Exercice 1 Exercice 2 est dite continue en x si pour Exercice 3

a b f : [a, b]→Rf(a)<0f(b)>0x0∈[a, b]
f(x0)=0
f:R→Rf f0(x)=0 x x0
f(x0)=0 f= 0
X X x0
f:N→Nn0f(n0)
n f(n0)≤f(n)
∀ > 0,∃α > 0,∀x|x−x0|< α |f(x)−f(x0)|< 
TX f :X→X x
V∈ T f(x)∈V f−1(V)∈ T f−1(V)
x f(x)∈V
x h f π R
x(f+R) = x(f) + R.x0(f) + R.π(R)R f +R∈h
n n = 2k k n
q r a =bq +r0≤r < b
a b d d a b d0≤d d0a b
x x =nx0n x0x0<1
IRx0∈I f I Rf x0
 > 0α > 0x∈I|x−x0|< α |f(x)−f(x0)|< 
S D S
D
∀n∈N, n ∈S n ∈D
∃!
C(x, y)C
x C y x →Cy

m C n n
C p m C p ∀m, n, p ∈N, m →Cn n →Cp m →Cp
1C0
0C
0C
0C
C
C
C
p C n n
Pk
i=1 ik
Q1≤k≤nk
Pp
n=1 1
Sn=Pn
k=1 k k n + 1 −k
Sn=n(n+ 1) −Sn
f(x) = Pn
i=0 aixig(x) = Pp j=0 bjxj
f(x)g(x)
∃k∈N, n = 2k
∀n, p ∈N,∃k, l ∈Nn= 2k p = 2l∃m∈Nn+p= 2m
∀a, b, c ∈C,∃z∈Caz2+bz +c= 0
≤N2(x, y)≤(x0, y0)x=x0y≤y0
∀x, y ∈N(x, y)≤(x, y)
∀x, y, x0, y0∈N(x, y)≤(x0, y0) (x0, y0)≤(x, y)x=x0y=y0
∀x, y, x0, y0, x00, y00 ∈N(x, y)≤(x0, y0) (x0, y0)≤(x00, y00) (x, y)≤(x00, y00)
∀x, y, x0, y0(x, y)≤(x0, y0) (x0, y0)≤(x, y)

0 1 2 3 4 5 6
π3
√8
n n!
0 1 2 3 4 5 6
p a p ap−1≡1(p)
0≤ρ < 1 1/(1 −ρ)
x y z x +z=y+z x =y
x y z xz =yz
x=y
a a a + 1 a+ 2
√2