ﺗﻤﺎرﻳﻦ ﺗﻤﺎرﻳﻦ ﻣﺤﻠﻮﻟﺔ 1ﺗﻤﺮﻳﻦ
lim+ x
x
;
x →0
x −3 lim x ln x →+∞ x
; lim
e
2x
x →0
−e x
x
ﺣﺪد اﻟﺤﻞ 2x
x
e −e x −3 ﻧﺤﺪد lim x ln ; lim x →+∞ x →0 x →0 x x e2 x − 1 e x − 1 e2 x − 1 e x − 1 e2 x − e x lim = lim − − = lim 2 = 2 −1 = 1 * x →0 x →0 x x x →0 2 x x x x
lim+ x
;
x=−
3 3 أيt = − ﻧﻀﻊ x t
ln (1 + t ) x −3 3 lim x ln = lim x ln 1 − = lim −3 = −3 x →+∞ t x x→+∞ x t →0 lim+ x
x →0
x
= lim+ e x →0
x ln x
= lim+ e 2 x →0
*
وﻣﻨﻪ x ln x
=1
*
------------------------------------------------------------------------------------------------------------
(
f ( x ) = ln e
2x
x
− 3e + 3
2ﺗﻤﺮﻳﻦ
)
: اﻟﻤﻌﺮﻓﺔ ﺑـx ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ﻟﻠﻤﺘﻐﻴﺮ اﻟﺤﻘﻴﻘﻲ
D f ﻋﻨﺪ ﻣﺤﺪاتf و ﻧﻬﺎﻳﺎتD f ﺣﺪد-1 f ( x ) ≥ 0 ﺣﻞ اﻟﻤﺘﺮاﺟﺤﺔ-2 lim ( f ( x ) − 2 x) ﺣﺪد-3
(
f ( x ) = ln e 2 x − 3e x + 3
)
x →+∞
اﻟﺤﻞ
D f ﻧﺤﺪد-4 x ∈ ﻟﺘﻜﻦ x ∈ D f ⇔ e2 x − 3e x + 3 ∀x ∈
e 2 x − 3e x + 3
0 و ﺑﺎﻟﺘﺎﻟﻲ
0
∆ = −3 و ﻣﻨﻪX 2 − 3 X + 3 ﻟﻴﻜﻦ ∆ ﻣﻤﻴﺰ D f = إذن
(
)
D f ﻋﻨﺪ ﻣﺤﺪاتf * ﻧﺤﺪد ﻧﻬﺎﻳﺎت
(
)
lim f ( x ) = lim ln e 2 x − 3e x + 3 lim ln e 2 x 1 − 3e − x + 3e−2 x = +∞ x→+∞ x→+∞
x→+∞
(
)
lim f ( x ) = lim ln e 2 x − 3e x + 3 = ln 3
x→−∞
(
)
f ( x ) ≥ 0 ⇔ ln e 2 x − 3e x + 3 ≥ 0 f ( x ) ≥ 0 ⇔ e2 x − 3e x + 3 ≥ 1 f ( x ) ≥ 0 ⇔ e2 x − 3e x + 2 ≥ 0
(
)(
)
f ( x) ≥ 0 ⇔ ex − 1 ex − 2 ≥ 0
1
x→−∞
f ( x ) ≥ 0 ﻧﺤﻞ اﻟﻤﺘﺮاﺟﺤﺔ-5