1 Limites 2 Continuité

f
x3−27
x−33; x2−5x+ 6 x+ 2 −2; √x
x+ 3√x0; sin(2x)
sin(3x)0;
(1 −x)(2 −x)(3 −x) + ∞;x2−1
2x3+ 1 +∞; cos x−3x+∞;sin x x+∞.
f
f(x) = 2 −1
x2−1
f
f(x) = 2x2−5x+ 7
x
Df+∞
f(x) = px2+ 2x+ 3,D:y=x+ 1
f
f(x) = (50 + x20)2−2500
x20
(50 + x20)2f(x)
x= 0,6; 0,4; 0,3; 0,2; 0,1; 0,01.
f0
f(x)x6= 0 f0
2f f(x) = ½x2−1x≤2
5−x x > 2
sin x= 1 −x
α α
x2+√x−3 = 0

x2+√x;−5
x2; 3xsin x; (2x−1)x2; cos3x; (x2+x+ 1)4;x3 x2−1;sin x+ cos x 1 + cos x
Cff x
f(x) = √x4
f x f(x) = −4
3x3+1 5x5
f:x7→ 2
5p25 −x2
x3−27
x−3=(x−3)(x2+ 3x+ 9) x−3=x2+ 3x+ 9 x→3
−−−→ 27
lim
x→−2(x2−5x+ 6) = 20 lim x→−2+
x2−5x+ 6
x+ 2 = +∞lim
x→−2−
x2−5x+ 6
x+ 2 =−∞
√x
x+ 3√x=1
√x+ 3 lim
x→0+
√x
x+ 3√x=1
30−)
sin(2x)
sin(3x)=sin(2x) 2x∗2x 3x∗3x sin(3x)=sin(2x) 2x∗2 3∗3x sin(3x)
x→0
−−−→ 2
3lim
x→0
sin(x)
x= 1
lim
x→+∞(1 −x)(2 −x)(3 −x) = −∞
x2−1
2x3+ 1 =1−1
x2
2x+1
x2
x→+∞
−−−−−→ 0
cos x−3x≤1−3xlim
x→+∞(1−3x) = −∞ lim x→+∞(cos x−3x) = −∞
|sin x
x| ≤ 1 xlim
x→+∞
1
x= 0 lim
x→+∞
sin x
x= 0 0 ≤ |sin x x| ≤ 1 x)
f f(x) = 2 −1
(x−1)(x+ 1)
lim
x→−1−
f(x) = −∞ lim
x→−1+f(x) = +∞
x=−1Cfx= 1
Cf
lim
x→+∞f(x) = 2 lim x→−∞ f(x) = 2
y= 2 Cf±∞

f(x) = 2x2−5x+ 7
x= 2x−5 + 7 x
f(x)−(2x−5) = 7
x
x→±∞
−−−−−→ 0
∆ : y= 2x−5Cf±∞
f(x)−(2x−5) = 7
xxCf∆x > 0
x < 0
x x2+ 2x+ 3 >0f IR
f(x)−(x+ 1) = px2+ 2x+ 3 −(x+ 1) = (px2+ 2x+ 3 −(x+ 1)) ∗√x2+ 2x+3+(x+ 1)
√x2+ 2x+3+(x+ 1)
=(x2+ 2x+ 3) −(x+ 1)2
√x2+ 2x+ 3 + x+ 1 =2 √x2+ 2x+3+x+ 1
x→+∞
−−−−−→ 0
D Cf+∞
f(x) = (50 + x20)2−2500
x20
(50 + x20)2= 2500 0,6 0,4 0,3f(x)≈100
0,2 0,1 0,01 f(x)=0
lim
0f= 0
(50 + x20)2−2500 = (50 + x20 −50)(50 + x20 + 50) = x20(100 + x20)
f(x) = x20(100 + x20)
x20 = 100 + x20 x→0
−−−→ 100
x40 x20
lim
x→2−
f(x) = f(2) = 3 = lim
x→2+f(x)
f2
f f(x) = sin x+x−1f(0) = −1<0f(π
2) = π 2>0
f IR 0∈]f(0), f(π
2)[
f(x) = 0 ⇔sin x= 1 −x]0,π
2[
f x ≥0f(x) = x2+√x−3
f IR∗
+x > 0
f0(x) = 2x+1
2√x>0

f IR+
f f(IR+) = [f(0),lim+∞f[= [−3,+∞[
f IR+0∈f(IR+)
f(x) = 0 α∈[0,+∞[
f(0) = −3<0f(2) = 1 + √2>0α∈]0,2[
]0,2[ f(1) = −1<0α∈]1,2[
x > 0
(x2+√x)0= 2x+1
2√x
x6= 0
(−5
x2)0= (−5) ∗(1 x2)0= (−5) ∗−2x x4=10 x3
x
(3xsin x)0= 3 sin x+ 3xcos x
x
((2x−1)x2)0= 2 ∗x2+ (2x−1) ∗2x= 6x2−2x
x
(cos3x)0= 3 ∗(−sin x)∗cos2x=−3 sin xcos2x
x
((x2+x+ 1)4)0= 4(2x+ 1)(x2+x+ 1)3
x x2−16= 0 x∈IR\{−1,1}
(x3
x2−1)0=3x2∗(x2−1) −x3∗2x (x2−1)2=x4−3x2 (x2−1)2
x1 + cos x6= 0 x∈IR\{2kπ, k ∈ZZ}
(sin x+ cos x
1 + cos x)0=(cos x−sin x)(1 + cos x)−(sin x+ cos x)(−sin x) (1 + cos x)2=1 + cos x−sin x (1 + cos x)2
∆Cf
y=f0(4)(x−4) + f(4)
f(4) = √4=2 f0(4) = 1
2√4=1
4f0(x) = 1
2√x
f0(4) = limx→4f(x)−f(4)
x−4= limx→41
√x+2 =1
4x > 0x=√x2
∆ : y=1
4x+ 1
f IR x
f0(x) = −4x2+x4=x2(−4 + x2) = x2(2 + x)(−2 + x)
f0(x) = 0 ⇔x∈ {−2,0,2}
f0(x)>0⇔x∈]− ∞,−2[∪]2,+∞[
f0(x)<0⇔x∈]−2,2[
f−2 2 f−2 2

f:x7→ 2
5p25 −x2
f x 25−x2≥0⇔(5+x)(5−x)≥0Df= [−5,5]
x∈[−5,5] −x∈[−5,5] f(−x) = 2
5p25 −(−x)2=f(x)
f[0,5]
f[0,5[ 25 −x2>0x[0,5[
f0(x) = 2
5∗(−2x)∗1
2√25 −x2=−2x 5√25 −x2
f0≤0 [0,5[ f[0,5[ f]−5,0]
f5
lim
x→5−
f(x)−f(5)
x−5
f(x)−f(5)
x−5=
2
5√25 −x2
x−5=2 5p(5 + x)(5 −x)
−p(5 −x)2=−2
5
√5 + x
√5−x
p(5 −x)2=|5−x|0≤x≤5x−5<0x−5 = −p(5 −x)2
lim
x→5−
f(x)−f(5)
x−5= lim
x→5−
2
5p(5 + x)(5 −x)
−p(5 −x)2=−∞
f5
[0,5] [−5,0] (0y)
f0(0) = 0 (0, f(0) = 2)
limx→5−
f(x)−f(5)
x−5=−∞ 5−5