Examen Partiel Analyse 1 - Université Aix-Marseille

f:R→Rx0∈R
lim
x→x0
f(x) = −∞
∀A∈R,∃α > 0,∀x∈R\ {x0},|x−x0|< α f(x)6A
A x ∈R\ {x0}
0<|x−x0|< α ∀A∈R,∃α > 0,∀x∈R,0<|x−x0|<
α f(x)< A
f x0
∀ > 0,∃α > 0,∀x∈R,|x−x0|< α |f(x)−f(x0)|< 
x x0
f: [a, b]→Rf
f f(a)f(b)f(a).f(b)60f a b
c∈[a, b]f(c)=0 f(a)f(b)f(a).f(b)<0c
a b c ∈]a, b[
f y f(a)f(b)
y∈[f(a), f(b)] [f(b), f(a)] f y a b x ∈[a, b]y=f(x)
y f(a)f(b)x a b
f
x a b y f(a)f(b)y=f(x)
f(x)f(a)f(b)
f:R+→Rflim
x→+∞f(x)=0
fR+
lim
x→+∞f(x) = 0
∀ > 0,∃A∈R,∀x∈R+, x >A|f(x)|< 
B∈R+x>B⇒ |f(x)|61
f+∞= 1
B x ∈R+x>B|f(x)|<1|f(x)|61x
x∈R+B B B 0
g: [0, B]→Rg(x) = f(x)x∈[0, B]g
g f [0, B]f[0,B]
g(x) = f(x)x∈[0, B]g[0, B]f
[0, B]g[0, B]
m M x ∈[0, B]m6g(x)6M g
C∈R|f(x)|6C x ∈[0, B]
g(x) = f(x) [0, B]m6f(x)6M x ∈[0, B]|f(x)|6
max(|m|,|M|)C= max(|m|,|M|)
a6c6b|c|6max(|a|,|b|)
a6c6b06a a b c |a|=a|b|=b|c|=c
|a|6|c|6|b|b60a b c |a|=−a|b|=−b|c|=−c|b|6|c|6|a|
a60b>0|a|6|b| −|b|6−|a|=a6c6b=|b| |c|6|b|
|b|6|a| −|a|=a6c6b=|b|6|a| |c|6|a|
|c|6|a| |c|6|b| |c|6max(|a|,|b|)

fR+
x∈[0, B]|f(x)|6C x >B|f(x)|61
x∈R+|x|6max(C, 1) fR+
f:R∗
+→R
f(x)=1−xE1
x,
t∈RE(t)E(t)6t <
E(t)+1
f0
t∈Rt−1< E(t)6t
E(t)6t < E(t)+1 t−1< E(t)
t−1< E(t)6t
x > 0x>f(x)>0
t= 1/x
1/x −1< E(1/x)61/x
1−x < xE(1/x)61x
x−1>−xE(1/x)>−1−1
x > 1−xE(1/x) = f(x)>−1
f0
lim
x→0f(x) = 0  > 0α=
0< x < α |f(x)|=f(x)f(x)>0|f(x)|< x < α =
f0 0
g:R+→Rg(x) = f(x)x > 0g(0) = lim
x→0f(x) = 0
x > 1f(x)=1
x > 1 0 61/x < 1E(1/x) = 0
f(x) = 1 −xE(1/x) = 1
f(1) f1
f(1) = 1 −1×E(1/1) = 0 E(1/1) = E(1) = 1 f(x) = 1 x > 1 lim x→1+f(x) =
16=f(1) f1
lim
x→+∞
x2+ 5 + exp(1/x)
(x+ 1)3
lim
x→+∞
x2+ 5 + exp(1/x)
(x+ 1)3= lim
x→+∞
x2
(x+ 1)3+ lim
x→+∞
5
(x+ 1)3+ lim
x→+∞
exp(1/x)
(x+ 1)3
x→+∞
(x+ 1)3→+∞5/(x+ 1)3→0
1/x →0e1/x →1e1/x/(x+ 1)3→0
x2
(x+ 1)3→0
x2
(x+ 1)3=x2 x3+ 3x2+ 3x+ 1 =x2 x2(x+ 3 + 3/x + 1/x2)=
1
x+ 3 + 3/x + 1/x21/x2→0 3/x →0 3 →3x→+∞x+3+3/x + 1/x2→+∞
1/(x+ 3 + 3/x + 1/x2)→0
lim
x→+∞
x2+ 5 + exp(1/x)
(x+ 1)3= 0
lim
x→+∞(px2+x+ 1 −x)

+∞x√x2=|x|=x(√x2+x+ 1)2=
|x2+x+ 1|=x2+x+ 1
px2+x+ 1 −x=(√x2+x+ 1 −x)(√x2+x+ 1 + x)
√x2+x+ 1 + x
=√x2+x+ 12−x2
px2(1 + 1/x + 1/x2) + x
=x+ 1
xp1 + 1/x + 1/x2+x
=x(1 + 1/x)
x(p1 + 1/x + 1/x2+ 1)
=1 + 1/x
p1 + 1/x + 1/x2+ 1
x→ ∞ lim
x→+∞
(px2+x+ 1−x) = 1
2
lim
x→0
(tan x)2
cos(2x)−1
cos(2x) = 2 cos2x−1 cos 2x−1 = 2 cos2x−2 = −2(1 −
cos2x) = −2 sin2x x 6= 0 0 x∈[−π/4, π/4]
tan2x
cos 2x−1=
sin2x
cos2x
−2 sin2x
=−sin2x
2 cos2xsin2x
=−1
2 cos2x
x→0 cos x→1 cos2x→1 lim
x→0
(tan x)2
cos(2x)−1=−1 2
lim
x→1+(x3−1) ln(2x2+x−3)
2x2+x−3x= 1
2x2+x−3x−1 2x2+x−3 = (x−1)(x+ 3) x3−1
x−1
x3−1 = (x−1)(x2+x+ 1)
(x3−1) ln(2x2+x−3) = (x−1)(x2+x+ 1) ln((x−1)(x+ 3))
= (x2+x+ 1)(x−1) ln(x−1) + (x2+x+ 1)(x−1) ln(x+ 3)
x→1+x x2+x+ 1 →3x−1→0x > 1
ln(x−1) → −∞ (x−1) ln(x−1) →0 (x2+x+ 1)(x−
1) ln(x−1) →0x+ 3 →4 (x−1) ln(x+ 3) →0 (x2+x+ 1)(x−1) ln(x+ 3) →0
lim
x→1+(x3−1) ln(2x2+x−3) = 0
n∈N∗x∈R+
fn(x) = −1 + x+x2+· · · +xn−1+xn.
n∈N∗fn
fnx7→ −1x7→ xk
k= 1,...,n R+RfnR+
R+x7→ xk
fn
fn(0) fn(1) xn∈[0,1] fn(xn) = 0
(xn)n∈N∗
fn(0) = −1 + 0 + 02+··· + 0n=−1fn(1) = −1+1+12+··· + 1n=−1 + n=n−1
fn(0) fn(1) fnR+[0,1]
xn∈[0,1] fn(xn) = 0 n > 1
fn(1) = n−1>0fn(0) = −1<0xnxn∈]0,1[

n∈N∗x∈R+fn+1(x)>fn(x)
fn+1(x) = −1 + x+x2+··· +xn+xn+1 =fn(x) + xn+1 x∈R+xn+1 >0
fn+1(x)>fn(x)
fn+1(xn)>0
x=xnfn+1(xn)>fn(xn)xn
fn(xn) = 0 fn+1(xn)>0
fn+1(xn)>fn+1(xn+1)xn>xn+1
fn+1(xn)>0xn+1 fn+1(xn+1) = 0 fn+1(xn)>fn+1(xn+1)
xn< xn+1 fn+1 fn+1(xn)< fn+1(xn+1)xn>xn+1
(xn)
(xn)xn∈
[0,1] n