Analyse 2 - Suites et intégrales

∼
x0∈Rb∈Rb > x0f]x0, b[
f l x0
f I a I
f a
f a
f I Ra∈I f
a
f I R
f a ∈I f a
f a f(a)6= 0 f
a
f g I x0∈I
f(x0)> g(x0)x0f
g
x−1
x2−1
√x−2
x2−5x+4
xE 1
x
x3+x2+5
5x3−x2+2 +∞
sin(x2)
x∞0
√1+x−√1−x
x0
x−√x
x+ln(x)+∞
ln(x) ln(ln(x)) 1

f0
∀x∈R, f(2x) = f(x).
x∈Rf(x) = f(x
2k)k∈N
f(x) = f(0) x
fR R
∀(x, y)∈R2, f(x+y) = f(x) + f(y).
f(0) f
fN Z Q f(1).
f0
f(x) = ax a ∈R
R0
∀x∈Rf(x) + f(2x) = 0
f:x7→ sin 1
x
R
f:R∗−→ R
x7→ sin xsin 1
x
g:R∗−→ R
x7→ 1
xln(ex+e−x 2)
h:R\{−1,1} −→ R
x7→ 1
1−x−2 (1−x)2
a b c R
f(x) = 
5x < −2
ax +b−2≤x < 1
ln x x ≥1
g(x) = (x−1)e2/x x < 0
3ex−c x ≥0

f[a, b]a < b
a, b ∈RF={f(x), x ∈[a, b]}
f[0,+∞[f(0) = limx→∞ f(x)=0
f
f(x) = 1x∈Q
0x /∈Qf(x) = E(x)f(x) = 1/x x ∈R∗ 36 x= 0
f[a, b]Ra∈Rb∈Ra<b
f f f
f f f
[a, b] [f(a), f(b)]
f[0,1] f(0) = f(1) g
g(x) = f(2x)x∈[0,1/2]
f(2x−1) x∈]1/2,1].
g[0,1]
a, b a < b f [a, b] [a, b]
f[a, b]
DR
f:R−→ Rf(x) = 0 x∈D f = 0
f:R−→ Rg:R−→ Rf(x) = g(x)
x∈D f =g

f g R|f|=|g|f
Rf=g f =−g
fR
∀(x, y)∈R2, f(x+y) = f(x)f(y).
fR
g I x0∈I
g x0
fR(x, y)∈R2
|f(x)−f(y)|≤|x−y|2. f
IRf I RI
(a, b)∈I2a < b f0(a)<0f0(b)>0
c∈]a, b[f0(c) = 0
f0(I)f0
fR+1f(xy) = f(x)f(y)
f(1) = 1
x0∈R∗
+h∈Rx0+h > 0
f(x0+h)−f(x0) = f(x0)[f(1 + h
x0
)−f(1)]
f R∗
+
f0(x)
f(x)=f0(1) x
x > 0
fDf(x) = ln(1+x)
√1+x−1
Df
1 0 x7→ ln(1 + x)
x7→ √1 + x
f0
l g [0,+∞[g(x) = f(x)
x > 0g(0) = l[0,+∞[