# Les fonctions en 2eme bac science math

Disponible uniquement sur Etudier
• Pages : 9 (2076 mots )
• Téléchargement(s) : 0
• Publié le : 19 décembre 2010

Aperçu du document
‫دروس اﻟﺪﻋﻢ و اﻟﺘﻘﻮﻳﺔ‬ ‫ﻗﺴﻢ : اﻟﺜﺎﻧﻴﺔ ﺑﺎآﺎﻟﻮرﻳﺎ ع.رﻳﺎﺿﻴﺔ‬
x → +∞

lim

x → +∞

[x lim ( x

2

+ x + 1 − (x + 1) = + 3x − 1 + mx =

]

‫درس : اﻹ ﺗﺼﺎل و اﻟﻨﻬﺎﻳﺎت‬ www.madariss.fr 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 lim− x 2 − 5x + 6 = (2 − x ) 2

2

)

. ‫اﻷﺳﺘﺎذ : ﻋﻠﻲ اﻟﺸﺮﻳﻒ‬ ‫اﻟﺨﻤﻴﺴﺎت‬ : 1 ‫اﻟﺘﻤﺮﻳﻦ رﻗﻢ‬ : ‫أ ﺣﺴﺐ اﻟﻨﻬﺎﻳﺎت اﻟﺘﺎﻟﻴﺔ‬ 1 2 3 4 5 6 7 8 9 10 11 12 13 14

x→2

⎛ ⎛ 1 ⎞⎞ lim x 2 ⎜1 − cos⎜ ⎟ ⎟ = ⎜ ⎟ x → +∞ ⎝ x ⎠⎠ ⎝

x → +∞

lim x 2 − x + 1 − x − 1 =

x2 − x 2x = lim x →1 (1 − x ) 2
x → −2
x 〈−2

x → +∞

lim

x +5 −x

=

lim

x2 + x − 2 = x →1 x −1

x →− 2

lim

x2 − 2 x+ 2

=

lim

x −1 = x+2

lim
x →0

x +1 −1 = x

1 ⎞ ⎛ 1 lim+ ⎜ 2 − 3 ⎟ = x →0 ⎝ x x ⎠
1 5 lim+ 2 − = x →0 x x
x + sin x = x → +∞ 2 x + 1 1 lim x2 . sin = x→ 0 x cos x = lim 2 x → +∞ x + x 2 + cos x = lim x → −∞ 2+x lim ⎛1⎞ lim x. sin (x ). cos⎜ ⎟ = x →0 ⎝x⎠

lim
x →0

x2 + x x2 − x x −1

= =

(x − 1)2 lim
x →1

sin (3x ) = x →0 7x sin (3x ) = lim x →0 tan (x ) cos(x ) + cos(2x ) = lim x → 0 cos(3x ) + 3 cos(4 x ) 2 tan (x ) + sin (x ) = lim x →0 x lim

lim
x→ π 4

1 − 2 cos(x ) 1 − 2 sin (x )

=

3x 2 − 5x = lim+ 2 x→ −1 x + 4 x + 3 lim
x →8

sin 2 (2 x ) lim = π x → 1 + sin (3x )
2

2x − 4 x +1 − 3

=

lim
x →0

1 − cos(x )

sin (2x )

=

x → −2

lim+

− 2x 2 − x + 6 = x 2 − 2x − 8

32

π x→ 3

5x − 1 = x → +∞ x − 4 x + 5 lim
2

33

3 cos(x ) − sin (x ) = π x− 3 π⎞ ⎛ sin ⎜ x − ⎟ 4⎠ lim ⎝ = π cos(2 x ) x→ lim
4

15

16

‫دروس اﻟﺪﻋﻢ و اﻟﺘﻘﻮﻳﺔ‬ ‫ﻗﺴﻢ : اﻟﺜﺎﻧﻴﺔ ﺑﺎآﺎﻟﻮرﻳﺎع.رﻳﺎﺿﻴﺔ‬ Arc tan (x ) = lim+ x →0 Arc tan x

‫درس : اﻹ ﺗﺼﺎل و اﻟﻨﻬﺎﻳﺎت‬ www.madariss.fr

. ‫اﻷﺳﺘﺎذ : ﻋﻠﻲ اﻟﺸﺮﻳﻒ‬ ‫اﻟﺨﻤﻴﺴﺎت‬ : 1 ‫ﺗﺎﺑﻊ اﻟﺘﻤﺮﻳﻦ رﻗﻢ‬ : ‫أ ﺣﺴﺐ اﻟﻨﻬﺎﻳﺎت اﻟﺘﺎﻟﻴﺔ‬

( )

49 50 51
3

⎛ 3 1+ x3 lim Arc tan⎜ x → −∞ ⎜ x ⎝
Arc sin (x ) − 1 lim = x →1 x2 −1

⎞ ⎟= ⎟ ⎠

lim
x →0

x +1 −1 = x

34 = 35

x → +∞

lim

x2 +1 − x 4x 2 − 1 − 2x
x +8 −2 = x
3

⎛1⎞ Arc tan (x) − Arc sin ⎜ ⎟ ⎝2⎠ = lim 3 3 x→ x− 3 3

52 53 54 55 56 57 58 59

3

lim
x →0

36 37 38

x → +∞

lim 4 x 3

(

4

x − 4 x −1 =
=

)

x → +∞

lim

2x 3 − x − 3 x 3 + 2x = x
3

x → +∞

lim

23 x + 1 − 15 x 3 33 x − 1 − 3 x

x → +∞ 3

lim
lim +

x

x + 3 x2
x+ x − x x
3

x → +∞

lim x 2 − x + 1 − x − 1 =
3

x →0

=
=

39 40 41 42 43 44 45 4647

lim
x →1

x −1

x + 1 − 3x + 1 x +3 x +4 x
5

= =

x → +∞ 6 4 x → +∞ 3

lim

x +1 − 4 x +1 x +1 + x +1
3

x → +∞

lim

lim

sin (2Arc tan (x )) = x → 0 tan (2Arc sin (x )) lim+ Arc sin x Arc sin 3 x = π 2 =

x+ x
6

lim

x +1 − 4 x x +1 − x

.12 x =
=

lim
x →1

3x 2 + 1 − 2 x + 3 − 3x − 2

x →0

lim

Arc sin (x ) −
x →1−

lim

60

1− x

22 − x 2 − 4x + 8 = x →2 x−2 Arc tan (2x 2 − x ) lim = x →0 x ⎛1⎞ lim x.Arc tan ⎜ ⎟ = x → −∞ ⎝x⎠
⎛ 3.x 2 + x − 1 ⎞ ⎟= lim Arc tan ⎜ ⎟ ⎜ x → +∞ x2 +1 ⎠ ⎝ Arc tan (sin (x )) lim = x →0 x. cos(x )

⎛1− x2 ⎞ π Arc sin ⎜ ⎟ ⎜1+ x2 ⎟ − 2 ⎠ ⎝ lim+ = 2 x → 0 Arc sin 3x − 4 x 3

(

)

61

⎛1⎞ E⎜ ⎟ + x x = lim ⎝ ⎠ x →0 + ⎛ 1 ⎞ E⎜ ⎟ − x ⎝x⎠

62

lim
x →1

Arc tan x − 1 = x −1

(

)48

‫درس : اﻹ ﺗﺼﺎل و اﻟﻨﻬﺎﻳﺎت‬ ‫اﻷﺳﺘﺎذ : ﻋﻠﻲ اﻟﺸﺮﻳﻒ .‬ ‫‪www.madariss.fr‬‬ ‫اﻟﺨﻤﻴﺴﺎت‬ ‫اﻟﺘﻤﺮﻳﻦ رﻗﻢ 4 :‬ ‫اﻟﺘﻤﺮﻳﻦ رﻗﻢ 2 :‬ ‫أدرس ﺁ ﺗﺼﺎل اﻟﺪاﻟﺔ ‪ f‬ﻓﻲ آﻞ ﺣﺎﻟﺔ ﻋﻨﺪ 0‪. x‬‬ ‫أ درس اﻟﺘﻤﺪﻳﺪ ﺑﺎﻹ ﺗﺼﺎل ﻟﻠﺪوال اﻟﺘﺎﻟﻴﺔ ﻋﻨﺪ 0 ‪. x‬‬ ‫⎧‬ ‫1− 1+ 2 ‪x‬‬ ‫‪1 − cos 2 x‬‬ ‫= )‪⎪f(x‬‬ ‫0 ≠ ‪;x‬‬ ‫1(‬ ‫= )‪x 0 = 0 ; f (x‬‬ ‫⎨ ; 0 = 0‪x‬‬ ‫1(‬ ‫2‬ ‫‪x‬‬ ‫‪3x‬‬ ‫0 = )0(‪⎪f‬‬ ‫⎩‬ ‫1‬ ‫) ‪cos( πx‬‬ ‫2(‬ ‫= 0‪x‬‬ ‫=)‪; f (x‬‬ ‫⎧‬ ‫2‬ ‫1 − ‪2x‬‬ ‫‪x‬‬ ‫0 ≠ ‪⎪f(x) = . x ; x‬‬ ‫‪1 + 3x‬‬ ‫⎨ ; 0 = 0‪x‬‬ ‫2(‬ ‫‪x‬‬ ‫3(‬ ‫2 = )‪x 0 = 0 ; f(x‬‬ ‫2 = )0(‪⎪f‬‬ ‫‪x +x‬‬ ‫⎩‬ ‫____________________________________‬ ‫⎧‬ ‫)1 − 2 ‪sin(x‬‬ ‫اﻟﺘﻤﺮﻳﻦ رﻗﻢ 5 :‬ ‫= )‪f(x‬‬ ‫[∞+,1] ∈ ‪; x‬‬ ‫⎪‬ ‫1− ‪x‬‬ ‫ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ‪ f‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ :‬ ‫⎪‬ ‫‪π‬‬ ‫2‬ ‫⎪‬ ‫5− 5+ ‪x +2 + x‬‬ ‫) ‪cos( x‬‬ ‫= ) ‪f (x‬‬ ‫⎪‬ ‫2‬ ‫= ) ‪x...