Fonction puissance entière Fonction puissance négative Fonction

(x→xn)n∈N∗
R+R+n
R R n
n x →+∞
0n
0n
n n
(x→x−n)n∈N∗
lim
x→+∞x−n= 0 lim
x→0+x−n= +∞0−n
x→+∞n
x→0++∞n
n(x7→ xn)R+R+
n
√·:R+→R+
x7→ n
√x.
∀x∈R+,∀y∈R+, xn=y⇔x=n
√y.
n(x7→ xn)R R
R
n
√·:R→R
x7→ n
√x.
x > 0y=n
√x⇐⇒ yn=x⇐⇒ nln(y) = ln(x)⇐⇒ ln(y) = 1
nln(x)⇐⇒ y= exp( 1 nln(x))
n
√x=x1
n.
(xp)1
q=x1
qp
=xp
q(−1)2= 1 6= (−1)1xa
x > 0 exp(aln(x))
n
√0 = 0 0
lim
x→+∞
n
√x= +∞n
0n
ln
(f, F )∈(F(I, R))2F f I F F 0=f.
(f, F, G)∈(F(I, R))3. F f I
G f I ⇐⇒ (∃k∈RG=F+k).

f∈ F(I, R). f I
I
f f I
f]a, +∞[
a f +∞+∞
(x→1
x)R+∗0x= 1.
ln(x) = x
1
1
tdt.
∀x > 0,ln0(x) := 1
x.
(x→ln(x))
∀(x, y)∈(R+∗)2,ln(x·y) = ln(x) + ln(y).
a∈R+∗(x→ln(ax)) h:h(x) = a
ax =1 x.
(x→1
x)
∃k∈R∀x∈R+∗,ln(ax) = ln(x) + k.
x= 1 ln(a) = k.
∀x∈R+∗,ln(ax) = ln(x) + ln(a). a
ln(1) = 0
∀x > 0,ln(x×1
x) = ln(1) = 0 ⇒ln(x) + ln( 1 x) = 0 ⇒ln( 1 x) = −ln(x).
∀x > 0,∀y > 0,ln(x
y) = ln(x)−ln(y).
∀x > 0,∀n∈Z,ln(xn) = nln(x).
a > 1 loga:]0,+∞[→R
∀x > 0,loga(x) := ln(x)
ln(a).
10 log log10
ln
d
dx(ln)(x) = 1 x>0,ln R+∗.
+∞ln(2n) = nln(2)
lim
n→∞ ln(2n) = +∞.
ln +∞+∞
X= 1/x lim
x→0ln(x) = −∞.
lim
x→+∞ln(x) = +∞,lim
x→0ln(x) = −∞.
φ(x) = x−1−ln(x)φ(x)≥0,∀x∈R+∗.
lim
x→1
ln(x)
x−1= lim
h→0
ln(1 + h)
h= 1.

ln R+∗R.R+∗
lim
x→0ln(x) = −∞ lim
x→+∞ln(x) = +∞
R+∗R. y x
x y
exp :] − ∞; +∞[−→]0; +∞[
y−→ xln(x) = y.
∀x∈R,∀y∈R+∗,(y= exp(x)⇔ln(y) = x).
∀x∈R,ln(exp(x)) = x.
∀y∈R+∗,exp(ln(y)) = y.
ln(1) = 0 ⇔exp(0) = 1.exp(1) = eln(e) = 1.
e= 2.718281828
∀x∈R, ex:= exp(x).
∀x∈N,ln(ex) = x
lim
x→−∞ ex= 0 lim x→+∞ex= +∞
∀(x, y)∈R2, ex+y=ex×ey.
∀y∈R,1
ey=e−y. x +y= 0
∀(x, y)∈R2, ex−y=ex
ey.
ex+yln(ex+y) = x+y exey
R
∀x∈R,exp0(x) = exp(x).
f:I→Ra∈I g :x7→ ef(x)a
d
dx(ef(x))(a) = g0(a) = f0(a)ef(a).
∀x∈R,1+ x≤ex
φ(x) = ex−1−x φ0(x) = ex−1>0x > 0φ0(x)<0x < 0φ(x)> φ(0) = 0
ax
x∈N, a ax=a×a×a··· × a
x
x∈Z−a6= 0 x < 0, ax=1
a×1 a×1 a··· × 1 a
−x
x∈Ra > 0ax= exp (xln (a))

x→axx→ax
(ax)
x=exp(xln(a)) x= ln(a)×exp(xln(a)) = ln(a)×ax
(a→ax)a a > 1.
lim
−∞ expa= lim x→−∞ exp(xln(a)) = 
0a > 1
+∞a < 1
1a= 1
f f(x) = u(x)v(x)
f u(x)>0; u v f
f(x) = exp (v(x)·ln(u(x)) ⇒d
dx(f(x)) = d dx[v(x) ln(u(x))] ·exp (v(x)·ln(u(x))
=d
dx[v(x) ln(u(x))] ·u(x)v(x)= [v0(x) ln(u(x)) + v(x)u0(x) u(x)]·u(x)v(x).
(x→xx).
f f(x) = u(x)v(x)d
dx[v(x) ln(u(x))]
f0(x). f(x) = exp(v(x) ln(u(x)))
g g exp ◦g
u(x)→1v(x)→ ∞
lim
x→+∞(1 + 1
x)xlim
x→+∞(1 + 1
x2)xlim
x→+∞(1 + 1
x)(x2)
lim
x→+∞
ln(x)
x=
∀(α, β)∈(R+∗)2,lim
x→+∞
(ln(x))α
xβ=
∀(α, β)∈(R+∗)2,lim
x→0+xβ(|ln(x)|)α=
∀a∈]1,+∞[,∀α∈R,lim
x→+∞
ax
xα=
∀a∈]1,+∞[,∀α∈R,lim
x→−∞ ax|x|α=
∀k∈N,∀x∈]−1,1[,lim
n→+∞nk×xn=
lim
x→+∞xxlim
x→0+xxlim
x→+∞1 + 3
x(5x)
lim
x→+∞
ln(x)
ex,lim
x→+∞
x3
e√x=
z∈Cz=a+ib z
eaeib ez
ez+z0=ezez0
a∈Cea=ea+2πi
ln(ρeiθ) = ln(ρ) + iθ θ 2π

f:D→RT∀x∈R, x ∈D⇒x+T∈D f (x+T) = f(x)
T
f(x+nT ) = f(x)
R2πsin0(x) = cos(x)
0 limx→0sin(x)
x= 1
R+∗∀x > 0,sin(x)< x
x=π
2+kπ k ∈Z
R2πcos0(x) = −sin(x) 0
lim
x→0
cos(x)−1
x2=−1 2
R\π
2+kπ |k∈Z=
k∈Z−π
2+kπ, π 2+kπ,tan x=sin x cos x.
tan0(x) = 1 + tan2(x) = 1
cos2(x)
0y=xlim
x→0
tan x
x= 1 ∀x∈]0, π/2[,tan(x)> x
t−π
2,π 2t
R
arctan : R→−π
2,π 2
x7→ arctan(x)
∀x∈−π
2,π 2,∀y∈R,tan(x) = y⇔x= arctan(y)
∀x∈−π
2,π 2,arctan(tan(x)) = x
∀y∈R,tan(arctan(y) = y.
arctan
y01
√31√3 +∞ −∞
arctan(y) 0 π
6
π
4
π
3π/2−π/2
∀x > 0,arctan(x) + arctan( 1
x) = π 2. x < 0