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f[0 ; 1] [0 ; 1]
x∈[0 ; 1] k∈N∗fk(x) = x x
f
f
limx→0√1+x−√1−x
x
limx→+∞x−√x
ln x+x
limx→0+ xx
limx→1+ ln x. ln(ln x)
limx→0(1 + x)1/x
limx→11−x
arccos x
limx→0x. sin 1
x
limx→+∞xcos ex
x2+1
limx→+∞ex−sin x
limx→+∞x+arctan x
x
limx→0xb1/xc
limx→+∞xb1/xc
limx→0+ b1/xclimx→0xb1/xclimx→0x2b1/xc
g:R→R+∞g
f, g :R→Rf+∞g f +g
g
a < b ∈¯ Rf: ]a;b[→R
x7→ limx+f
R
f:x7→ bxc+px− bxc
f:R→R
f(x) = (1x∈Q
0
f
f:x7→ sup
n∈N
xn
n!
R+
f:R→Rlim−∞ f=−1 lim+∞f= 1 f

f: [0 ; 1] →[0 ; 1] f
f:R→R
f
f: [0 ; +∞[→R
lim
x→+∞
f(x)
x=` < 1
α∈[0 ; +∞[f(α) = α
R Z
f:I→Rg:I→R
∀x∈I, |f(x)|=|g(x)| 6= 0
f=g f =−g
f: [0 ; +∞[→R|f| −→
+∞+∞
f−→
+∞+∞f−→ +∞−∞
f: [a;b]→Rp, q ∈R+
c∈[a;b]
p.f(a) + q.f(b)=(p+q).f(c)
f:I→R
∃(x1, y1)∈I2, x1< y1f(x1)≥f(y1)∃(x2, y2)∈I2, x2< y2f(x2)≤f(y2)
ϕ: [0 ; 1] →R
ϕ(t) = f((1 −t)x1+tx2)−f((1 −t)y1+ty2)
z∈C7→ zexp(z)∈C
R
f:R→Rg:R→R
g◦f f ◦g
f, g : [a;b]→R
∀x∈[a;b], f(x)< g(x)
α > 0
∀x∈[a;b], f(x)≤g(x)−α

f:R→R
lim
+∞f= lim −∞ f= +∞
f
f:R→Ry∈R
f
y∈R
f:R→R
f(x) = x
1 + |x|
fR]−1 ; 1[
y∈]−1 ; 1[ f−1(y)
f(x)
a<b∈Rf: ]a;b[→R
f f(]a;b[) = ]limaf; limbf[
α1/e
f∈ C1(R,R)
∀x∈R, f0(x) = αf(x+ 1)
α= 1/e
f: [0 ; +∞[→[0 ; +∞[
f◦f= Id
f
f:R→R
∀x∈R, f(x) = f(x2)
f
f:R→R∀x∈R
fx+ 1
2=f(x)
f
f:R→R
∀x∈R, f(2x) = f(x) cos x
f
f:R→R
∀x, y ∈R, f(x+y) = f(x) + f(y)
f(0) x∈Rf(−x) = −f(x)
n∈Zx∈Rf(nx) = nf(x)
r∈Qf(r) = ar a =f(1)
x∈Rf(x) = ax
f:R→Rx, y ∈R
f(x+y) = f(x) + f(y)
f x0∈R
f

f:R→R
∀x, y ∈R, f x+y
2=1 2(f(x) + f(y))
f f(0) = f(1) = 0
f
∀x∈R,2f(x) = f(2x)
f
f
f:R→R
∀x, y ∈R, f x+y
2=1 2(f(x) + f(y))
f(0) = 0
∀x, y ∈R, f (x+y) = f(x) + f(y)
f
x∈R|sin x|≤|x|
x7→ sin x
f:R→Rk k ∈[0 ; 1[ f(0) = 0
a∈R(un)
u0=a∀n∈N, un+1 =f(un)
un→0
f, g : [0 ; 1] →R
ϕ(t) = sup
x∈[0;1]
(f(x) + tg(x))
ϕR

f(x)> x f
fk(x)≥fk−1(x)≥. . . ≥f(x)> x
f(x)< x
f(0) ≥0f(1) ≤1
(an) (bn)
f(an)≥anf(bn)≤bn
(an) (bn)a0= 0 b0= 1
anbnm= (an+bn)/2
f(m)≥m an+1 =m bn+1 =bn
an+1 =anbn+1 =m
(an) (bn)
c an≤c≤bn
f(an)≤f(c)≤f(bn)
an≤f(c)≤bn
(an) (bn)c
f(c) = c
f
c= sup {x∈[0 ; 1], f(x)≥x}
x→0
√1 + x−√1−x
x=1 + x−(1 −x)
x√1 + x+√1−x=2
√1 + x+√1−x→1
x→+∞x−√x
ln x+x=1−1/√x
ln x
x+ 1 →1
x→0+
xx= exln x= eX
X=xln x→0xx→1
x→1+
ln x. ln(ln x) = Xln X
X= ln x→0 ln x. ln(ln x)→0
x→0
(1 + x)1/x = e 1
xln(1+x)= eX
X=ln(1+x)
x→1 (1 + x)1/x →e
x→1
1−x
arccos x=1−cos y y=2 sin2(y/2) y= sin(y/2)sin(y/2) y/2
y= arccos x→0 sin y/2→0sin y/2
y/2→11−x arccos x→0
x→0
xsin 1
x≤ |x| → 0
x→+∞
xcos ex
x2+ 1 ≤x x2+ 1 →0
x→+∞ex−sin x≥ex−1→+∞
x→+∞
x+ arctan x
x−1≤arctan x x≤π 2x→0
x→0
1/x −1≤ b1/xc ≤ 1/x
|b1/xc − 1/x| ≤ 1
|xb1/xc − 1|≤|x| → 0