Les fonctions en 2eme bac science math
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
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( )
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