+∞fn1 + α−ε
nlim
n→+∞fn1 + α+ε
n
n01+ α−ε
n6un61+ α+ε n
lim
n→+∞n(un−1)
(fn)
I
•(fn)f∀x∈Ilim
n→+∞fn(x) = f(x)
•(fn)f∀ε > 0∃n0∈N∀n>n0∀x∈I
|fn(x)−f(x)|6ε
(fn)f(fn)f
fnI f
fn(x)f(x) 0