‫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 x 2 . 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 ﺑﺎآﺎﻟﻮرﻳﺎ ع.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