الفصل الثالث-الدوال الحقيقية
-3ﺍﻟﺪﻭﺍﻝ ﺍﳊﻘﻴﻘﻴﺔ ﳌﺘﻐﲑ ﺣﻘﻴﻘﻲ
ﻟﻴﻜﻦ Χ
ﳎﻤﻮﻋﺔ ﻋﺪﺩﻳﺔ ،ﺃﻱ:
Χ⊂ℜ
ﺗﻌﺮﻳﻒ :ﺇﺫﺍ ﺃﳊﻘﻨﺎ ﺑﻜﻞ ﻋﺪﺩ xﻣﻦ ، Χﺑﻮﺍﺳﻄﺔ ﻗﺎﻧﻮﻥ ﻣﺎ ﻋﺪﺩﺍ yﻣﻦ ، ℜﻓﺈﻧﻨﺎ ﻧﻘﻮﻝ ﻗﺪ ﻋﺮﻓﻨﺎ ﻋﻠﻰ Χﺩﺍﻟﺔ ﺣﻘﻴﻘﻴﺔ
ﳌﺘﻐﲑ ﺣﻘﻴﻘﻲ .ﻳﺮﻣﺰ ﳍﺬﺍ ﺍﻟﻘﺎﻧﻮﻥ ﺑﺎﻟﺮﻣﺰ ، fﻭﻧﻜﺘﺐ:
x∈ Χ
y = f ( x) ,
ﺗﻨﺒﻴﻪ :
ﺃ -ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﻌﺪﺩ yﻟﻴﺲ ﻭﺣﻴﺪﺍﹰ،ﻓﺈﻥ ﺍﻟﺪﺍﻟﺔ fﺗﺴﻤﻰ ﺑﺎﻟﺪﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ.
ﺏ -ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﻌﺪﺩ yﻭﺣﻴﺪﺍﹰ ،ﻓﺈﻥ ﺍﻟﺪﺍﻟﺔ fﺗﺴﻤﻰ ﺑﺎﻟﺪﺍﻟﺔ ﻭﺣﻴﺪﺓ ﺍﻟﻘﻴﻤﺔ.
ﺝ -ﰲ ﻛﻞ ﺩﺭﺍﺳﺘﻨﺎ ﺇﺫﺍ ﻗﻠﻨﺎ ﺩﺍﻟﺔ،ﻧﻌﲏ ﺎ ﺩﺍﻟﺔ ﻭﺣﻴﺪﺓ ﺍﻟﻘﻴﻤﺔ ،ﺇﻻ ﺇﺫﺍ ﺃﺷﲑ ﺇﱃ ﻋﻜﺲ ﺫﻟﻚ.
wﺍﻤﻮﻋﺔ fﺗﺴﻤﻰ ﳎﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ ﺍﻟﺪﺍﻟﺔ ، fﻭﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ ) ، D( fﻭﺗﺴﻤﻰ ﳎﻤﻮﻋﺔ ﺍﻟﻘﻴﻢ yﺍﻟﱵ ﺗﺄﺧﺬﻫﺎ ﻫﺬﻩ
ﺍﻟﺪﺍﻟﺔ ﲟﺠﻤﻮﻋﺔ ﻗﻴﻤﻬﺎ،ﻭﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ ) . Ε( f
ﻋﻨﺪﻫﺎ ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﺗﻄﺒﻴﻖ ﻣﻦ ) D( fﻋﻠﻰ ) . Ε( f
ﺇﺫﺍ ﻛﺎﻧﺖ Ε( f ) ⊂ Υﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ
f
ﺗﻄﺒﻴﻖ ﻣﻦ ) D( fﰲ . Υ
ﺃﻣﺜﻠﺔ :
-1ﺩﺍﻟﺔ ﺩﺭﳜﻠﻲ :
D( f ) = ℜ
،
ﻻﺣﻆ ﺃﻥ:
-2ﺍﻟﺪﺍﻟﺔ ﺍﻹﺷﺎﺭﺓ :
}Ε( f ) = {0,1
0 , x ∉ Q
y = f ( x) =
1 , x ∈ Q
1 , x > 0
y = f ( x) = sgn x = 0 , x = 0
− 1 , x < 0
ﻭﻧﻘﺮﺃ yﺗﺴﺎﻭﻱ ﺇﺷﺎﺭﺓ
ﻻﺣﻆ ﺃﻥΕ( f ) = {− 1,0,+1} ، D( f ) = ℜ :
-3ﺩﺍﻟﺔ ﺍﳉﺰﺀ ﺍﻟﺼﺤﻴﺢ :
x
]y = f ( x) = [x
ﻭﻧﻘﺮﺃ yﺗﺴﺎﻭﻱ ﺍﳉﺰﺀ ﺍﻟﺼﺤﻴﺢ ﻟﻠﻌﺪﺩ
Ε( f ) = Ζ ، D ( f ) = ℜ
ﻻﺣﻆ ﺃﻥ:
x
.
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-4ﺍﻟﺪﺍﻟﺔ ﻋﺎﻣﻠﻲ :
!y = f (n ) = n
ﻻﺣﻆ ﺃﻥ، D( f ) = Ν :
Ε( f ) = Ν
-1.3ﺍﶈﺪﻭﺩﻳﺔ :
ﻟﺘﻜﻦ Aﳎﻤﻮﻋﺔ ﻏﲑ ﺧﺎﻟﻴﺔ ﻣﻦ ) ، D( fﻭ
B
ﺻﻮﺭﺓ Aﻭﻓﻖ ﺍﻟﺪﺍﻟﺔ . f
ﺗﻌﺎﺭﻳﻒ :
-1ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺳﻔﻞ ﻋﻠﻰ ﺍﻤﻮﻋﺔ، Aﺇﺫﺍ ﻭﺟﺪ ﻋﺪﺩ c1ﲝﻴﺚ ﻣﻬﻤﺎ ﻳﻜﻦ xﻣﻦ Aﻳﻜﻮﻥ
، f (x) ≥ c1ﺃﻱ :
∃c1 ∈ ℜ / ∀x ∈ A → f ( x) ≥ c1
-2ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ ﻋﻠﻰ ﺍﻤﻮﻋﺔ، Aﺇﺫﺍ ﻭﺟﺪ ﻋﺪﺩ c 2ﲝﻴﺚ ﻣﻬﻤﺎ ﻳﻜﻦ xﻣﻦ Aﻳﻜﻮﻥ
، f (x) ≤ c2ﺃﻱ:
∃c 2 ∈ ℜ / ∀x ∈ A → f ( x) ≤ c 2
-3ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Aﺇﺫﺍ ﻛﺎﻧﺖ ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺳﻔﻞ ﻭﻣﻦ ﺍﻷﻋﻠﻰ ﻋﻠﻰ ﺍﻤﻮﻋﺔ، Aﺃﻱ:
∃c > 0 / ∀x ∈ A → f ( x) ≤ c
)(1
ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﳌﺘﺮﺍﺟﺤﺔ ) (1ﳏﻘﻘﺔ ﻣﻬﻤﺎ ﺗﻜﻦ xﻣﻦ ، fﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ .
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-4ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ
f
ﳏﺪﻭﺩﺓ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Aﺇﺫﺍ ﲢﻘﻖ ﺍﻟﺸﺮﻁ )، (1ﺃﻱ :
∃c > 0 / ∀xc ∈ A → f ( xc ) > c
-5ﻳﻌﺮﻑ ﺍﳊﺪ ﺍﻷﻋﻠﻰ ﻟﻠﺪﺍﻟﺔ fﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Aﺑﺄﻧﻪ ﺍﳊﺪ ﺍﻷﻋﻠﻰ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Bﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ). sup f (x
x∈A
-6ﻳﻌﺮﻑ ﺍﳊﺪ ﺍﻷﺩﱏ ﻟﻠﺪﺍﻟﺔ
f
. inf
ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Aﺑﺄﻧﻪ ﺍﳊﺪ ﺍﻷﺩﱏ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Bﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ )f (x
x∈A
-7ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﺗﺄﺧﺬ ﻗﻴﻤﺔ ﻋﻈﻤﻰ ﰲ ﺍﻟﻘﻄﺔ x0ﻣﻦ ، Aﺇﺫﺍ ﻛﺎﻥ :
) ∀x ∈ A → f ( x) ≤ f ( x0
. f (xx ) = max
ﻧﺮﻣﺰ ﻟﻠﻘﻴﻤﺔ ﺍﻟﻌﻈﻤﻰ ﻟﻠﺪﺍﻟﺔ fﻋﻠﻰ ﺍﻤﻮﻋﺔ Aﺑﺎﻟﺮﻣﺰ ) ، max f ( xﻭﻧﻜﺘﺐ )f ( x
x∈A
x∈A
. sup f (x) = max
ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ )f ( x
x∈A
x∈A
-8ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﺗﺄﺧﺬ ﻗﻴﻤﺔ ﺻﻐﺮﻯ ﰲ ﺍﻟﻘﻄﺔ x0ﻣﻦ ، Aﺇﺫﺍ ﻛﺎﻥ:
) ∀x ∈ A → f ( x) ≥ f ( x0
. f (xx ) = min
ﻧﺮﻣﺰ ﻟﻠﻘﻴﻤﺔ ﺍﻟﺼﻐﺮﻯ ﻟﻠﺪﺍﻟﺔ fﻋﻠﻰ ﺍﻤﻮﻋﺔ Aﺑﺎﻟﺮﻣﺰ )، min f ( xﻭﻧﻜﺘﺐ )f ( x
x∈A
x∈A
. sup f (x) = min
ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ )f ( x
x∈A
x∈A
wﺍﻟﻘﻴﻤﺔ ﺍﻟﻌﻈﻤﻰ،ﻭﺍﻟﻘﻴﻤﺔ ﺍﻟﺼﻐﺮﻯ ﻳﺴﻤﻴﺎ ﻗﻴﻤﺎﹰ ﻗﺼﻮﻯ.
-2.3ﺍﻟﺮﺗﺎﺑﺔ :
ﺍﻟﺪﺍﻟﺔ
f
ﺗﺴﻤﻰ ﻋﻠﻰ ﺍﻤﻮﻋﺔ Aﻣﻦ ) : D( f
ﺃ -ﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ ،ﺇﺫﺍ ﻛﺎﻥ :
) ∀x1 ∈ A , ∀x2 ∈ A ; x1 ≤ x2 → f ( x1 ) ≤ f ( x2
ﺏ -ﻣﺘﺰﺍﻳﺪﺓ ،ﺇﺫﺍ ﻛﺎﻥ :
) ∀x1 ∈ A , ∀x2 ∈ A ; x1 < x2 → f ( x1 ) < f ( x2
ﺝ -ﻟﻴﺴﺖ ﻣﺘﺰﺍﻳﺪﺓ ،ﺇﺫﺍ ﻛﺎﻥ :
) ∀x1 ∈ A , ∀x2 ∈ A ; x1 < x2 → f ( x1 ) ≥ f ( x2
ﺩ -ﻣﺘﻨﺎﻗﺼﺔ ،ﺇﺫﺍ ﻛﺎﻥ :
) ∀x1 ∈ A , ∀x2 ∈ A ; x1 < x2 → f ( x1 ) > f ( x2
ﻩ -ﺭﺗﻴﺒﺔ ،ﺇﺫﺍ ﻛﺎﻧﺖ ﻟﻴﺴﺖ ﻣﺘﺰﺍﻳﺪﺓ ﺃﻭ ﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ.
ﻭ -ﺭﺗﻴﺒﺔ ﲤﺎﻣﺎﹰ ،ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺘﺰﺍﻳﺪﺓ ﺃﻭ ﻣﺘﻨﺎﻗﺼﺔ.
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wﻧﻔﺲ ﺍﻟﺘﻌﺎﺭﻳﻒ ﰲ ﺣﺎﻟﺔ ) . A= D( f
-3.3ﺍﻟﻨﻬﺎﻳﺎﺕ :
ﻟﺘﻜﻦ
ﺩﺍﻟﺔ ﺣﻘﻴﻘﺔ ﳌﺘﻐﲑ ﺣﻘﻴﻘﻲ ﻭ aﻋﺪﺩﺍﹰ ﻣﻦ . ℜ
f
-1.3.3ﺗﻌﺎﺭﻳﻒ :
•
-1ﻳﻌﺮﻑ − εﺟﻮﺍﺭ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ ، aﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ) ، Vε (aﺑﺄﻧﻪ ﳎﻤﻮﻋﺔ ﺍﻟﻨﻘﻂ xﺍﳌﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ :
} Vε (a ) = {x ∈ ℜ / x − a < ε , x ≠ a } = {x ∈ ℜ / 0 < x − a < ε
•
ﺃﻱ ﺃﻥ:
•
} Vε (a ) = Vε (a ) − {a
-2ﺍﻟﻨﻬﺎﻳﺔ ﺳﺤﺐ ﻛﻮﺷﻲ :ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ
aﺑﺈﻧﺸﺎﺀ ﳏﺘﻤﻞ ﻟـ ، aﺃﻭ ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ . aﻭﻣﻦ ﺃﺟﻞ ﻛﻞ (ε > 0) , εﻳﻮﺟﺪ (δ > 0) , δ
ﲝﻴﺚ ﻣﻦ ﺃﺟﻞ ﻛﻞ xﳛﻘﻖ ﺍﻟﺸﺮﻁ ، 0 < x − a < εﺗﺘﺤﻘﻖ ﺍﳌﺘﺮﺍﺟﺤﺔ . f ( x) − l < ε
ﺃﻱ :
(lim
) f ( x) = l ) ⇔ (∀ε > 0, ∃δ > 0 / (∀x ∈ D( f ) / 0 < x − a < δ ) → f ( x) − l < ε
x→ a
ﺃﻭ
(lim f (x) = l) ⇔ ∀ε > 0, ∃δ > 0 / ∀x ∈ V (a ) ⊂ D( f ) → f (x) ∈ V (l)
•
δ
ε
x→ a
-3ﺍﻟﻨﻬﺎﻳﺔ ﺣﺴﺐ ﻗﺎﻳﻦ :ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ
ﻟﻠﻨﻘﻄﺔ . aﺃﻱ :
•
) ∃δ 0 > 0 / Vδ 0 (a ) ⊂ D( f
ﻭﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ ، aﲝﻴﺚ:
•
) ∀n ≥ 1 → a n ∈ Vδ 0 (a
ﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n ≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ . l
ﻧﻈﺮﻳﺔ :ﺗﻌﺮﻳﻒ ﺍﻟﻨﻬﺎﻳﺔ ﺣﺴﺐ ﻛﻮﺷﻲ ،ﻳﻜﺎﻓﺊ ﺗﻌﺮﻳﻔﻬﺎ ﺣﺴﺐ ﻗﺎﻳﻦ.
ﺍﻟﱪﻫﺎﻥ :
]⇐[ ﻋﻨﺪﻧﺎ
lim f ( x) = l
x→ a
،ﺣﺴﺐ ﻛﻮﺷﻲ ﺗﻌﲏ :
•
) ∃δ 0 > 0 / Vδ 0 (a ) ⊂ D( f
)*(
ﻟﺘﻜﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ
•
) ∀ε > 0, ∃δ ∈ ]0, δ 0 ] / ∀x ∈ Vδ (a ) → f ( x) ∈ Vε (l
(a n )n≥1ﻣﺘﺘﺎﻟﻴﺔ ﻛﻴﻔﻴﺔ ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻨﻘﻄﺔ ، aﺣﻴﺚ:
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•
) ∀n ≥ 1 → a n ∈ Vδ 0 (a
)
•
= a ⇔ ∀ε > 0 ∃nε ∈ N / ∀n ≥ nε → a n ∈ Vε (a )
n
(lim a
∞→ n
•
ﻻﺣﻆ ﻣﻦ ﺃﺟﻞ ε = δﺣﻴﺚ δﻣﻌﺮﻓﺔ ﰲ )*( ﻳﻮﺟﺪ ، nδﲝﻴﺚ ﻣﻦ ﺃﺟﻞ ﻛﻞ n ≥ nδﻳﻜﻮﻥ ) . a n ∈Vε (a
ﻫﺬﺍ ﻳﻌﲏ ﺣﺴﺐ )*( ﺃﻥ ) ، f (a n ) ∈Vε (lﻭﻣﻨﻪ :
) ∀ε > 0 , ∃nε = nδ ∈ Ν / ∀n ≥ nε → f (a n ) ∈ Vε (l
ﺃﻱ
lim f (a n ) = l
∞→ n
]⇒[ lim f ( x) = l
x→ a
،ﻭﺑﺎﻟﺘﺎﱄ ﺍﻟﻌﺪﺩ lﻫﻮ ﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻘﻄﺔ aﺣﺴﺐ ﻗﺎﻳﻦ.
ﺣﺴﺐ ﻗﺎﻳﻦ.
ﻧﻔﺮﺽ ﺃﻥ lﻟﻴﺴﺖ ﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻘﻄﺔ aﺣﺴﺐ ﻛﻮﺷﻲ ،ﺃﻱ ﺃﻥ ﺍﻟﺼﻴﻐﺔ )*( ﻏﲑ ﺻﺤﻴﺤﺔ،ﺃﻱ:
•
∃ε 0 > 0 , ∀δ ∈ ]0, δ 0 ] , ∃xδ ∈ V δ (a ) / f ( xδ ) − l ≥ ε 0
)∗ ∗(
ﻻﺣﻆ ﺍﻟﺼﻴﻐﺔ )∗ ∗( ﺻﺤﻴﺤﺔ ﻣﻦ ﺃﺟﻞ
δ0
n
= δ
، n ≥ 1,ﺃﻱ ﻳﻮﺟﺪ )، a δﻧﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ
0
•
،( a nﻣﻦ ) Vδ 0 (a
n
n
ﻣﻦ ﺃﺟﻠﻪ ﻳﺘﺤﻘﻖ:
)∗ ∗ ∗(
f (a n ) − l ≥ ε 0
ﻭﺑﺎﻟﺘﺎﱄ ﻣﻦ ﺃﺟﻞ ﻛﻞ (n ≥ 1 ), nﻳﻜﻮﻥ :
•
•
) a n ∈ Vδ 0 (a ) ⊂ Vδ 0 (a
ﻭ
n
ﺍﻟﱵ
ﺗﻌﲏ lim a n = a
∞→ n
δ0
n
< 0 < an − a
.
ﻟﻜﻦ ﺍﳌﺘﺮﺍﺟﺤﺔ f (a n ) − l ≥ ε 0ﺗﻌﲏ ﺃﻥ lﻟﻴﺲ ﺎﻳﺔ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ ، ( f (a n ))n≥1ﻭﻫﺬﺍ ﻣﻨﺎﰲ ﻟﻜﻮﻥ lﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ
ﺍﻟﻘﻄﺔ aﺣﺴﺐ ﻗﺎﻳﻦ ﻭﺑﺎﻟﺘﺎﱄ ﺍﻟﻔﺮﺽ ﻏﲑ ﺻﺤﻴﺒﺢ.
ﻣﺜﺎﻝ :
ﺃ -ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ . f (x) = x
ﻧﺘﺄﻛﺪ ﺣﺴﺐ ﻗﺎﻳﻦ ﺃﻥlim f ( x) = a :
x→ a
. a ∈ ℜ,
fﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻛﻞ ℜﻭﺑﺎﻟﺘﺎﱄ ﻫﻲ ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ ، aﻻﺣﻆ ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ﻫﺬﺍ
limﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﻧﻔﺲ ﺍﻟﻨﻘﻄﺔ ، aﻷﻥ ﻣﻦ ﺃﺟﻞ ﻛﻞ (n ≥ 1 ), n
ﺍﳉﻮﺍﺭ ﺣﻴﺚ a n = a
∞→ n
. lim
ﻋﻨﺪﻧﺎ ، f (a n ) = a nﻭﺑﺎﻟﺘﺎﱄf ( x) = a :
x→ a
ﺏ -ﺍﻟﺪﺍﻟﺔ
1
x
f ( x) = sin
ﻟﻴﺴﺖ ﳍﺎ ﺎﻳﺔ ﰲ ﺍﻟﺼﻔﺮ.
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ﻧﺄﺧﺬ ﺍﳌﺘﺘﺎﻟﻴﺘﲔ ، (bn )n≥1 ، (a n )n≥1ﺣﻴﺚ :
1
nπ
ﻻﺣﻆ ﺃﻥ
2
(4n − 3)π
= , an
= ∀n ≥ 1 , bn
lim bn = lim a n = 0
∞→ n
∞→ n
، limﺫﻟﻚ ﻷﻥ:
ﺑﻴﻨﻤﺎ ) f (a n ) ≠ lim f (bn
∞→n
∞→n
1
f (a n ) = f
= sin π = 0 ⇒ lim f (a n ) = 0
∞→ n
nπ
1
4n − 3
= sin
f (bn ) = f
π = 1 ⇒ lim f (bn ) = 1
∞→ n
2
(4n − 3)π
ﻭ
ﻭﻣﻨﻪ ﺍﻟﺪﺍﻟﺔ
f
ﻟﻴﺴﺖ ﳍﺎ ﺎﻳﺔ ﰲ ﺍﻟﻨﻘﻄﺔ . x = 0
ﻧﺘﻴﺠﺔ :
ﺃ -ﺍﻟﻨﻘﻄﺔ aﻗﺪ ﻻ ﺗﻨﺘﻤﻲ ﺇﱃ ﳎﺎﻝ ﺗﻌﺮﻳﻒ ﺍﻟﺪﺍﻟﺔ.
ﺏ -ﺎﻳﺔ ﺍﻟﺪﺍﻟﺔ fﺇﺫﺍ ﻭﺟﺪﺕ ﺗﻜﻮﻥ ﻭﺣﻴﺪﺓ.
-2.3.3ﺃﻧﻮﺍﻉ ﺍﻟﻨﻬﺎﻳﺎﺕ :
.1ﺍﻟﻨﻬﺎﻳﺔ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭﺍﻟﻨﻬﺎﻳﺔ ﻣﻦ ﺍﻟﻴﺴﺎﺭ :
•
•
ﻣﻦ ﺃﺟﻞ ﻛﻞ (ε > 0), εﻳﺮﻣﺰ ﺇﱃ ﺍﻟﻨﺼﻒ ﺍﻷﳝﻦ ﻟﻠﺠﻮﺍﺭ ) Vε (aﺑﺎﻟﺮﻣﺰ ) ، Vε (a + 0ﻭﺇﱃ ﻧﺼﻔﻪ ﺍﻷﻳﺴﺮ ﺑﺎﻟﺮﻣﺰ
•
)، Vε (a − 0ﺃﻱ :
•
[ Vε (a − 0) = ]a − ε , a
،
•
[ Vε (a + 0) = ]a , a + ε
ﺗﻌﺎﺭﻳﻒ :
ﺃ -ﺣﺴﺐ ﻛﻮﺷﻲ :
-1ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﻣﻦ ﺍﻟﻴﺴﺎﺭ )،ﻳﺴﺮﻯ( ،ﻟﻠﺪﺍﻟﺔ
f
ﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻥ :
∀ε > 0 , ∃δ > 0 / ∀x ∈ ]a − δ , a [ ⊂ D( f ) → f ( x) − l < ε
-2ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﻣﻦ ﺍﻟﻴﻤﲔ )،ﳝﲎ ( ﻟﻠﺪﺍﻟﺔ
f
ﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻥ :
∀ε > 0 , ∃δ > 0 / ∀x ∈ ]a , a + δ [ ⊂ D( f ) → f ( x) − l < ε
ﺏ -ﺣﺴﺐ ﻗﺎﻳﻦ :
-1ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﻳﺴﺮﻯ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻭﺟﺪ (δ ≥ 0) , δﲝﻴﺚ ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ fﻣﻌﺮﻓﺔ ﰲ
•
)، Vδ (a − 0ﻭﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ) Vδ (a − 0ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ aﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n≥1
•
ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ . l
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-2ﺍﻟﻌﺪﺩ lﻳﺴﻤﻰ ﺎﻳﺔ ﳝﲎ ﻟﻠﺪﺍﻟﺔ
ﰲ ﺍﻟﻘﻄﺔ aﺇﺫﺍ ﻭﺟﺪ (δ ≥ 0) , δﲝﻴﺚ ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ
f
f
ﻣﻌﺮﻓﺔ ﰲ
) ، Vδ (a + 0ﻭﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ) Vδ (a + 0ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ aﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n≥1ﻣﺘﻘﺎﺭﺑﺔ
ﳓﻮ ﺍﻟﻌﺪﺩ . l
ﻳﺮﻣﺰ ﻟﻠﻨﻬﺎﻳﺔ ﺍﻟﻴﺴﺮﻯ ﺑﺎﻟﺮﻣﺰ ) lim f ( xﺃﻭ ) ، f (a − 0ﻭﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻴﻤﲎ ﺑﺎﻟﺮﻣﺰ ) lim f (xﺃﻭ ) f (a + 0
x→ a + 0
x→ a − 0
•
•
ﻧﺘﻴﺠﺔ :
-1ﺃ ⇔ 1-ﺏ1-
،ﻭ
ﺃ ⇔ 2-ﺏ2-
.
-2ﺗﻜﻮﻥ ﻟﻠﺪﺍﻟﺔ fﺎﻳﺔ lﰲ ﺍﻟﻘﻄﺔ ، aﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻭﺟﺪﺕ ﻟﻠﺪﺍﻟﺔ fﺎﻳﺔ ﳝﲎ ﻭﺎﻳﺔ ﻳﺴﺮﻯ ﰲ ﺍﻟﻘﻄﺔ aﻣﺘﺴﺎﻭﻳﺘﺎﻥ،
. f (a − 0) = lim
ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥf ( x) = f (a + 0 ) :
x→ a
ﻣﻼﺣﻈﺔ :ﰲ ﺣﺎﻟﺔ " " a = 0ﺗﻜﺘﺐ ﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻴﺴﺮﻯ ﻭﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻴﻤﲎ ﻋﻠﻰ ﺍﻟﺸﻜﻞ :
)lim f ( x
x→ −0
ﺃﻭ )، f (− 0ﻭ
)lim f ( x
x→ +0
ﺃﻭ ) f (+ 0ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ .
ﻣﺜﺎﻝ :ﻟﺘﻜﻦ fﺩﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻙ ﺍﻟﺘﺎﱄ:
x + 2 , x > 0
f ( x) = 3
, x≤ 0
x
ﻧﺒﺤﺚ ﻋﻦ ﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻴﻤﲎ ﻭﺍﻟﻴﺴﺮﻯ ﻟﻠﺪﺍﻟﺔ ﰲ ﺍﻟﺼﻔﺮ
ﻻﺣﻆ ﺃﻥ:
f (+ 0) = lim ( x + 2 ) = 2
x→ +0
f (− 0) = lim x 3 = 0
x→ −0
ﻭﻣﻨﻪ ) ، f (+ 0) ≠ f (− 0ﺃﻱ ﺃﻥ ﺍﻟﺪﺍﻟﺔ
f
ﻟﻴﺲ ﳍﺎ ﺎﻳﺔ ﰲ ﺍﻟﻨﻘﻄﺔ . x = 0
ﺗﻨﺒﻴﻪ :ﳝﻜﻦ ﻧﻘﻞ ﻣﻔﻬﻮﻡ ﺍﻟﻴﻤﲔ ﻭﺍﻟﻴﺴﺎﺭ ﺇﱃ ﳎﻤﻮﻋﺔ ﻗﻴﻢ ﺍﻟﺪﺍﻟﺔ fﻛﺎﻟﺘﺎﱄ :
(lim f (x) = l + 0) ⇔ ∀ε > 0, ∃δ > 0 / ∀x ∈ V (a ) ⊂ D( f ) → f (x) ∈ [l, l + ε [ = V (l + 0)
(lim f (x) = l − 0) ⇔ ∀ε > 0, ∃δ > 0 / ∀x ∈ V (a ) ⊂ D( f ) → f (x) ∈ ]l − ε , l] = V (l − 0)
•
δ
ε
x→ a
•
δ
ε
x→ a
ﻣﺜﺎﻝ :ﻟﺘﻜﻦ fﺩﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ:
, x< 0
, x=0
x>0
ﻻﺣﻆ ﺃﻥ
,
1 − x
f ( x) = 2
1 + x
lim f ( x) = 1 + 0
x→0
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-2ﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻼﻣﺘﻨﻬﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﻧﻘﻄﺔ ﺛﺎﺑﺘﺔ :
ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ
f
ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ . a
ﺗﻌﺎﺭﻳﻒ :
ﺃ -ﺣﺴﺐ ﻛﻮﺷﻲ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ
f
، limﺇﺫﺍ ﻛﺎﻥ
ﳍﺎ ﺎﻳﺔ ﻏﲑ ﻣﻨﺘﻬﻴﺔ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﻭﻧﻜﺘﺐ ∞ = )f ( x
x→ a
•
∀ε > 0, ∃δ > 0 / ∀x ∈ Vδ (a ) ⊂ D( f ) → f ( x) > ε
•
) ∞( ∀ε > 0, ∃δ > 0 / ∀x ∈ Vδ (a ) ⊂ D( f ) → f ( x) ∈ Vε
ﺃﻱ:
ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ
•
> 0 / ∀x ∈ Vδ1 (a ) ⊂ D( f ) → f ( x) ∈ Vε (+ ∞ )
•
> 0 / ∀x ∈ Vδ 2 (a ) ⊂ D( f ) → f ( x) ∈ Vε (− ∞ )
(lim f (x) = +∞) ⇔ ∀ε > 0 , ∃δ
(lim f (x) = −∞) ⇔ ∀ε > 0 , ∃δ
1
x→ a
2
x→ a
ﺣﻴﺚ ) ∞( Vε (+ ∞ ) , Vε (− ∞ ) , Vεﻫﻲ − εﺟﻮﺍﺭ ﻟـ ∞ + ∞ , − ∞ ,ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ .ﺍﻧﻈﺮ ﻓﻘﺮﺓ . 2.4.2
، limﺇﺫﺍ ﻛﺎﻧﺖ ﻣﻦ ﺃﺟﻞ
ﺏ -ﺣﺴﺐ ﻗﺎﻳﻦ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﳍﺎ ﺎﻳﺔ ﻏﲑ ﻣﻨﺘﻬﻴﺔ ﰲ ﺍﻟﻨﻘﻄﺔ aﻭﻧﻜﺘﺐ ∞ = )f ( x
x→ a
، limﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n≥1ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ
ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ﺟﻮﺍﺭ ﺍﻟﺘﻌﺮﻳﻒ ﺍﳌﺜﻘﻮﺏ ﺣﻴﺚ a n = a
∞→ n
. lim
ﺍﻟﻜﱪ ،ﺃﻱf (a n ) = ∞ :
∞→ n
ﻧﺘﻴﺠﺔ :
-1ﺃ ⇔ ﺏ
-2ﻛﻤﺎ ﰲ ﺍﻟﺘﻨﺒﻴﻪ ﺍﻟﺴﺎﺑﻖ ،ﳝﻜﻦ ﺃﻥ ﻧﻌﺮﻑ ﺍﻟﻨﻬﺎﻳﺔ ﻏﲑ ﻣﻨﺘﻬﻴﺔ ﻟﻠﺪﺍﻟﺔ fﻋﻠﻰ ﻳﺴﺎﺭ ﺍﻟﻨﻘﻄﺔ aﻭﻋﻠﻰ ﳝﻴﻨﻬﺎ،ﺍﻟﱵ ﻧﻜﺘﺒﻬﺎ
xlimﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ
ﻋﻠﻰ ﺍﻟﺸﻜﻞ lim f (x) = ∞ :ﻭ ∞ = )f (x
→a + 0
x→ a − 0
ﺗﻨﺒﻴﻪ :ﺍﻟﺪﻭﺍﻝ ﺍﻟﱵ ﺎﻳﺘﻬﺎ ﻣﺎ ﻻﺎﻳﺔ ﺗﺴﻤﻰ ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﻜﱪ،ﻭﺍﻟﺪﻭﺍﻝ ﺍﻟﱵ ﺎﻳﺘﻬﺎ ﺍﻟﺼﻔﺮ ﺗﺴﻤﻰ ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ
ﻣﺜﺎﻝ :
f ( x) = lg x2
lim f ( x) = −∞ ,
= )f ( x
lim f ( x) = +∞ ,
-3ﺍﻟﻨﻬﺎﻳﺔ ﰲ ﻣﺎ ﻻ ﺎﻳﺔ :
ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ
f
1
x4
x→0
x→ 0
ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻤﺎ ﻻ ﺎﻳﺔ ،ﺃﻱ :
) ∃δ > 0 / Vδ (∞ ) ⊂ D( f
ﺗﻨﺒﻴﻪ ﺃﻥ
•
) ∞( Vδ (∞ ) = Vδ
.
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ﺗﻌﺎﺭﻳﻒ :
ﺃ -ﺣﺴﺐ ﻛﻮﺷﻲ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﻌﺪﺩ lﻫﻮ ﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ
، limﺇﺫﺍ ﻛﺎﻥ :
ﰲ ﺍﳌﺎ ﻻ ﺎﻳﺔ ﻭﻧﻜﺘﺐ f ( x) = l
∞→x
f
) ∀ε > 0، ∃δ > 0 / ∀x ∈ Vδ (∞ ) ⊂ D( f ) → f ( x) ∈ Vε (l
ﺃﻱ :
)
)) (− ∞ ) ⊂ D( f ) → f (x) ∈ V (l
) > 0 / ∀x ∈ Vδ1 (+ ∞ ) ⊂ D( f ) → f ( x) ∈ Vε (l
1
> 0 / ∀x ∈ Vδ 2
2
ε
( lim f (x) = 0) ⇔ (∀ε > 0، ∃δ
( lim f (x) = 0) ⇔ (∀ε > 0 ، ∃δ
∞x→ +
∞x→ −
ﺏ -ﺣﺴﺐ ﻗﺎﻳﻦ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﻌﺪﺩ lﺎﻳﺔ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﳌﺎ ﻻ ﺎﻳﺔ ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ
) ∞( ، Vδﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﻜﱪ ﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ . l
) ﰲ ﺣﺎﻟﺔ ∞ x → +ﺣﺪﻭﺩ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺗﻜﻮﻥ ﰲ ) ∞ ، Vδ (+ﻭﰲ ﺣﺎﻟﺔ ∞ x → −ﺣﺪﻭﺩ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺗﻜﻮﻥ ﰲ ) ∞ ( Vδ (−
ﻧﺘﻴﺠﺔ :ﺃ ⇔ ﺏ
ﻣﺜﺎﻝ :
3 − 2x
x +1
= )f ( x
3 − 2x
= +2 + 0
x +1
-4ﺍﻟﻨﻬﺎﻳﺔ ﺍﻟﻼﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﳌﺎ ﻻ ﺎﻳﺔ :
3 − 2x
= −2 − 0 ,
x +1
lim
∞x→ +
lim
∞x→ −
∞ = )lim f ( x
∞→x
ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ﺗﻜﺘﺐ ﻛﺎﻟﺘﺎﱄ :
) ∞( ∀ε > 0، ∃δ > 0 / ∀x ∈ Vδ (∞ ) ⊂ D( f ) → f ( x) ∈ Vε
ﺍﻟﺘﻔﺎﺻﻴﻞ ﺗﺴﺘﻨﺘﺞ ﻣﺒﺎﺷﺮﺓ ﻣﻦ ﺃﻧﻮﺍﻉ ﺍﻟﻨﻬﺎﻳﺎﺕ ﺍﻟﺴﺎﺑﻘﺔ.
-3.3.3ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳊﺴﺎﺑﻴﺔ ﻋﻠﻰ ﺍﻟﻨﻬﺎﻳﺎﺕ ﻭﺧﺼﺎﺋﺺ ﺍﻟﻨﻬﺎﻳﺎﺕ :
-1ﺇﺫﺍ ﻛﺎﻧﺖ
ﺃ-
lim g ( x) = l 2 , lim f ( x) = l 1
x→ a
lim( f ( x) ± g ( x)) = l 1 ± l 2
x→ a
lim f ( x) ⋅ g ( x) = l 1 ⋅ l 2
ﺏ-
ﺝ-
x→ a
،ﻓﺈﻥ :
l2 ≠ 0
،
f ( x) l 1
=
g ( x) l 2
x→ a
lim
x→ a
-2ﳎﻤﻮﻉ ﻭﻓﺮﻕ ﺩﺍﻟﺘﲔ ﻻ ﻣﺘﻨﺎﻫﻴﺘﲔ ﰲ ﺍﻟﺼﻐﺮ،ﻫﻮ ﺩﺍﻟﺔ ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ.
-3ﺟﺪﺍﺀ ﺩﺍﻟﺔ ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ،ﻭﺩﺍﻟﺔ ﳏﺪﻭﺩﺓ ،ﻫﻮ ﺩﺍﻟﺔ ﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ.
، limﻓﺈﻧﻪ :
-4ﺇﺫﺍ ﻛﺎﻥ f (x) = l
x→ a
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∃δ > 0 / ∀x ∈ Vδ (a ) → f ( x) ≤ c
cﺛﺎﺑﺖ
، limﻓﺈﻧﻪ:
-5ﺇﺫﺍ ﻛﺎﻥ f ( x) = l ≠ 0
x→ a
∃δ 1 > 0 / ∀x ∈ Vδ1 (a ) → f ( x)l > c
ﺃ-
cﺛﺎﺑﺖ
ﺏ-
-6ﺇﺫﺍ ﻭﺟﺪ
>0
1
≤c
)f ( x
→ ) ∃δ 2 > 0 / ∀x ∈ Vδ 2 (a
، δﲝﻴﺚ ﻣﻦ ﺃﺟﻞ ﻛﻞ xﻣﻦ ) ، Vδ (aﻳﻜﻮﻥ:
)g ( x) ≤ f ( x) ≤ h( x
)g ( x) ≤ f ( x) ≤ h( x
ﻓﺈﻥ:
lim f ( x) = l
x→ a
-7ﺇﺫﺍ
ﻭﺟﺪ > 0
ﻭ
lim g ( x) = lim h( x) = l
x→ a
x→ a
.
، δﲝﻴﺚ ﻣﻦ ﺃﺟﻞ ﻛﻞ xﻣﻦ ) ، Vδ (aﻳﻜﻮﻥ:
) f ( x) ≤ g ( xﻭ , lim f ( x) = l 1
x→ a
*
lim g ( x) = l 2
x→ a
ﻓﺈﻥ. l 1 ≤ l 2 :
ﻣﻼﺣﻈﺔ :ﺍﳋﺼﺎﺋﺺ 7 ← 1ﺻﺤﻴﺤﺔ ﺃﻳﻀﺎﹰ ﰲ ﺣﺎﻟﺔ:
. x → +∞ , x → −∞ , x → a + 0 , x → a − 0
ﻣﺜﺎﻝ :
sin x
=1
x
-1
lim
x→ 0
ﻟﱪﻫﺎﻥ ﺍﻟﻨﻬﺎﻳﺔ ﻧﱪﻫﻦ ﺻﺤﺔ ﺍﳌﺘﺮﺍﺟﺤﺔ
)(1
ﺣﻴﺚ :
sin x
<1
x
< cos x
π π
0 ≠ x ∈ − ,
2 2
ﻧﺄﺧﺬ ﺩﺍﺋﺮﺓ ﺍﻟﻮﺣﺪﺓ ،ﺃﻱ. R = 1 :
>
>
>
ﻣﻦ ﺍﻟﺸﻜﻞ ﻧﻼﺣﻆ ﺃﻥ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ AOBﺃﻗﻞ ﻣﻦ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ AOBﺍﻟﱵ ﺃﻗﻞ ﻣﻦ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ ، AODﺃﻱ:
ﲟﺎﺃﻥ
)*(
1
(OA)2 sin x < 1 (OA)2 x < 1 OA. AD
2
2
2
, OA = 1 , sin x = AB
ﳛﺼﺮﻫﺎ OB
، tgx = ADﺣﻴﺚ xﻗﻴﺎﺱ ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﱵ
ﻣﻊ
OA
ﺑﺮﺍﺩﻳﺎﻥ،ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ :
1
1
1
sin x < x < tgx
2
2
2
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: ﻭﻣﻨﻪ ﻳﻜﻮﻥ
sin x > 0 ، sin x < x < tg x
1<
x
1
<
sin x cos x
ﻭﺑﺎﻟﺘﺎﱄ ﳓﺼﻞ ﻋﻠﻰ ﺍﳌﺘﺮﺍﺟﺤﺔ
. cos x <
sin x
<1
x
:ﺃﻱ
sin x
π π
x ≠ 0 , − , ﻣﻦx ﳏﻘﻘﺔ ﻣﻦ ﺃﺟﻞ ﻛﻞ-1- ﻓﺈﻥ ﺍﳌﺘﺮﺍﺟﺤﺔ،ﺯﻭﺟﻴﺔ
ﻭcos x ﲟﺎ ﺃﻥ ﺍﻟﺪﺍﻟﺔ
x
2 2
: ﻳﻜﻮﻥ، ﳊﺴﺎﺏ ﺍﻟﻨﻬﺎﻳﺔ ﺍﳌﻄﻠﻮﺑﺔ-1- ﻧﻄﺒﻴﻖ ﺍﳌﺘﺮﺍﺟﺤﺔ
lim cos x < lim
x→ 0
x→ 0
sin x
<1
x
, 1 < lim
x→ 0
sin x
<1
x
sin x
=1
x→ 0
x
x
1
-2
lim
1 − = e
x→ ∞
x
n
1
lim 1 + = e ﻣﻌﻠﻮﻡ ﺃﻥ
n → +∞
n
, lim
x∈ℜ
n∈Ν
1
1 1
< <
n +1 x n
: ﺃﻱ، n ≤ x ≤ n + 1 ﻳﻜﻮﻥn = [x] ﺑﻮﺿﻊ. x > 1 ﻓﺈﻥ، x → +∞ ﲟﺎﺃﻥ-ﺃ
ﻭﺍﺿﺢ ﺃﻥ ﺍﳌﺘﺮﺍﺟﺤﺔ
n
x
1 1 1
1 +
< 1 + < 1 +
n + 1 x n
n +1
. ﺻﺤﻴﺤﺔ
(x → +∞ ) ⇔ (n → +∞ ) ﻭﺍﺿﺢ ﺃﻳﻀﺎﹰ ﺃﻥ
ﻋﻨﺪﻧﺎ
1
lim
1 +
n →∞
n
n +1
n
1
1
= lim
1
+
lim
1 + = e ⋅ 1 = e
n →∞
n n →∞ n
n +1
1
lim1 +
n
n →∞
1
e
n +1
lim
1
+
=
=
=e
n→∞
1
1
n + 1
lim
1 +
n→∞
n + 1
ﻭ
ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺍﳊﺼﺮ ﻳﻜﻮﻥ
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x
1
lim
1 + = e
∞x→ +
x
ﺏ -ﲟﺎﺃﻥ ∞، x → −ﻓﺈﻥ . x < −1ﺑﻮﺿﻊ y = − xﳓﺼﻞ ﻋﻠﻰ:
−y
y
−y
y −1
y
= lim
=
= lim
∞y → +
∞y → +
y
y −1
y −1
1
1
lim 1 +
= e.1 = e
= lim 1 +
∞y → +
y − 1 y→ +∞
y − 1
ﺃﻱ ﺃﻥ:
1
1
lim 1 + = lim 1 −
∞x→ −
∞y → +
y
x
x
y
1
= lim 1 +
∞y → +
y − 1
x
1
lim 1 + = e
∞x→ −
x
-4.3.3ﺎﻳﺔ ﺍﻟﺪﻭﺍﻝ ﺍﻟﺮﺗﻴﺒﺔ :
ﻧﻈﺮﻳﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﻌﺮﻓﺔ ﻭﺭﺗﻴﺒﺔ ﻋﻠﻰ ﺍﺎﻝ ]، [a, bﻓﺈﺎ ﰲ ﻛﻞ ﻧﻘﻄﺔ x0ﻣﻦ ﺍﺎﻝ [ ]a, bﲤﻠﻚ ﺎﻳﺔ
ﻳﺴﺮﻯ ﻭﺎﻳﺔ ﳝﲎ ﻣﻨﺘﻬﻴﺔ،ﻭﰲ ﺍﻟﻨﻘﻄﺔ b, aﺎﻳﺔ ﳝﲎ ،ﻳﺴﺮﻯ ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ.
ﺍﻟﱪﻫﺎﻥ :ﻧﱪﻫﻦ
-1ﻧﱪﻫﻦ ﺃﻥ ﻟﻠﺪﺍﻟﺔ fﺎﻳﺔ ﳝﲎ ﰲ ﻛﻞ ﻧﻘﻄﺔ ، x0ﺣﻴﺚ . a ≤ x0 < b
-2ﻧﱪﻫﻦ ﺃﻥ ﻟﻠﺪﺍﻟﺔ fﺎﻳﺔ ﻳﺴﺮﻯ ﰲ ﻛﻞ ﻧﻘﻄﺔ ، x0ﺣﻴﺚ . a < x0 ≤ b
ﻧﺄﺧﺬ ﺍﳊﺎﻟﺔ ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ fﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ )،ﺣﺎﻟﺔ ﻟﻴﺴﺖ ﻣﺘﺰﺍﻳﺪﺓ ﺗﱪﻫﻦ ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ (.
-1ﻋﻨﺪﻧﺎ:
)∀x ∈ ]x0 , b] → f ( x0 ) ≤ f ( x
ﻧﻼﺣﻆ ﺃﻥ ﳎﻤﻮﻋﺔ ﻗﻴﻢ ﺍﻟﺪﺍﻟﺔ fﻋﻠﻰ ﺍﺎﻝ ] ]x0 , bﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﺳﻔﻞ ،ﺃﻱ:
∃m = inf
) f ( x) ≥ f ( x0
] ]
x∈ x0 ,b
ﻣﻦ ﺧﺼﺎﺋﺺ ﺍﻟﻌﺪﺩ m
∀x ∈ ]x0 , b] → f ( x) ≥ m > m − ε
w
) , ∃xε ∈ ]x0 , b] / m + ε > f ( xε
w
ﻧﺮﻣﺰ ﻟﻠﻔﺮﻕ xε − x0ﺑﺎﻟﺮﻣﺰ ، δﻭﺍﺿﺢ ﺃﻥ . δ > 0
∀ε > 0
,
∀ε > 0
)(1
)(2
ﻻﺣﻆ
) ∀x ∈ ]x0 , xε [ = ]x0 , x.0 + δ [ → f ( x) ≤ f ( xε
)(3
ﻣﻦ ) (3), (2), (1ﻳﻜﻮﻥ :
51
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•
∀ε > 0, ∃δ > 0 / ∀x ∈ ]x0 , x.0 + δ [ = Vδ ( x0 + 0 ) → f ( x) − m < ε
f ( x0 + 0 ) = lim f ( x) = m
ﺃﻱ ﺃﻥ
x→ x0 + 0
-2ﻋﻨﺪﻧﺎ:
) ∀x ∈ [a , x0 [ → f ( x) ≤ f ( x0
ﻧﻼﺣﻆ ﺃﻥ ﳎﻤﻮﻋﺔ ﻗﻴﻢ ﺍﻟﺪﺍﻟﺔ fﻋﻠﻰ ﺍﺎﻝ [ [a , x0ﳏﺪﻭﺩﺓ ﻣﻦ ﺍﻷﻋﻠﻰ ،ﺃﻱ :
) ∃M = sup f ( x) ≤ f (x0
[ x∈[a , x0
ﻣﻦ ﺧﺼﺎﺋﺺ ﺍﻟﻌﺪﺩ : M
)(1
∀x ∈ [a , x0 [ → f ( x) ≤ M < M + ε ,
∀ε > 0
w
∀ε > 0 , ∃xε ∈ [a , x0 [ / f ( xε ) > M − ε
w
ﻧﺮﻣﺰ ﻟﻠﻔﺮﻕ x0 − xεﺑﺎﻟﺮﻣﺰ ، δﻭﺍﺿﺢ ﺃﻥ . 0 < δ
)(3
) ∀x ∈ ]xε , x0 [ = ]x0 − δ , x.0 [ → f ( x) ≥ f ( xε
ﻻﺣﻆ
ﻣﻦ ) (3), (2), (1ﻳﻜﻮﻥ :
)(2
•
∀ε > 0, ∃δ > 0 / ∀x ∈ ]x0 − δ , x.0 [ = Vδ ( x0 + 0) → f ( x) − M < ε
f ( x0 − 0 ) = lim f ( x) = M
ﺃﻱ ﺃﻥ:
x→ x0 − 0
ﻧﺘﻴﺠﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ
-1ﺇﺫﺍ ﻛﺎﻧﺖ fﻣﻌﺮﻓﺔ ﻭﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ] ، [a, bﻓﺈﻥ :
)∀x ∈ [a , b] → f ( x − 0 ) ≤ f ( x) ≤ f ( x + 0
-2ﺇﺫﺍ ﻛﺎﻧﺖ fﻣﻌﺮﻓﺔ ﻭﻟﻴﺴﺖ ﻣﺘﺰﺍﻳﺪﺓ ﻋﻠﻰ ] ، [a, bﻓﺈﻥ :
)∀x ∈ [a , b] → f ( x + 0 ) ≤ f (x) ≤ f ( x − 0
-3ﺍﻟﻨﻈﺮﻳﺔ ﺻﺤﻴﺤﺔ ﰲ ﺣﺎﻟﺔ ﺍﺎﻝ ] [a, bﻛﻴﻔﻲ ،ﺃﻱ ﻣﻔﺘﻮﺡ ﺃﻭ ﻏﲑ ﻣﻨﺘﻪ .
-5.3.3ﻣﻌﻴﺎﺭ ﻛﻮﺷﻲ ﻟﻨﻬﺎﻳﺎﺕ ﺍﻟﺪﻭﺍﻝ :
ﺗﻌﺮﻳﻒ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﲢﻘﻖ ﺷﺮﻁ ﻛﻮﺷﻲ ﰲ ﺍﻟﻨﻘﻄﺔ،ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ aﻭﲢﻘﻖ ﺍﻟﺸﺮﻁ:
•
∀ε > 0, ∃δ > 0 / ∀x′, x′′ ∈ Vδ (a ) → f ( x′) − f ( x′′) < ε
ﻧﻈﺮﻳﺔ :ﺗﻜﻮﻥ ﻟﻠﺪﺍﻟﺔ fﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﲢﻘﻖ ﺷﺮﻁ ﻛﻮﺷﻲ ﰲ ﺗﻠﻚ ﺍﻟﻨﻘﻄﺔ.
، limﺃﻱ :
ﺍﻟﱪﻫﺎﻥ [⇐] :ﻧﻔﺮﺽ ﺃﻥ f ( x) = l
x→ a
−1−
ε
2
•
< ∀ε > 0, ∃δ > 0 / ∀x ∈ Vδ (a ) ⊂ D( f ) → f (x) − l
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•
ﺇﺫﺍ ﻛﺎﻧﺖ x′′ , x′ﻧﻘﻄﺘﲔ ﻛﻴﻔﻴﺘﲔ ﻣﻦ ) ،Vδ (aﻓﺈﻧﻪ ﻣﻦ −1 −ﻳﻨﺘﺞ
ε ε
+ =ε
2 2
< f ( x′) − f ( x′′) = ( f ( x′) − l ) − ( f ( x′′) − l ) ≤ f ( x′) − l + f ( x′′) − l
ﻭﻣﻨﻪ ﺷﺮﻁ ﻛﻮﺷﻲ ﳏﻘﻖ .
]⇒[ ﻧﻔﺮﺽ ﺃﻥ fﲢﻘﻖ ﺷﺮﻁ ﻛﻮﺷﻲ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺃﻱ:
−2−
•
∀ε > 0, ∃δ > 0 / ∀x′, x′′ ∈ Vδ (a ) ⊂ D( f ) → f ( x′) − f ( x′′) < ε
•
ﻭﻣﻨﻪ ﻳﻮﺟﺪ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ) Vδ (aﻟﻠﻨﻘﻄﺔ aﺗﻜﻮﻥ ﻓﻴﻪ ﺍﻟﺪﺍﻟﺔ ﻣﻌﺮﻓﺔ.
•
ﻟﺘﻜﻦ (a n )n≥1ﻣﺘﺘﺎﻟﻴﺔ ﻣﻦ ) ،Vδ (aﺣﻴﺚ:
. lim
an = a
∞→n
ﻧﱪﻫﻦ ﺃﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ، ( f (a n ))n ≥1ﲤﻠﻚ ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻏﲑ ﻣﺘﻌﻠﻘﺔ ﺑﺎﺧﺘﻴﺎﺭ . (a n )n≥1
ﻋﻨﺪﻧﺎ a n = a
limﺗﻌﲏ :
∞→n
∀ε > 0, ∃nε ≥ 1 / ∀n ≥ nε → a n − a < ε
ﻣﻦ ﺃﺟﻞ ε = δ εﺣﻴﺚ δﻣﻌﺮﻓﺔ ﰲ ﺍﻟﺼﻴﻐﺔ −1 −ﻳﻜﻮﻥ :
ε = δ , ∃nδ ε ≥ 1 / ∀n ≥ nδ ε → a n − a < δ ε
−3−
ﻫﺬﺍ ﻳﻌﲏ ﺃﻧﻪ
•
•
) , ∀m ≥ nδε → a n ∈Vδ ε (a ) , a m ∈Vδ ε (a
ﻣﻦ ﻫﻨﺎ ﺣﺴﺐ
ﺍﻟﺼﻴﻐﺔ − 3 − , − 2 −
∀n ≥ nδ ε
ﻳﻜﻮﻥ :
∀ε > 0, ∃nε′ = nδ ε / ∀n ≥ nε′ , ∀m ≥ nε′ → f (a n ) − f (a m ) < ε
ﺃﻱ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ( f (a n ))n ≥1ﻟﻜﻮﺷﻲ .
∃l = lim
ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﻛﻮﺷﻲ ﻓﻘﺮﺓ 1.8.2ﺍﳌﺘﺘﺎﻟﻴﺔ ( f (a n ))n ≥1ﺗﻜﻮﻥ ﻣﺘﻘﺎﺭﺑﺔ ،ﺃﻱ f (a n ) :
∞→n
•
ﻧﱪﻫﻦ ﺃﻥ lﻣﺴﺘﻘﻞ ﻋﻦ ﺍﺧﺘﻴﺎﺭ ﺍﳌﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ) . Vδ (a
•
ﻧﻔﺮﺽ ﺍﻟﻌﻜﺲ ،ﺃﻱ ﺗﻮﺟﻮﺩ ﻣﺘﺘﺎﻟﻴﺔ ﺃﺧﺮﻯ (bn )n ≥1ﻣﻦ ) Vδ (aﲢﻘﻖ:
lim
b = lim
an = a
n→∞ n
∞→ n
lim
، limﺣﻴﺚf (a n ) = l 1 ≠ l 2 :
ﻭ f (bn ) = l 2
∞→n
∞→ n
ﻣﻌﻠﻮﻡ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ (c n )n ≥1ﺍﳌﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ :
}= {c1 , c 2 , c3 ,........} ≡ {a 1 , b1 , a 2 , b2 , a 3 ........
) (c
n ≥1
n
•
. lim
ﻣﻦ ) ،Vδ (aﻭﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ ، aﺃﻱc = a :
n→∞ n
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ﻣﻦ ﻫﻨﺎ ﻭﻣﻦ ﺷﺮﻁ ﻛﻮﺷﻲ ،ﻭﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ ، (a n )n≥1ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ ( f (c n ))n ≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﻋﺪﺩ l 3ﺃﻱ:
lim
f (cn ) = l 3
∞→n
ﺍﳌﺘﺘﺎﻟﻴﺔ ( f (c n ))n ≥1ﻫﻲ ﺍﳌﺘﺘﺎﻟﻴﺔ
}{ f (a ), f (b ), f (a ), f (b ),...........
ﻫﺬﺍ ﻳﻌﻦ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ )) ( f (b )) , ( f (aﺟﺰﺋﻴﺘﺎﻥ ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ )) . ( f (c
2
n ≥1
n
n ≥1
2
1
1
n ≥1
n
n
ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﻓﻘﺮﺓ 7.2ﻳﻜﻮﻥ :
l 1 = l 3ﻭ ، l 2 = l 3ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ. l 1 = l 2 :
•
ﻭﻣﻨﻪ ﺎﻳﺔ ( f (a n ))n ≥1ﻣﺴﺘﻘﻠﺔ ﻋﻦ ﺍﺧﺘﻴﺎﺭ ﺍﳌﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ) ،Vδ (aﺃﻱ ﺃﻥ fﳍﺎ ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﰲ ﺍﻟﻨﻘﻄﺔ . a
ﻣﻼﺣﻈﺔ :ﻧﻈﺮﻳﺔ ﻛﻮﺷﻲ ﺻﺤﻴﺤﺔ ﻋﻨﺪﻣﺎ ﻧﺴﺘﺒﺪﻝ ﺍﻟﻜﺘﺎﺑﺔ ، x → aﺑﺎﻟﻜﺘﺎﺑﺔ:
x→ a +0
،ﺃﻭ
x→ a −0
،ﺃﻭ
∞→x
-4.3ﺍﻻﺳﺘﻤﺮﺍﺭ :
ﻟﺘﻜﻦ fﺩﺍﻟﺔ ﺣﻘﻴﻘﻴﺔ ﳌﺘﻐﲑ ﺣﻘﻴﻘﻲ ﻭ aﻋﺪﺩﺍﹰ ﻣﻦ . ℜ
-1.4.3ﺗﻌﺎﺭﻳﻒ :
-1ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻧﺖ :
wﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ a
f ( x) = f (a ) w
lim
x→ a
ﺃﻱ ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﲢﻘﻘﺖ ﺍﻟﺸﺮﻭﻁ ﺍﻟﺘﺎﻟﻴﺔ :
ﺃ-
ﺏ-
ﺝ-
•
) ∃δ 0 > 0 / Vδ 0 (a ) ⊂ D( f
∃ lim
f ( x) = l
x→ a
f (a ) = l
ﻧﺘﻴﺠﺔ :ﻣﻦ ﺍﻟﺘﻌﺮﻳﻒ ﺍﻟﺴﺎﺑﻖ ﻭﻣﻦ ﺗﻌﺮﻳﻒ ﺍﻟﻨﻬﺎﻳﺔ ﺣﺴﺐ ﻗﺎﻳﻦ ﻭ ﻛﻮﺷﻲ ﻧﺴﺘﻨﺘﺞ ﺗﻌﺮﻳﻔﲔ ﻟﻼﺳﺘﻤﺮﺍﺭ.
ﺃ -1ﺣﺴﺐ ﻛﻮﺷﻲ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﲢﻘﻖ :
•
) ∃δ 0 > 0 / Vδ 0 (a ) ⊂ D( f
w
)) ∀ε > 0 , ∃δ ∈ ]0, δ 0 ] / ∀x ∈ Vδ (a ) → f ( x) ∈ Vε ( f (a
w
ﺏ -1ﺣﺴﺐ ﻗﺎﻳﻦ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﲢﻘﻖ :
w
•
) ∃δ 0 > 0 / Vδ 0 (a ) ⊂ D( f
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wﻣﻦ ﺃﺟﻞ ﻛﻞ ﻣﺘﺘﺎﻟﻴﺔ (a n )n≥1ﻣﻦ ) Vδ (aﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ ﺍﻟﻌﺪﺩ ، aﺗﻜﻮﻥ ﻣﺘﺘﺎﻟﻴﺔ ﺍﻟﺼﻮﺭ ( f (a n ))n ≥1ﻣﺘﻘﺎﺭﺑﺔ ﳓﻮ
0
ﺍﻟﻌﺪﺩ ) . f (a
-2ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ
ﺃ -2ﻋﻠﻰ ﻳﺴﺎﺭ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻧﺖ :
w
w
•
) ∃δ 0 > 0 / Vδ 0 (a − 0) ⊂ D( f
) lim f ( x) = f (a
x→ a − 0
ﺏ -2ﻋﻠﻰ ﳝﲔ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﻛﺎﻧﺖ :
w
w
•
) ∃δ 0 > 0 / Vδ 0 (a + 0) ⊂ D( f
) lim f ( x) = f (a
x→ a + 0
ﻣﻼﺣﻈﺔ :
-1ﻣﻦ ﺍﻟﺘﻌﺎﺭﻳﻒ ﺃ ، 1ﺃ ، 2ﺏ 2ﻧﺴﺘﺨﻠﺺ ﺗﻌﺮﻳﻒ ﻛﻮﺷﻲ ﻟﻼﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻴﻤﲔ ،ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ.
-2ﻣﻦ ﺍﻟﺘﻌﺎﺭﻳﻒ ﺏ ، 1ﺃ ، 2ﺏ 2ﻧﺴﺘﺨﻠﺺ ﺗﻌﺮﻳﻒ ﻗﺎﻳﻦ ﻟﻼﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻴﻤﲔ ،ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ.
ﻧﺘﺎﺋﺞ :
-1ﺃ ⇔ 1ﺏ
f -2ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ، Aﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺴﺘﻤﺮﺓ ﰲ ﻛﻞ ﻧﻘﻄﺔ ﻣﻦ . A
f -3ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺴﺘﻤﺮﺓ ﻣﻦ ﺍﻟﻴﻤﲔ،ﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ ﰲ ﺍﻟﻨﻘﻄﺔ . a
1
)
(
( lim
f ( x) = f lim
f ) -4ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ f ) ⇔ ( aﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻟﻠﻨﻘﻄﺔ aﻭ x
x→ a
x→ a
-5ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ]، [a, bﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺴﺘﻤﺮﺓ ﰲ ﻛﻞ ﻧﻘﻄﺔ ﻣﻦ [، ]a, bﻭﻣﺴﺘﻤﺮﺓ ﻣﻦ ﺍﻟﻴﻤﲔ ﰲ
ﺍﻟﻨﻘﻄﺔ aﻭﻣﻦ ﺍﻟﻴﺴﺎﺭ ﰲ ﺍﻟﻨﻘﻄﺔ . b
ﺃﻣﺜﻠﺔ :
-1ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ ، fﺣﻴﺚ :
a =0
ﻭﺍﺿﺢ ﺃﻥ ، D( f ) = ℜﻭ
1
≤ x
x
1
, x≠0
x sin
f ( x) =
x
0
, x=0
∀x ∈ ℜ → 0 ≤ f ( x) − f (0) = f ( x) = x sin
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) ∃δ 0 > 0 / Vδ 0 (0 ) ⊂ D( f
ﺃﻱ
ﻭ
ﻳﻜﻔﻲ ﺃﺧﺬ
∀ε > 0, ∃δ > 0 / ∀x ∈ Vδ (0 ) ⊂ D( f ) → f ( x) < ε
δ =ε
،ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ
) lim f ( x) = 0 = f (0
x→0
ﺃﻱ ﺃﻥ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ . a = 0
-2ﺗﺄﺧﺬ ] f (x) = [xﺍﻧﻈﺮ ﻣﺜﺎﻝ 3ﺑﺪﺍﻳﺔ ﺍﻟﻔﺼﻞ.
lim f ( x) = f (1 + 0 ) = f (1) = 1
ﻣﻦ ﺷﻜﻞ ﺍﻟﺪﺍﻟﺔ ﻭﺍﺿﺢ ﺃﻥ
x→1+ 0
)lim f ( x) = f (1 − 0 ) = 0 ≠ f (1
ﻭ
x→1−0
ﺃﻱ ﺃﻥ
f
ﻣﺴﺘﻤﺮﺓ ﻣﻦ ﺍﻟﻴﻤﲔ ﰲ ﺍﻟﻨﻘﻄﺔ ، a = 1ﻟﻜﻦ ﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﰲ ﻧﻔﺲ ﺍﻟﻨﻘﻄﺔ.
-2.4.3ﻧﻘﻂ ﻋﺪﻡ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻭﺍﻟﺘﻤﺪﻳﺪ ﺑﺎﻻﺳﺘﻤﺮﺍﺭ :
ﺗﻌﺮﻳﻒ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﻨﻘﻄﺔ aﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻟﻠﺪﺍﻟﺔ
-1
-2
f
،ﺇﺫﺍ ﱂ ﻳﺘﺤﻘﻖ ﺃﺣﺪ ﺍﻟﺸﺮﻭﻁ ﺍﻟﺘﺎﻟﻴﺔ :
) a ∈ D( f
)∃l ∈ ℜ / l = lim f ( x
x→ a
) l = f (a
-3
ﺗﺼﻨﻒ ﻧﻘﻂ ﻋﺪﻡ ﺍﻻﺳﺘﻤﺮﺍﺭ ﺇﱃ ﺛﻼﺙ ﺃﺻﻨﺎﻑ ﻫﻲ:
ﺃ -ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ :
ﻧﻘﻮﻝ ﻋﻦ ﺍﻟﻨﻘﻄﺔ aﺇﺎ ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺪﺍﻟﺔ ، fﺇﺫﺍ ﲢﻘﻖ ﺍﻟﺸﺮﻁ -2-ﻭﱂ ﻳﺘﺤﻘﻖ ﺃﺣﺪ ﺍﻟﺸﺮﻭﻁ
. -3- ، -1ﻧﺘﻴﺠﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻨﻘﻄﺔ aﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ ﻟﻠﺪﺍﻟﺔ
ﳓﺼﻞ ﻋﻠﻰ ﺩﺍﻟﺔ
*f
x→ a
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﺗﻜﻮﻥ ﻣﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ :
f ( x) , x ≠ a
f ∗ ( x) =
x=a
ﻓﻌﻼﹰ ﻷﻥ
، fﺃﻱ )∃l = lim f ( x
،ﻓﺈﻥ ﺑﻮﺿﻌﻨﺎ
) l = f ∗ (a
l
,
) a ∈ D( f ) , lim f ∗ ( x) = lim f ( x) = l = f ∗ (a
x→ a
x→ a
ﻭﺑﺎﻟﺘﺎﱄ ﻧﻜﻮﻥ ﻗﺪ ﻣﺪﺩﻧﺎ fﺑﺎﻻﺳﺘﻤﺮﺍﺭﻳﺔ ﰲ ﺍﻟﻨﻘﻄﺔ . a
ﻋﻨﺪﻫﺎ ﻧﻘﻮﻝ ﺇﻥ ﻛﻞ ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ ﳝﻜﻦ ﺟﻌﻠﻬﺎ ﻧﻘﻄﺔ ﺍﺳﺘﻤﺮﺍﺭ.
ﻣﺜﺎﻝ : 1ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ . f (x) = x sin 1
x
ﻻﺣﻆ ﺃﻥ
1
=0
x
lim x sin
x→ 0
،ﻟﻜﻦ ﺍﻟﻌﺒﺎﺭﺓ
1
x sin
x
ﻟﻴﺲ ﳍﺎ ﻣﻌﲎ ﰲ ﺍﻟﺼﻔﺮ ،ﻭﺑﺎﻟﺘﺎﱄ ﺍﻟﻨﻘﻄﺔ a = 0ﻫﻲ ﻧﻘﻄﺔ ﻋﺪﻡ
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ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ.
ﻟﻠﺒﺤﺚ ﻋﻦ ﲤﺪﻳﺪ
f
ﺑﺎﻻﺳﺘﻤﺮﺍﺭ ﰲ ﺍﻟﻨﻘﻄﺔ a = 0ﻳﻜﻔﻲ ﺃﻥ ﳒﻌﻞ
f ( x) , x ≠ a
f ∗ ( x) =
x=a
ﻻﺣﻆ
0
,
ﺃﻥ ، 0 ∈ D( f ), lim f ∗ ( x) = lim f ( x) = 0 = f ∗ (0 ) :ﺃﻱ ∗ f
x→ 0
x→ 0
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﺼﻔﺮ.
ﺏ -ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ :
ﻧﻘﻮﻝ ﻋﻦ ﺍﻟﻨﻘﻄﺔ aﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ ﻟﻠﺪﺍﻟﺔ ، fﺇﺫﺍ ﻭﺟﺪﺕ ﺎﻳﺔ ﻣﻨﺘﻬﻴﺔ ﻣﻦ ﺍﻟﻴﻤﲔ ﻭﺃﺧﺮﻯ ﻣﻨﺘﻬﻴﺔ
ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻟﻠﺪﺍﻟﺔ fﰲ ﺍﻟﻨﻘﻄﺔ aﳐﺘﻠﻔﺘﲔ ،ﺃﻱ :
∃f (a − 0) ∈ ℜ , ∃f (a + 0 ) ∈ ℜﻭ )f (a − 0) ≠ f (a + 0
ﺍﻟﻔﺮﻕ ) ، α = f (a + 0) + f (a − 0ﻳﺴﻤﻰ ﻗﻔﺰﺓ ﺍﻟﺪﺍﻟﺔ fﰲ ﺍﻟﻨﻘﻄﺔ . a
ﻣﺜﺎﻝ : 2ﺍﻟﺪﺍﻟﺔ ﺍﻹﺷﺎﺭﺓ ﺍﻧﻈﺮ ﻣﺜﺎﻝ -2-ﺑﺪﺍﻳﺔ ﺍﻟﻔﺼﻞ
1 , x > 0
f ( x) = sgn x = 0 , x = 0
− 1 , x < 0
ﻻﺣﻆ ﺃﻥ:
ﺃﻱ ﺃﻥ
ﺍﻟﻨﻘﻄﺔ a = 0
∃ lim f ( x) = f (+ 0) = 1
x→ +0
) ⇒ f (+ 0) ≠ f (− 0
∃ lim f ( x) = f (− 0) = −1
x→ −0
ﻫﻲ ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ .ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻗﻔﺰﺓ ﺍﻟﺪﺍﻟﺔ ﻫﻲ . α = 1 − (− 1) = 2
ﺝ -ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻟﺜﺎﱐ :
ﻧﻘﻮﻝ ﻋﻦ ﺍﻟﻨﻘﻄﺔ aﻫﻲ ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻟﺜﺎﱐ ﻟﻠﺪﺍﻟﺔ ، fﺇﺫﺍ ﻛﺎﻧﺖ ﰲ ﻫﺬﻩ ﺍﻟﻨﻘﻄﺔ ﻋﻠﻰ ﺍﻷﻗﻞ ﺃﺣﺪ
ﺍﻟﻨﻬﺎﻳﺘﲔ ﺍﻟﻴﻤﲎ ﺃﻭ ﺍﻟﻴﺴﺮﻯ ﻏﲑ ﻣﻮﺟﻮﺩﺓ ﺃﻭ ﺗﺴﺎﻭﻱ ﺍﳌﺎﻻﺎﻳﺔ.
ﻣﺜﺎﻝ :ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ fﺣﻴﺚ:
ﺑﺎﺳﺘﻌﻤﺎﻝ ﺗﻌﺮﻳﻒ ﻗﺎﻳﻦ ﻟﻠﻨﻬﺎﻳﺔ ﳓﺼﻞ ﻋﻠﻰ
ﻟﻜﻦ ﺗﻮﺟﺪ ﺎﻳﺔ ﻣﻦ ﺍﻟﻴﻤﲔ ﻟﻠﺪﺍﻟﺔ
f
1
x cos x , x < 0
f ( x) = 0
, x=0
1
cos
, x>0
x
lim f (x) = 0
ﰲ ﺍﻟﻨﻘﻄﺔ
x→ −0
a =0
.
.
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ﻫﺬﺍ ﻳﻌﲎ ﺃﻥ ﺍﻟﻨﻘﻄﺔ
a =0
ﻫﻲ ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻟﺜﺎﱐ ﻟﻠﺪﺍﻟﺔ . f
ﻧﻈﺮﻳﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﻌﺮﻓﺔ ﻋﻠﻰ ] [a, bﻭﺭﺗﻴﺒﺔ ﻋﻠﻴﻪ ،ﻓﺈﻧﻪ ﺇﺫﺍ ﻛﺎﻧﺖ ﳍﺎ ﻧﻘﻂ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﰲ ﺍﺎﻝ [ ]a, bﻓﺈﺎ
ﺗﻜﻮﻥ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ.
ﺍﻟﱪﻫﺎﻥ :ﻧﻔﺮﺽ ﺃﻥ
f
ﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ.
ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﻓﻘﺮﺓ ،4.3.3ﻟﻠﺪﺍﻟﺔ fﰲ ﻛﻞ ﻧﻘﻄﺔ ﻣﻦ ﺍﺎﻝ [ ]a, bﺎﻳﺔ ﳝﲎ ﻣﻨﺘﻬﻴﺔ،ﻭﻳﺴﺮﻯ ﻣﻨﺘﻬﻴﺔ ﻭﻳﻜﻮﻥ ﻋﻨﺪﻫﺎ :
) ∀x0 ∈ ]a , b[ → f ( x0 − 0) ≤ f ( x0 ) ≤ f ( x0 + 0
ﻫﺬﺍ ﻳﻌﲎ
، ∃f (x0 + 0 ),ﺃﻱ ﺃﻥ ﺍﻟﻨﻘﻄﺔ x0ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ.
) ∃f ( x0 − 0
ﺇﺫﺍ ﻛﺎﻧﺖ ) ، f (x0 + 0) = f (x0 − 0ﻓﺈﻥ:
)∃ lim f ( x) = f ( x0 + 0) = f ( x0 − 0
x→ x0
ﻭﺑﺎﻟﺘﺎﱄ ﺗﻜﻮﻥ x0ﻧﻘﻄﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻌﺰﻭﻝ ،ﻭﻫﻲ ﺗﺆﻭﻝ ﺇﱃ ﻧﻘﻄﺔ ﺍﺳﺘﻤﺮﺍﺭ ﻛﻤﺎ ﻋﺮﻓﻨﺎ ﰲ ﺍﻟﻨﺘﻴﺠﺔ ﻓﻘﺮﺓ .2.4.3
ﰲ ﺣﺎﻟﺔ fﻟﻴﺴﺖ ﻣﺘﺰﺍﻳﺪﺓ ﺗﱪﻫﻦ ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ.
-3.4.3ﺍﻻﺳﺘﻤﺮﺍﺭ ﺑﺎﻷﺟﺰﺍﺀ :
ﺗﻌﺮﻳﻒ ﺃ :ﻧﻘﻮﻝ ﺇﻥ fﻣﺴﺘﻤﺮﺓ ﺑﺎﻷﺟﺰﺍﺀ ﻋﻠﻰ ﺍﺎﻝ ]، [a, bﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﺍﺎﻝ [ ]a, bﺑﺎﺳﺘﺜﻨﺎﺀ ﳏﺘﻤﻞ ﻟﻌﺪﺩ
ﻣﻨﺘﻪ ﻣﻦ ﻧﻘﻂ ﺍﺎﻝ [، ]a, bﺍﻟﱵ ﺗﻜﻮﻥ ﻧﻘﻂ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ .ﻭﺑﺎﻹﺿﺎﻓﺔ ﺇﱃ ﺫﺍﻟﻚ ﲤﻠﻚ ﺎﻳﺔ ﻳﺴﺮﻯ ﰲ
ﺍﻟﻨﻘﻄﺔ ، bﻭﺎﻳﺔ ﳝﲎ ﰲ ﺍﻟﻨﻘﻄﺔ . a
ﺗﻌﺮﻳﻒ ﺏ :ﻧﻘﻮﻝ ﺇﻥ fﻣﺴﺘﻤﺮﺓ ﺑﺎﻷﺟﺰﺍﺀ ﻋﻠﻰ ، ℜﺇﺫﺍ ﻛﺎﻧﺖ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻷﺟﺰﺍﺀ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ] [a, bﻣﻦ . ℜ
ﻣﺜﺎﻝ :ﺍﻟﺪﺍﻟﺔ ﺍﳉﺰﺀ ﺍﻟﺼﺤﻴﺢ ]. f (x) = [x
ﻣﺴﺘﻤﺮﺓ ﺑﺎﻷﺟﺰﺍﺀ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ] [a, bﻣﻦ ، ℜﻭﺑﺎﻟﺘﺎﱄ ﻫﻲ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻷﺟﺰﺍﺀ ﻋﻠﻰ ﻛﻞ ، ℜﻷﻧﻪ ﰲ ﻛﻞ ﻋﺪﺩ ﺻﺤﻴﺢ
qﲤﻠﻚ ﺍﻟﺪﺍﻟﺔ fﺎﻳﺔ ﳝﲎ ﻣﻨﺘﻬﻴﺔ ﻭﻳﺴﺮﻯ ﻣﻨﺘﻬﻴﺔ ،ﻭ ) ، f (q + 0 ) ≠ f (q − 0ﺃﻱ ﰲ ﻛﻞ ﻧﻘﻄﺔ qﻣﻦ Ζﲤﻠﻚ
ﺍﻟﺪﺍﻟﺔ ﻋﺪﻡ ﺍﺳﺘﻤﺮﺍﺭ ﻣﻦ ﺍﻟﻨﻮﻉ ﺍﻷﻭﻝ.
-4.4.3ﺧﻮﺍﺹ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺴﺘﻤﺮﺓ :
-1.4.4.3ﺧﻮﺍﺹ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺴﺘﻤﺮﺓ ﰲ ﻧﻘﻄﺔ :
-1ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ
f
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ aﻓﺈﺎ ﺗﻜﻮﻥ ﳏﺪﻭﺩﺓ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ ، aﺃﻱ:
∀x ∈ Vδ (a ) → f ( x) ≤ c
/
∃c > 0 , ∃δ > 0
-2ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﻭ ، f (a ) ≠ 0ﻓﺈﻧﻪ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ ، aﺗﻜﻮﻥ ﺇﺷﺎﺭﺓ ﺍﻟﺪﺍﻟﺔ fﻣﻦ
ﻧﻔﺲ ﺇﺷﺎﺭﺓ ﺍﻟﻌﺪﺩ ) ، f (aﺃﻱ :
) ∀x ∈ Vδ (a ) → signf ( x) = signf (a
/
∃δ > 0
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-3ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺘﺎﻥ
f ⋅ g,
f ±g
g, f
f
,
g (a ) ≠ 0 ,
g
-4ﻧﻈﺮﻳﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ
) y0 = ϕ ( x0
ﻣﺴﺘﻤﺮﺗﲔ ﰲ ﺍﻟﻨﻘﻄﺔ ، aﻓﺈﻥ ﺍﻟﺪﻭﺍﻝ :
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ . a
)z = f ( y
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، y0ﻭﺍﻟﺪﺍﻟﺔ
،ﻓﺈﻧﻪ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻟﻠﻨﻘﻄﺔ x0ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺮﻛﺒﺔ
)y = ϕ (x
h = f oϕ
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، x0ﺣﻴﺚ
ﻣﻌﺮﻓﺔ ﻭﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ . x0
ﺍﻟﱪﻫﺎﻥ :
ﻧﻀﻊ f ( y0 ) = z0
ﻋﻨﺪﻧﺎ
f
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ y0ﻳﻌﲏ
) / ∀y ∈ Vδ1 ( y0 ) ⊂ D( f ) → f ( y) ∈ Vε ( z0
−1−
∀ε > 0 , ∃δ 1 > 0
ﻋﻨﺪﻧﺎ ϕﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ x0ﻳﻌﲏ
) / ∀x ∈ Vδ 0 ( x0 ) ⊂ D(ϕ ) → ϕ ( x) ∈ Vε ( y0
−2−
∀ε > 0 , ∃δ 0 > 0
ﻣﻦ ﺃﺟﻞ ε = δ 1ﺣﻴﺚ δ 1ﻣﻌﺮﻓﺔ ﰲ ﺍﻟﺼﻴﻐﺔ −1 −ﻳﻜﻮﻥ
) / ∀x ∈ Vδ ′ ( x0 ) ⊂ D(ϕ ) → ϕ ( x) ∈ Vδ1 ( y0
−3−
ε = δ 1 , ∃δ ′ > 0
ﻻﺣﻆ ﰲ ﺍﻟﺼﻴﻐﺔ y = ϕ (x)∈Vδ ( y0 ) − 3 −ﺣﺴﺐ ﺍﻟﺼﻴﻐﺔ −1 −ﻳﻜﻮﻥ:
) ، f ( y) = f (ϕ (x)) ∈Vε (z0ﺣﻴﺚ ) ، z0 = f (ϕ (x0 )) = f ( y0ﻫﺬﺍ ﻳﻌﲏ ﺃﻧﻪ ) ، Vδ ′ (x0 ) ⊂ D( f o ϕﻭ
1
))) ∀ε > 0, ∃δ ′ > 0 / ∀x ∈ Vδ ′ ( x0 ) → f (ϕ ( x)) ∈ Vε ( f (ϕ ( x0
ﺃﻱ ﺃﻥ ﺍﻟﺪﺍﻟﺔ hﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ x0
-2.4.4.3ﺧﻮﺍﺹ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﻤﻮﻋﺔ :
-1ﻧﻈﺮﻳﺔ ) :ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﻭﱃ ﻟﻔﲑﺷﺘﺮﺍﺹ ( :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﺍﺎﻝ ] [a, bﻓﻬﻲ ﳏﺪﻭﺩﺓ ﻋﻠﻴﻪ ،ﺃﻱ:
/ ∀x ∈ [a , b] → f ( x) ≤ c
∃c > 0
ﺍﻟﱪﻫﺎﻥ :ﻧﻔﺮﺽ ﺍﻟﻌﻜﺲ ،ﺃﻱ :
−1−
/ ∀x ∈ [a , b] → f ( x) > c
∃c > 0
ﻣﻦ ﺍﻟﺼﻴﻐﺔ −1 −ﻧﺴﺘﻨﺘﺞ ﺃﻧﻪ
−2−
∀n ∈ Ν , ∃xn ∈ [a , b] / f ( xn ) > n
ﻻﺣﻆ ﺍﳌﺘﺘﺎﻟﻴﺔ (xn )n≥1ﳏﺪﻭﺩﺓ ﻷﻥ
ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺑﻮﻟﺰﺍﻧﻮ-ﻓﲑﺷﺘﺮﺍﺹ ﻓﻘﺮﺓ 1.7.2ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ (xn )n≥1ﳝﻜﻦ ﺇﺧﺮﺍﺝ ﻣﺘﺘﺎﻟﻴﺔ ﺟﺰﺋﻴﺔ (xk )n ≥1ﻣﺘﻘﺎﺭﺑﺔ،ﺃﻱ:
, a ≤ xn ≤ b
∀n ≥ 1
n
/ ξ = lim xkn
∞→n
∃ξ ∈ ℜ
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ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺍﳊﺼﺮ . a ≤ ξ ≤ b
ﲟﺎﺃﻥ fﻣﺴﺘﻤﺮﺓ،ﻓﺎﻧﻪ ﺣﺴﺐ ﺍﻟﻨﺘﻴﺠﺔ 4ﻓﻘﺮﺓ 1.4.3ﻳﻜﻮﻥ:
) lim f (xk ) = f (ξ
−3−
∞→ n
n
ﻋﻨﺪﻧﺎ ﺍﻟﺼﻴﻐﺔ − 2 −ﳏﻘﻘﺔ ﻣﻦ ﺃﺟﻞ ﻛﻞ (n ≥ 1), nﻭﻣﻨﻪ ﻣﻦ ﺃﺟﻞ knﳏﻘﻘﺔ ﺃﻳﻀﺎ ،ﺃﻱ:
) (
−4−
ﺍﳌﺘﺮﺍﺟﺤﺔ -4-ﺗﻌﲏ ﺃﻥ:
) (
f xkn > kn
∞ = lim f xkn
∞→ n
ﻫﺬﺍ ﻣﻨﺎﻑ ﻟﻠﻤﺴﺎﻭﺍﺓ − 3 −ﻭﻣﻨﻪ ﺍﻟﻔﺮﺽ ﻏﲑ ﺻﺤﻴﺢ ،ﺃﻱ ﺃﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ ﻓﻌﻼﹰ ﻋﻠﻰ ]. [a, b
ﻗﻀﻴﺔ :ﺣﺎﻟﺔ ﺍﺎﻝ ﻏﲑ ﻣﻐﻠﻖ ،ﺍﻟﻨﻈﺮﻳﺔ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ﻏﲑ ﺻﺤﻴﺤﺔ.
ﻣﺜﺎﻝ :ﺍﻟﺪﺍﻟﺔ ، f (x) = 1ﻭﺍﺿﺢ ﺃﺎ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ [ ، ]0,1ﻟﻜﻦ ﻟﻴﺴﺖ ﳏﺪﻭﺩﺓ ﻋﻠﻴﻪ.
x
-2ﻧﻈﺮﻳﺔ ) ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻟﻔﲑﺷﺘﺮﺍﺹ ( :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] ، [a, bﻓﺈﺎ ﻋﻠﻰ ﻫﺬﺍ ﺍﺎﻝ ﺗﺼﻞ ﺇﱃ
ﺣﺪﻫﺎ ﺍﻻﻋﻠﻰ،ﻭﺍﻟىﺤﺪﻫﺎ ﺍﻷﺩﱏ ،ﺃﻱ:
)f ( x
f (ς ) = sup
∃ς ∈ [a , b ] /
)f (ς ′) = inf f ( x
∃ς ′ ∈ [a , b] /
] x∈[a ,b
] x∈[a ,b
ﺍﻟﱪﻫﺎﻥ :ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] [a, bﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻜﻮﻥ ﳏﺪﻭﺩﺓ ﻋﻠﻴﻪ ،ﺃﻱ :
,
)∃ inf f ( x
)∃ sup f ( x
] x∈[a ,b
] x∈[a ,b
ﻧﱪﻫﻦ ﺃﻥ
)f (ς ) = sup f ( x
] x∈[a ,b
ﺑﻮﺿﻊ
)M = sup f (x
] x∈[a ,b
ﺃ-
ﺏ-
∃ς ∈ [a , b] /
ﻳﺘﺤﻘﻖ :
∀x ∈ [a , b] → f ( x) ≤ M
∀ε > 0, ∃x(ε ) ∈ [a , b] / f ( x(ε )) > M − ε
ﻣﻦ ﺃﺟﻞ ، ε = 1ﻣﻦ –ﺏ -ﻧﺴﺘﻨﺘﺞ ﺃﻧﻪ:
n
ﺏ-
ﻣﻦ ﺃ ،ﺏ ﻧﺴﺘﻨﺞ ﺃﻧﻪ :
ﺣﻴﺚ
) n ≥ 1 , xn = x( 1n
1
n
f (x( 1n )) > M −
∀n ≥ 1, ∃x( 1n )∈ [a , b] /
1
1
< f ( xn ) ≤ M < M +
n
n
∀n ≥ 1 → M −
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. lim
ﻫﺬﺍ ﻳﻌﲎ ﺃﻥ f ( xn ) = M
∞→ n
ﻣﻦ ﺏ ﻳﻜﻮﻥ
], xn ∈ [a , b
، ∀n ≥ 1ﺃﻱ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ (xn )n≥1ﳏﺪﻭﺩﺓ،ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺑﻮﻟﺰﺍﻧﻮ-ﻓﲑﺷﺘﺮﺍﺹ ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ
(xn )n≥1ﳝﻜﻦ ﺇﺧﺮﺍﺝ ﻣﺘﺘﺎﻟﻴﺔ ﺟﺰﺋﻴﺔ
)
n ≥1
n
(xkﻣﺘﻘﺎﺭﺑﺔ،ﺃﻱ:
∃ς ∈ ℜ
/ ς = lim xkn
∞→n
ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺍﳊﺼﺮ . a ≤ ς ≤ b
ﲟﺎﺃﻥ fﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، ξﻓﺎﻧﻪ ﻳﻜﻮﻥ :
ﻻﺣﻆ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ
ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﳉﺰﺋﻴﺔ
))
n ≥1
))
n ≥1
) (
) lim f xkn = f (ς
∞→ n
( f (xkﺟﺰﺋﻴﺔ ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺔ ( f (xn ))n ≥1ﺍﳌﺘﻘﺎﺭﺑﺔ ﳓﻮ ، Mﺣﺴﺐ ﺧﺼﺎﺋﺺ ﺍﳌﺘﺘﺎﻟﻴﺔ ﺍﳉﺰﺋﻴﺔ
n
( f (xkﻣﺘﻘﺎﺭﺑﺔ ﺃﻳﻀﺎﹰ ﳓﻮ ، Mﺃﻱ ﺃﻥ ) ، M = f (ςﻭﺑﺎﻟﺘﺎﱄ :
)∃ς ∈ [a , b] / ς = sup f ( x
n
] x∈[a ,b
ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﻧﱪﻫﻦ
)f (ς ′) = inf f ( x
] x∈[a ,b
ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻧﻜﺘﺐ
)sup f ( x) = min f ( x
] x∈[a ,b
] x∈[a ,b
∃ς ′ ∈ [a , b] /
,
)inf f ( x) = min f ( x
] x∈[a ,b
] x∈[a ,b
ﻗﻀﻴﺔ :ﰲ ﺣﺎﻟﺔ ﺍﺎﻝ ﻏﲑ ﻣﻐﻠﻖ ،ﺍﻟﻨﻈﺮﻳﺔ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ﻏﲑ ﺻﺤﻴﺤﺔ.
ﻣﺜﺎﻝ :
ﺃ -ﺍﻟﺪﺍﻟﺔ fﺣﻴﺚ ، f ( x) = x 2ﻋﻠﻰ ﺍﺎﻝ [ ]0,1ﻻﺗﺼﻞ ﺇﱃ ﺣﺪﻫﺎ ﺍﻷﻋﻠﻰ،ﻭﻻ ﺇﱃ ﺣﺪﻫﺎ ﺍﻷﺩﱏ.
ﺫﻟﻚ ﻷﻥ inf f ( x) = 0 , sup f ( x) = 1 :
[x∈]0 ,1
[x∈]0 ,1
ﺏ -ﻫﺬﺍ ﺍﳌﺜﺎﻝ ﻳﺒﲔ ﺃﻥ ﺷﺮﻁ ﺍﻻﺳﺘﻤﺮﺍﺭ ﺃﺳﺎﺳﻲ ﻟﺼﺤﺔ ﺍﻟﻨﻈﺮﻳﺔ .
ﰲ ﺣﺎﻝ fﻣﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ :
ﻻﺣﻆ ﺃﻥ:
sup f ( x) = 1
]x∈[0 ,1
x2 , 0 < x < 1
f ( x) = 1
, x = 0 , x =1
2
inf f ( x) = 0 ,
]x∈[0 ,1
ﻋﻠﻰ ﺍﺎﻝ ] [0,1ﺍﻟﺪﺍﻟﺔ ﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ،ﻭﻫﻲ ﻻ ﺗﺼﻞ ﻻ ﺇﱃ ﺣﺪﻫﺎ ﺍﻷﻋﻠﻰ ﻭﻻ ﺇﱃ ﺣﺪﻫﺎ ﺍﻷﺩﱏ.
-3ﻧﻈﺮﻳﺔ)ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﻭﱃ ﻟﺒﻮﻟﺰﺍﻧﻮ-ﻛﻮﺷﻲ(:ﺇﺫﺍ ﻛﺎﻧﺖ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] [a, bﻭﲤﻠﻚ ﰲ
) ، ( f (a ) ⋅ f (b ) < 0ﻓﺈﺎ ﻋﻠﻰ ] [a, bﲤﻠﻚ ﻋﻠﻰ ﺍﻷﻗﻞ ﺻﻔﺮﺍ ﻭﺍﺣﺪﺍ ،ﺃﻱ:
f (α ) = 0
b, a
ﺇﺷﺎﺭﺗﲔ ﳐﺘﻠﻔﺘﲔ،
∃α ∈ [a , b] /
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ﺍﻟﱪﻫﺎﻥ :ﻧﻘﺴﻢ ﺍﺎﻝ ] [a, bﺇﱃ ﻧﺼﻔﲔ.ﻟﺘﻜﻦ dﻭﺳﻂ ﺍﺎﻝ.
ﺇﺫﺍ ﻛﺎﻧﺖ f (d ) = 0ﻓﻬﻮ ﺍﳌﻄﻠﻮﺏ.
ﺇﺫﺍ ﻛﺎﻧﺖ ، f (d ) ≠ 0ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﺇﻣﺎ ، f (a ) ⋅ f (d ) < 0ﻭﺇﻣﺎ . f (b ) ⋅ f (d ) < 0
ﻧﺮﻣﺰ ﻟﻠﻤﺠﺎﻝ ﺍﻟﺬﻱ ﺗﺄﺧﺬ ﺍﻟﺪﺍﻟﺔ ﰲ ﺎﻳﺘﻴﻪ ﺇﺷﺎﺭﺗﲔ ﳐﺘﻠﻔﺘﲔ ﺑﺎﻟﺮﻣﺰ ] . ∆1 = [a1 ,b1
ﻧﻘﺴﻢ ﺍﺎﻝ ∆ 1ﺇﱃ ﻧﺼﻔﲔ ،ﻟﺘﻜﻦ d1ﻧﺼﻒ ﺍﺎﻝ ، ∆ 1ﺇﺫﺍ ﻛﺎﻧﺖ f (d1 ) = 0ﻓﻬﻮ ﺍﳌﻄﻠﻮﺏ.
ﺇﺫﺍ ﻛﺎﻧﺖ ، f (d1 ) ≠ 0ﻓﺎﻧﻪ ﻧﻔﺲ ﺍﻟﺸﻲﺀ ﺍﻟﺪﺍﻟﺔ fﰲ ﺎﻳﱵ ﺃﺣﺪ ﺍﺎﻟﲔ ] ، [d1 ,b1ﺃﻭ ] [a1 ,d1ﲤﻠﻚ ﺇﺷﺎﺭﺗﲔ
ﳐﺘﻠﻔﺘﲔ ،ﻧﺮﻣﺰ ﻟﻠﻤﺠﺎﻝ ﺍﻟﺬﻱ ﲤﻠﻚ ﺍﻟﺪﺍﻟﺔ fﰲ ﺎﻳﺘﻴﺘﻴﻪ ﺇﺷﺎﺭﺗﲔ ﳐﺘﻠﻔﺘﲔ ﺑﺎﻟﺮﻣﺰ ] . ∆ 2 = [a 2 ,b2
ﻧﻮﺍﺻﻞ ﺍﻟﻌﻤﻠﻴﺔ ﳓﺼﻞ ﻋﻠﻰ ﺍﺣﺘﻤﺎﻟﲔ ﺇﻣﺎ ﻳﻮﺟﺪ ﻋﺪﺩ αﻣﻦ ] ، [a, bﲝﻴﺚ ﻳﻜﻮﻥ f (α ) = 0ﻭﻣﻨﻪ ﻓﻬﻮ ﺍﳌﻄﻠﻮﺏ.
ﻭﺇﻣﺎ ﳓﺼﻞ ﻋﻠﻰ ﻣﺘﺘﺎﻟﻴﺔ ﻣﻦ ﺍﺎﻻﺕ ، (∆ n )n≥1ﺣﻴﺚ ] . ∀n ≥ 1 , ∆ n = [a n , bnﲢﻘﻖ ﻣﻦ ﺃﺟﻞ ﻛﻞ (n ≥ 1), n
ﺍﳌﺘﺮﺍﺟﺤﺔ . f (a n ) ⋅ f (bn ) < 0
ﻻﺣﻆ ﺃﻥ ﺍﳌﺘﺘﺎﻟﻴﺔ (∆ n )n≥1ﻫﻲ ﻣﺘﺘﺎﻟﻴﺔ ﻣﻦ ﺍﺎﻻﺕ ﺍﳌﻐﻠﻘﺔ ﻭﺍﶈﺪﻭﺩﺓ ﺍﳌﺘﺪﺍﺧﻠﺔ ﻭﻗﻄﺮﻫﺎ ﻳﺆﻭﻝ ﺇﱃ ﺍﻟﺼﻔﺮ،ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ
ﳛﻮﻯ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ . α
-2ﻓﻘﺮﺓ 2.4.1ﻳﻜﻮﻥ ﺗﻘﺎﻃﻊ ﻛﻞ ﺍﺎﻻﺕﻧﱪﻫﻦ ﺃﻥ . f (α ) = 0
ﻧﻔﺮﺽ ﺍﻟﻌﻜﺲ ﺃﻱ ، f (α ) > 0ﺃﻭ . f (α ) < 0
n ≥1 , ∆n
ﰲ ﺣﺎﻟﺔ f (α ) > 0ﺣﺴﺐ ﺍﳋﺎﺻﻴﺔ -2-ﻓﻘﺮﺓ 1.4.4.3ﻳﺘﺤﻘﻖ:
∀x ∈ Vδ (α ) → f ( x) > 0
)∗(
/
∃δ > 0
ﻋﻨﺪﻧﺎ ، bn − a n = b −n aﺃﻱ :
2
∀n ≥ nε → bn − a n < ε
/
∀ε > 0 , ∃nε ∈ Ν
ﻣﻦ ﺃﺟﻞ ε = δﺗﻮﺟﺪ nδﲝﻴﺚ :
∀n ≥ nδ → bn − a n < δ
ﻭﻣﻨﻪ
bnδ − a nδ < δ
ﲟﺎ ﺃﻥ ] ، α ∈ [a n , bnﻭ ، bn − a n < δﻓﺈﻥ
ﻣﻦ ﻫﻨﺎ ﻭﻣﻦ )∗( ﻳﻜﻮﻥ:
δ
δ
δ
δ
/
∃nδ ∈ Ν
) ∆ nδ ⊂ Vδ (α
.
∀x ∈ ∆ nδ → f (x) > 0
ﻫﺬﺍ ﻳﻨﺎﻗﺾ ﻛﻮﻥ
f
ﲤﻠﻚ ﺇﺷﺎﺭﺗﲔ ﳐﺘﻠﻔﺘﲔ ﰲ ﺎﻳﱵ ﻛﻞ ﳎﺎﻝ ﻣﻦ ﺍﻟﻨﻮﻉ ∆ nﲟﺎ ﻓﻴﻬﺎ ﺍﺎﻝ ، ∆ nﻭﻋﻠﻴﻪ ﺍﻟﻔﺮﺽ
δ
f (α ) > 0ﻣﺴﺘﺤﻴﻞ.
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ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﻧﱪﻫﻦ ﺃﻥ ﺍﻟﻔﺮﺽ f (α ) < 0ﻣﺴﺘﺤﻴﻞ ،ﻭﻋﻠﻴﻪ ﻳﻜﻮﻥ . f (α ) = 0
ﻣﺜﺎﻝ :ﻧﱪﻫﻦ ﺃﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ 2 x = 4 xﺟﺬﻭﺭ.
ﻧﻼﺣﻆ ﺃﻥ x = 4ﺟﺬﺭ ﻭﺍﺿﺢ ﻟﻠﻤﻌﺎﺩﻟﺔ.
ﻟﺘﻌﲔ ﻣﻮﺿﻊ ﺍﳉﺬﻭﺭ ﺍﻷﺧﺮﻯ ﻧﻀﻊ . f (x) = 2 x − 4 x
ﻭﺍﺿﺢ ﺃﻥ ﺟﺬﻭﺭ ﺍﳌﻌﺎﺩﻟﺔ
ﻋﻨﺪﻧﺎ
2 −2<0
2 x − 4x = 0
= )( 2
ﻫﻲ ﺃﺻﻔﺎﺭ ﺍﻟﺪﺍﻟﺔ . f
f (0) = 1 − 0 = 1 > 0,
f 1
،ﻭ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ
1
0, 2
،ﻭﻣﻨﻪ ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺴﺎﺑﻘﺔ،
ﻟﻠﺪﺍﻟﺔ fﻋﻠﻰ ﺍﻷﻗﻞ ﺻﻔﺮ ﰲ ﺍﺎﻝ . 0, 1
2
-4ﻧﻈﺮﻳﺔ)ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻟﺒﻮﻟﺰﺍﻧﻮ-ﻛﻮﺷﻲ( :ﺇﺫﺍ ﻛﺎﻧﺖ
ﻗﻴﻤﺔ
λ
ﺑﲔ ) f (aﻭ ) f (bﺗﻮﺟﺪ ﻧﻘﻄﺔ
ﺍﻟﱪﻫﺎﻥ :ﻧﻀﻊ
f (a ) = α
ξ
f (b ) = β
,
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ]، [a, bﻭ ) ، f (a ) ≠ f (bﻓﺈﻧﻪ ﻣﻦ ﺃﺟﻞ ﺃﻱ
f
ﻣﻦ ] [a, bﲝﻴﺚ ﻳﻜﻮﻥ ) . C = f (ξ
,
α≠β
ﺍﳊﺎﻟﺔ : α < β
ﻧﱪﻫﻦ ﺃﻧﻪ:
∀λ ∈ [α , β ] , ∃ξ ∈ [a , b] / f (ξ ) = λ
ﰲ ﺣﺎﻟﺔ
λ =α
ﺃﻭ
ﰲ ﺍﳊﺎﻟﺔ
ﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ :
α <λ <β
λ=β
ﻧﻀﻊ
ﻧﺄﺧﺬ
ξ =a
ﺃﻭ
ξ =b
ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ .
x ∈ [a , b] , ϕ ( x) = f ( x) − λ
ϕ (a ) = f (a ) − λ = α − λ < 0
ϕ (b ) = f (b ) − λ = β − λ > 0
)∗(
ﺍﻟﺪﺍﻟﺔ ) ϕ (xﲢﻘﻖ ﺷﺮﻭﻁ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﻭﱃ ﻟﺒﻮﻟﺰﺍﻧﻮ-ﻛﻮﺷﻲ ،ﻭﺑﺎﻟﺘﺎﱄ :
∃ξ ∈ [a , b] / ϕ (ξ ) = 0
ﺃﻱ :
ﻭﺑﺎﻟﺘﺎﱄ . f (ξ ) = λ
ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﰲ ﺣﺎﻟﺔ
f (ξ ) − λ = 0
∃ξ ∈ [a , b] /
.β <α
ﻧﺘﻴﺠﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] ، [a, bﻓﺈﻥ ﳎﻤﻮﻋﺔ ﻗﻴﻢ ﺍﻟﺪﺍﻟﺔ ﻋﻠﻰ ] [a, bﻫﻲ ] ، [m, Mﺣﻴﺚ:
)m = inf f ( x
,
)M = sup f ( x
] x∈[a ,b
] x∈[a ,b
ﺍﻟﱪﻫﺎﻥ :ﻭﺍﺿﺢ ﺃﻥ:
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∀x ∈ [a , b] → m ≤ f ( x) ≤ M
ﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻟﻔﲑﺷﺘﺮﺍﺹ ،ﺍﻟﺪﺍﻟﺔ fﺗﺄﺧﺬ ﺍﻟﻘﻴﻤﺔ ، mﻭ Mﻋﻠﻰ ] [a, bﻭﺣﺴﺐ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺜﺎﻧﻴﺔ ﻟﺒﻮﻟﺰﺍﻧﻮ-
ﻛﻮﺷﻲ ،ﺍﻟﺪﺍﻟﺔ fﺗﺄﺧﺬ ﻛﻞ ﻗﻴﻢ ﺍﺎﻝ ] ، [m, Mﻭﺑﺎﻟﺘﺎﱄ ﻓﺈﻥ ) Ε( fﻫﻲ ] ، [m, Mﺣﻴﺚ ]. x ∈ [a , b
ﻣﻼﺣﻈﺔ :ﰲ ﺣﺎﻟﺔ
m= M
ﺗﻜﻮﻥ fﺛﺎﺑﺘﺔ.
-5.3ﺍﻻﺳﺘﻤﺮﺍﺭ ﺑﺎﻧﺘﻈﺎﻡ ،ﻣﻌﻴﺎﺭ ﺍﻻﺳﺘﻤﺮﺍﺭ :
-1.5.3ﺍﻻﺳﺘﻤﺮﺍﺭ ﺑﺎﻧﺘﻈﺎﻡ :
ﻣﻦ ﺍﻟﺘﻌﺎﺭﻳﻒ ﺍﳌﻮﺟﻮﺩﺓ ﰲ ﻓﻘﺮﺓ 1.4.3ﻧﺴﺘﻄﻴﻊ ﺃﻥ ﻧﻘﻮﻝ :
f
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﺎﻝ ﻣﺎ Iﻣﻦ ، ℜﺇﺫﺍ ﲢﻘﻖ ﻣﻦ ﺃﺟﻞ ﻛﻞ aﻣﻦ Iﺍﻟﺸﺮﻁ:
x − a < δ ) → f ( x) − f (a ) < ε
/
(∀x ∈ I
/
∀ε > 0, ∃δ > 0
ﻣﻦ ﻫﺬﺍ ﺍﻟﺘﻌﺮﻳﻒ ﻭﺍﺿﺢ ﺃﻧﻪ ﻛﻠﻤﺎ ﺗﻐﲑﺕ aﻣﻦ ، Iﻓﺈﻥ δﺗﺘﻐﲑ ،ﻭﻛﻠﻤﺎ ﺗﻐﲑﺕ εﺃﻳﻀﺎﹰ δﺗﺘﻐﲑ ،ﺃﻱ ﺃﻥ
) ، δ = δ (ε , aﻭﺣﱴ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﻓﻴﻬﺎ εﻣﻌﻄﻰ ) ﺛﺎﺑﺖ ( ﻓﺈﻥ δﺗﺘﻐﲑ ﺑﺘﻐﲑ . a
ﻣﺜﺎﻝ :ﺍﻟﺪﺍﻟﺔ f (x) = 1ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ [∞]0,+
x
ﻻﺣﻆ ﺃﻥ:
ε1 = ε 2
) δ 1 ≠ δ 2 ⇒ δ = δ (a
a1 ≠ a 2
ﻭﺑﺎﻟﺘﺎﱄ ﰲ ﺣﺎﻟﺔ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻳﻜﻮﻥ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ) δ = δ (ε , a
ﻭﺍﻵﻥ ﻳﻄﺮﺡ ﺳﺆﺍﻝ :ﻫﻞ ﺑﺎﻣﻜﺎﻧﻨﺎ ﺇﳚﺎﺩ ، δﻳﻜﻮﻥ ﻣﻼﺋﻤﺎ ﻟﻜﻞ aﻣﻦ ، Iﺃﻱ ) . δ = δ (ε
ﻻﺣﻆ ﰲ ﺣﺎﻟﺔ Iﻣﻨﺘﻪ ﻳﻜﻮﻥ ﻋﺪﺩ δﺑﻌﺪﺩ aﻣﻦ Iﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺃﺻﻐﺮ ﺍﻷﻋﺪﺍﺩ δﻳﻜﻮﻥ ﻣﻼﺋﻤﺎ ﻟﻜﻞ aﻣﻦ . I
ﺃﻣﺎ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ،ﺇﺫﺍ ﻭﺟﺪ δﻟﻜﻞ aﻣﻦ ، Iﻧﻘﻮﻝ ﺇﻥ ﻫﺬﻩ ﺍﻟﺪﺍﻟﺔ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ . I
ﺗﻌﺮﻳﻒ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ ﳎﺎﻝ Iﻣﻦ ، ℜﺇﺫﺍ ﻛﺎﻥ ﻣﻦ ﺃﺟﻞ ﻛﻞ (ε > 0), εﺗﻮﺟﺪ (δ > 0), δ
ﲝﻴﺚ ﻣﻦ ﺍﺟﻞ ﻛﻞ ﻧﻘﻄﺘﲔ x′, x′′ﻣﻦ Iﳛﻘﻘﺎﻥ ﺍﻟﻌﻼﻗﺔ
x′ − x′′ < δ
،ﻳﻜﻮﻥ ، f (x′) − f (x′′) < εﺃﻱ:
∀ε > 0, ∃δ > 0 / (∀x′, x′′ ∈ I / x′ − x′′ < δ ) → f ( x′) − f ( x′′) < ε
ﺃﻱ ﺃﻥ. δ = δ (ε ) :
ﻧﺘﻴﺠﺔ :
ﺃ -ﺍﻻﺳﺘﻤﺮﺍﺭ ﺑﺎﻧﺘﻈﺎﻡ ⇐ ﺍﻻﺳﺘﻤﺮﺍﺭ .
ﺏ -ﻋﻜﺲ -ﺃ -ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ ﻏﲑ ﺻﺤﻴﺢ.
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ﻣﺜﺎﻝ :
ﺃ -ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ
1
x
= )، f ( xﻭ [∞I = [1,+
ﻻﺣﻆ ﺃﻥ:
1
1
x′ − x′′
−
=
≤ x′ − x′′
x′ x′′
x′ ⋅ x′′
ﻭﺑﺎﻟﺘﺎﱄ ﻣﻦ ﺃﺟﻞ ﻛﻞ
ε
= )f ( x′) − f ( x′′
(ε > 0),ﳝﻜﻦ ﺃﺧﺬ ، δ = εﻭﻣﻨﻪ ﻧﻜﺘﺐ:
x′ − x′′ < δ ) → f ( x′) − f ( x′′) < ε
ﻭﺑﺎﻟﺘﺎﱄ
f
/
/ (∀x′, x′′ ∈ I
∀ε > 0, ∃δ = ε
ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ . I
ﺏ -ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ
f
→ ∀x′, x′′ ∈ I
1
x
= )، f ( xﻭ ]I = ]0,1
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ Iﻟﻜﻦ ﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻵﻥ
∃ε 0 > 0 / ∀δ > 0 , (∃x1 x2 ∈ I / x1 − x2 < δ ) / f ( x1 ) − f ( x2 ) > ε 0
ﻭﻫﺬﺍ ﻭﺍﺿﺢ ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﻟﺴﺎﺑﻖ ،ﻭﺍﳌﺘﺮﺍﺟﺤﺔ ﺍﻟﺘﺎﻟﻴﺔ
1 1
x′ − x′′
−
=
≥ x′ − x′′
x′ x′′
x′ ⋅ x′′
ﻧﻈﺮﻳﺔ)ﻧﻈﺮﻳﺔ ﻛﺎﻧﺘﻮﺭ( :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ
f
= )f ( x′) − f (x′′
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﺍﺎﻝ ﺍﳌﻐﻠﻖ ﻭﺍﶈﺪﻭﺩ ] [a, bﻓﺈﺎ ﺗﻜﻮﻥ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ
ﻋﻠﻴﻪ.
ﺍﻟﱪﻫﺎﻥ :ﻧﻔﺮﺽ ﺃﻥ fﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ ] ، [a, bﺃﻱ:
∃ε 0 > 0 / ∀δ > 0, (∃x′, x′′ ∈ [a , b ] / x′ − x′′ < δ ) / f ( x′) − f ( x′′) ≥ ε 0
ﻻﺣﻆ ﻣﻦ ﺃﺟﻞ
1
n
= n ∈ Ν ,δ
ﻳﻮﺟﺪ x′n′, x′nﳛﻘﻘﺎﻥ :
)∗(
ﻟﻜﻦ
x′n − x′n′ < δ
(∗)′
ﻣﻦ ]، [a, bﻫﺬﺍ ﻳﻌﲎ ﺃﻥ ﻛﻞ ﻣﻦ ﺍﳌﺘﺘﺎﻟﻴﺘﲔ , ( xn′ )n≥1
f ( x′n ) − f ( xn′′ ) ≥ ε 0
n ≥ 1 , x′n′ , x′n
(xn′′ )n≥1ﳏﺪﻭﺩﺓ.
ﻧﺄﺧﺬ ﺍﳌﺘﺘﺎﻟﻴﺔ (x′n )n≥1ﺍﶈﺪﻭﺩﺓ ،ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺑﺎﻟﺰﺍﻧﻮ-ﻓﺸﺘﺮﺍﺹ ،ﻫﺬﻩ ﺍﳌﺘﺘﺎﻟﻴﺔ ﲤﻠﻚ ﻣﺘﺘﺎﻟﻴﺔ ﺟﺰﺋﻴﺔ
ﲟﺎ ﺃﻥ ، n ≥ 1 , a ≤ x′k ≤ bﻓﺈﻧﻪ:
)* *(
∃ξ ∈ [a , b ] / ξ = lim xk′
∞→ n
)
n ≥1
(xk′ﻣﺘﻘﺎﺭﺑﺔ.
n
n
n
ﻋﻨﺪﻧﺎ ﺍﳌﺘﺮﺍﺟﺤﺔ
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)* * *(
x′k′n − ξ = xk′′n − x′kn + x′kn − ξ ≤ xk′′n − x′kn + xk′ n − ξ
ﳏﻘﻘﺔ
ﻣﻦ )*( ﻭ )* *( ﳓﺼﻞ ﺑﺈﺩﺧﺎﻝ ﺍﻟﻨﻬﺎﻳﺔ ﻋﻠﻰ )* * *( ،ﻋﻠﻰ:
lim
x′k′ = ξ
∞→n
n
ﲟﺎ ﺃﻥ
ﻣﺴﺘﻤﺮﺓ ﰲ ﺍﻟﻨﻘﻄﺔ ، ξﻓﺈﻧﻪ ﺣﺴﺐ ﺗﻌﺮﻳﻒ ﻗﺎﻳﻦ ﻟﻼﺳﺘﻤﺮﺍﺭ ﻳﻜﻮﻥ :
) lim f (x′k′ ) = lim f (x′k ) = f (ξ
∞→ n
∞→ n
f
n
ﻫﺬﺍ ﻳﻌﲏ
)) ( ) ( (
ﺃﻥ lim f x′k′n − f x′kn = 0
∞→ n
n
،ﻭﻫﻮ ﻣﻨﺎﻑ ﻟﻜﻮﻥ
) ( ) (
∀n ≥ 1 → f xk′ n − f xk′′n ≥ ε 0
ﻫﺬﺍ ﲜﻌﻞ knﰲ ﻣﻜﺎﻥ nﰲ ﺍﻟﺼﻴﻐﺔ . (∗)′
ﻭﺑﺎﻟﺘﺎﱄ ﺍﻟﻔﺮﺽ ﻏﲑ ﺻﺤﻴﺢ ،ﺃﻱ ﺃﻥ
f
ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ ]. [a, b
ﻧﺘﻴﺠﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ]، [a, bﻓﺈﻧﻪ ﻣﻦ ﺃﺟﻞ ﻛﻞ
(ε > 0 ), εﻳﻮﺟﺪ > 0 ), δ
(δﲝﻴﺚ ﻳﻜﻮﻥ ﺗﺬﺑﺬﺏ
ﺍﻟﺪﺍﻟﺔ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﻣﻐﻠﻮﻕ ﳏﺘﻮﻯ ﰲ ﺍﺎﻝ ] [a, bﻭﻃﻮﻟﻪ ﺃﻗﻞ ﻣﻦ ، δﺃﻗﻞ ﻣﻦ . ε
ﺍﻟﺘﺬﺑﺬﺏ ﻋﻠﻰ ﻛﻞ ﺍﺎﻝ ] [a, bﻳﻌﺮﻑ ﻛﺎﻟﺘﺎﱄ w = M − mﺣﻴﺚm = inf f ( x) , M = sup f ( x) :
] x∈[a ,b
] x∈[a ,b
ﺍﻟﱪﻫﺎﻥ :ﻭﺍﺿﺢ ﺃﻥ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﻣﻐﻠﻖ ﳏﺘﻮﻯ ﰲ ﺍﺎﻝ ] [a, bﺗﻜﻮﻥ fﻣﺴﺘﻤﺮﺓ .
ﺗﺄﺧﺬ ﳎﺎﻝ ﻛﻴﻔﻲ ] [a 1 ,b1ﻣﻦ ] [a, bﺣﻴﺚ b1 − a 1 < δ
f
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] [a 1 ,b1ﻳﻌﲏ
)∃x′ ∈ [a 1 , b1 ] / f ( x′) = m1 = inf f ( x
] x∈[a1 ,b1
)∃x′′ ∈ [a 1 , b1 ] / f ( x′′) = M 1 = sup f ( x
] x∈[a1 ,b1
ﻭﺍﺿﺢ ﺃﻥ
f
x′ − x′′ < δ
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ] ، [a 1 ,b1ﺃﻱ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻴﻪ ،ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ ، f (x′) − f (x′′) < εﻭﻣﻨﻪ
ﻳﻜﻮﻥw = M − m :
-2.5.3ﻣﻌﻴﺎﺭ ﺍﻻﺳﺘﻤﺮﺍﺭ :
ﻟﺘﻜﻦ
f
ﻣﻌﺮﻓﺔ ﻭ ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳎﺎﻝ Iﻣﻦ . ℜ
ﺗﻌﺮﻳﻒ :ﻣﻦ ﺃﺟﻞ ﻛﻞ
> 0 ), δ
(δﻳﻌﺮﻑ ﻣﻌﻴﺎﺭ ﺍﺳﺘﻤﺮﺍﺭ ﺍﻟﺪﺍﻟﺔ fﻋﻠﻰ ، Iﺑﺄﻧﻪ ﺍﳊﺪ ﺍﻷﻋﻠﻰ
ﻟﻸﻋﺪﺍﺩ )f ( x′) − f ( x′′
ﺣﻴﺚ x′′, x′ﻣﻦ Iﲢﻘﻖ . x′ − x′′ ≤ δ
ﻳﺮﻣﺰ ﳌﻌﻴﺎﺭ ﺍﺳﺘﻤﺮﺍﺭ ﺍﻟﺪﺍﻟﺔ fﻋﻠﻰ Iﺑﺎﻟﺮﻣﺰ ) ، w( f ; I , δﺃﻭ ) ، w( f ; δﻭﻧﻜﺘﺐ:
)*(
} w( f ; δ ) = sup{ f ( x′) − f ( x′′) / x′ − x′′ ≤ δ , x′, x′′ ∈ I
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ﺗﻨﺒﻴﻪ :ﰲ ﺍﻟﺼﻴﻐﺔ )∗( ﳝﻜﻦ ﺍﻻﺳﺘﻐﻨﺎﺀ ﻋﻦ ﺍﻟﻘﻴﻤﺔ ﺍﳌﻄﻠﻘﺔ ،ﺃﻱ ﻧﻜﺘﺐ:
} w( f ; δ ) = sup{( f ( x′) − f ( x′′)) / x′ − x′′ ≤ δ , x′, x′′ ∈ I
ﺧﻮﺍﺹ ﻣﻌﻴﺎﺭ ﺍﻻﺳﺘﻤﺮﺍﺭ :
-1ﻣﻌﻴﺎﺭ ﺍﻻﺳﺘﻤﺮﺍﺭ ﺩﻭﻣﺎﹰ ﻣﻮﺟﺐ ،ﺃﻱ. w( f ;δ ) ≥ 0 :
w( f ; δ ) -2ﺩﺍﻟﺔ ﻟﻴﺴﺖ ﻣﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ﻧﺼﻒ ﺍﳌﺴﺘﻘﻴﻢ . δ > 0
ﻣﺜﺎﻝ :ﻟﺘﻜﻦ fﺩﺍﻟﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﺍﺎﻝ ]، [0,1ﻛﺎﻟﺘﺎﱃ:
ﺣﺴﺎﺏ ﻣﻌﻴﺎﺭ ﺍﺳﺘﻤﺮﺍﺭ ﺍﻟﺪﺍﻟﺔ
f
f ( x) = x 2
.
} ]w( f ; δ ) = sup{( f (x′) − f ( x′′)) / x′ − x′′ ≤ δ , x′, x′′ ∈ [0,1
ﰲ ﺣﺎﻟﺔ ، δ = 1ﻭﺍﺿﺢ ﺃﻥ
ﰲ ﺣﺎﻟﺔ : 0 < δ < 1ﻋﻨﺪﻧﺎ
w( f ;1) = 1
0 ≤ x′′ − δ ≤ x′ ≤ x′′ < 1 ⇔ x′ − x′′ ≤ δ
ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
2
x′′ 2 − x′ 2 ≤ x′′ 2 − ( x′′ − δ ) = 2 x′′δ − δ 2 ≤ 2δ − δ 2
ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ:
)∗( (x′ 2 − x′′ 2 ) ≤ 2δ − δ 2
w( f ; δ ) = sup{( f (x′) − f ( x′′)) / x′ − x′′ ≤ δ , x′, x′′ ∈ [0,1] } = sup
x′− x′′ ≤δ
ﻻﺣﻆ ﰲ ﺍﳊﺎﻟﺔ ﺍﳋﺎﺻﺔ ﻋﻨﺪﻣﺎ ، x′ = 1 − δ , x′′ = 1ﻳﻜﻮﻥ:
)* *(
(x′′2 − x′2 ) = 1 − (1 − δ )2 = 2δ − δ 2
ﻣﻦ )*( (* *) ,ﻳﻜﻮﻥ:
ﺍﻟﱪﻫﺎﻥ :
x′′− x′ ≤δ
w( f , δ ) = 2δ − δ 2
ﻧﻈﺮﻳﺔ :ﺗﻜﻮﻥ fﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ ﺍﺎﻝ ، Iﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ :
f
w( f , δ ) = sup
lim w( f , δ ) = 0
δ → +0
ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ،ﺃﻱ :
∀ε > 0, ∃δ ε > 0, (∀x′, x′′ ∈ I / x′ − x′′ < δ ε ) → f ( x′) − f (x′′) < ε
)(1
ﺍﻟﺼﻴﻐﺔ ) (1ﳝﻜﻦ ﻛﺘﺎﺑﺘﻬﺎ ﻛﺎﻟﺘﺎﱄ :
)(2
ε
2
< )∀ε > 0, ∃δ ε > 0 / (∀x′, x′′ ∈ I / x′ − x′′ < δ ε ) → f ( x′) − f ( x′′
ﻣﻦ ) (2ﻧﺴﺘﻨﺘﺞ ﺃﻧﻪ ﻣﻦ ﺃﺟﻞ ﻛﻞ ، δﺣﻴﺚ 0 < δ < δ εﻳﺘﺤﻘﻖ ﻣﺎ ﻳﻠﻲ:
ﺃﻱ ﺃﻥ:
ε
<ε
2
≤ } , x′, x′′ ∈ I
w( f , δ ) = sup{ f ( x′) − f (x′′) / x′ − x′′ ≤ δ
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∀ε > 0, ∃δ ε > 0 / ∀δ < δ ε , (δ > 0 ) → w( f , δ ) < ε
ﻭﺑﺎﻟﺘﺎﱄ :
lim w( f , δ ) = 0
]⇒[ ﻟﺘﻜﻦ
δ → +0
lim w( f , δ ) = 0
δ → +0
،ﺃﻱ:
∀ε > 0, ∃δ ε > 0 / ∀δ < δ ε , (δ > 0 ) → w( f , δ ) < ε
ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ:
∀x′, x′′ ∈ I / x′ − x′′ ≤ δ < δ ε → f ( x′) − f ( x′′) < ε
ﻭﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ :
∀ε > 0, ∃δ ε > 0 / (∀x′, x′′ ∈ I / x′ − x′′ < δ ε ) → f ( x′) − f ( x′′) < ε
ﻭﺑﺎﻟﺘﺎﱄ fﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ . I
ﻣﺜﺎﻝ :
ﺃ -ﻋﺮﻓﻨﺎ ﰲ ﺍﳌﺜﺎﻝ ﻓﻘﺮﺓ ، 2.5.3ﺃﻥ ، w( f , δ ) = 2δ − δ 2ﺣﻴﺚ f (x) = x2ﻣﺄﺧﻮﺫﺓ ﻋﻠﻰ ﺍﺎﻝ ]. [0,1
. δlimﻭﺑﺎﻟﺘﺎﱄ ﺍﻟﺪﺍﻟﺔ f (x) = x 2ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ ]. [0,1
ﻭﺍﺿﺢ ﺃﻥ w( f , δ ) = 0
→ +0
ﺏ -ﻧﺄﺧﺬ ﺍﻟﺪﺍﻟﺔ f (x) = x2ﻋﻠﻰ ﺍﺎﻝ [∞. ]− ∞,+
ﺗﺄﻛﺪ ﻣﻦ ﺃﻥ ∞. − ∞ < x < +∞ , w( f , δ ) = +
، δlimﺃﻱ ﺍﻟﺪﺍﻟﺔ fﻟﻴﺴﺖ ﻣﺴﺘﻤﺮﺓ ﺑﺎﻧﺘﻈﺎﻡ ﻋﻠﻰ [∞. ]− ∞,+
ﻭﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ∞w( f , δ ) = +
→ +0
-6.3ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ :
ﻣﻔﻬﻮﻡ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ :
ﻟﺘﻜﻦ
f
ﺩﺍﻟﺔ ﻋﺪﺩﻳﺔ
،ﺃﻱ x ∈ D(ℜ) ⊂ ℜ
، y = f (x) ,ﺣﻴﺚ ﻣﻦ ﺃﺟﻞ
ﻛﻞ x0
ﻣﻦ ) D( fﻳﻮﺟﺪ y0ﻭﺣﻴﺪ ﳛﻘﻖ
) y0 = f ( x0 ) ∈ Ε( f
ﻻﺣﻆ ﺃﻧﻪ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﻌﺎﻣﺔ
ﺍﳌﻌﺎﺩﻟﺔ ، f ( x) = y0ﺣﻴﺚ ) y0 ∈ Ε( f
،ﲤﻠﻚ ﺃﻛﺜﺮ ﻣﻦ ﺣﻞ.
ﻣﺜﺎﻝ :
ﺃ -ﻣﻦ ﺃﺟﻞ ( y0 > 0) ، y0ﺛﺎﺑﺖ ﻳﻜﻮﻥ ﻛﻞ ﻣﻦ x2 = − y0 ، x1 = y0ﺣﻼ ﻟﻠﻤﻌﺎﺩﻟﺔ ، f (x) = y0ﺣﻴﺚ . f (x) = x2
ﻻﺣﻆ ﻟﻮ ﻧﺄﺧﺬ ﺍﻟﺪﺍﻟﺔ f (x) = x2ﰲ ﺍﺎﻝ [ ، ]− ∞,0ﺃﻭ ﺍﺎﻝ [∞ [0,+ﻳﻜﻮﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ f (x) = y0ﺣﻞ ﻭﺣﻴﺪ ﻫﻮ
، y0ﺃﻭ − y0ﻋﻠﻰ ﺍﻟﺘﻮﺍﱄ.
ﺏ -ﻟﺘﻜﻦ ﺍﻟﺪﺍﻟﺔ ، f ( x) = x + 1ﺣﻴﺚ D( f ) = ℜ
ﻭﺍﺿﺢ ﺃﻧﻪ ﻣﻦ ﺃﺟﻞ ﻛﻞ y0ﻳﻮﺟﺪ x0ﻭﺣﻴﺪ،ﺣﻴﺚ . (x0 = y0 − 1) ، f (x0 ) = y0
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ﺗﻌﺮﻳﻒ :
ﺃ -ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﻓﻴﻬﺎ ﻟﻠﻤﻌﺎﺩﻟﺔ y0 ∈ Ε( f ) ، f (x) = y0ﺃﻛﺜﺮ ﻣﻦ ﺣﻞ ،ﻳﺴﻤﻲ ﺍﻟﻘﺎﻧﻮﻥ ﺍﻟﺬﻱ ﻳﺮﻓﻖ ﺑﻜﻞ y
ﻣﻦ ) x ، Ε( fﻣﻦ ) D( fﺑﺎﻟﺼﻮﺭﺓ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ، fﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ، fﻭﻧﻜﺘﺐ:
) f : Ε( f ) → D ( f
y → f ( y) = x
ﺍﻟﺼﻮﺭﺓ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ fﺗﺴﻤﻰ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ﻟﻠﺪﺍﻟﺔ . f
ﺏ -ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﻓﻴﻬﺎ ﻟﻠﻤﻌﺎﺩﻟﺔ y0 ∈ Ε( f ) ، f (x) = y0ﺣﻞ ﻭﺣﻴﺪ ،ﻳﺴﻤﻰ ﺍﻟﻘﺎﻧﻮﻥ ﺍﻟﺬﻱ ﻳﺮﻓﻖ ﺑﻜﻞ yﻣﻦ
) x ، Ε( fﻭﺣﻴﺪ ﻣﻦ ) D( fﺑﺎﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ، fﻭﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ . f −1
) f −1 : Ε ( f ) → D ( f
y → f −1 ( y) = x
ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﰲ ﺍﻷﺻﻞ،ﺗﺴﻤﻰ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻭﺣﻴﺪﺓ ﺍﻟﻘﻴﻤﺔ ﻟﻠﺪﺍﻟﺔ . f
ﺧﺼﺎﺋﺺ :
fﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ . f −1
ﺃ-
) D ( f −1 ) = Ε ( f ) , Ε ( f −1 ) = D ( f
ﺏ-
ﺝ-
f ( f −1 ( x)) = x ، ∀x ∈ D( f ) → f −1 ( f ( x)) = x
ﺩ -ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ f −1ﻣﺘﻨﺎﻇﺮ ﻣﻊ ﺑﻴﺎﻥ fﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻨﺼﻒ ﺍﻷﻭﻝ.
ﻧﺘﻴﺠﺔ :ﻣﻦ ﺍﻟﺘﻌﺮﻳﻒ ﺏ -ﻟﻀﻤﺎﻥ ﻭﺟﻮﺩ ﺍﻟﺪﺍﻟﺔ
ﻧﻈﺮﻳﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﺍﻟﺔ
ﺍﻟﻌﻜﺴﻴﺔ f −1
f : D ( f ) → Ε( f ) ، f
→ ) ∀x ∈ Ε( f
،ﻧﺸﺘﺮﻁ ﺃﻥ ﺗﻜﻮﻥ fﺭﺗﻴﺒﺔ ﲤﺎﻣﺎﹰ ﻋﻠﻰ ) . D( f
ﺭﺗﻴﺒﺔ ﲤﺎﻣﺎﹰ ﻭﻣﺴﺘﻤﺮﺓ ،ﻓﺈﻧﻪ ﺗﻮﺟﺪ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻌﻜﺴﻴﺔ ، f −1
) ، f −1 : Ε( f ) → D( fﻭﺗﻜﻮﻥ ﺭﺗﻴﺒﺔ ﲤﺎﻣﺎﹰ)،ﺍﻟﺮﺗﺎﺑﺔ ﻣﻦ ﻧﻔﺲ ﺭﺗﺎﺑﺔ ،( fﻭﻣﺴﺘﻤﺮﺓ.
ﺍﻟﱪﻫﺎﻥ :ﻧﺄﺧﺬ ﺍﳊﺎﻟﺔ ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ fﻣﺘﺰﺍﻳﺪﺓ.
ﺍﻟﺪﺍﻟﺔ f −1ﻣﻮﺟﻮﺩﺓ ﺣﺴﺐ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺴﺎﺑﻘﺔ .
ﻧﱪﻫﻦ ﺃﻥ ) f −1 : Ε( f ) → D( fﻣﺘﺰﺍﻳﺪﺓ ﺃﻳﻀﺎﹰ ،ﺃﻱ ﻧﱪﻫﻦ :
)*(
ﻧﻀﻊ ) f −1 ( y2 ), x1 = f −1 ( y1
ﻧﻔﺮﺽ ﺃﻥ . x1 ≥ x2
) ∀y1 , y2 ∈ Ε( f ) , y1 < y2 ⇒ f −1 ( y1 ) < f −1 ( y2
= x2
ﻋﻨﺪﻧﺎ fﻣﺘﺰﺍﻳﺪﺓ ﻳﻌﲏ ) ، f (x1 ) ≥ f (x2ﺃﻱ ، y1 ≥ y2ﻭﻫﺬﺍ ﻳﻨﺎﰲ )∗( ،ﻭﻣﻨﻪ ، x1 < x2ﺃﻱ:
ﻧﱪﻫﻦ ﺃﻥ ) f −1 : Ε( f ) → D( fﻣﺴﺘﻤﺮﺓ ،ﺃﻱ ﻣﻦ ﺃﺟﻞ ﻛﻞ y0ﻛﻴﻔﻴﺔ ﻣﻦ ) Ε( fﻳﺘﺤﻘﻖ :
) f −1 ( y2
< ) . f −1 ( y1
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∀ε > 0, ∃δ > 0 / (∀y ∈ Ε( f ) / y − y0 < δ ) → f −1 ( y) − f −1 ( y0 ) < ε
ﻋﻨﺪﻧﺎ fﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ) D( fﻫﺬﺍ ﻳﻌﲏ ﺃﻥ ) Ε( fﻋﺒﺎﺭﺓ ﻋﻦ
ﳎﺎﻝ ﺎﻳﺘﻪ Mﻭ ، mﺣﻴﺚ:
)m = inf f ( x), M = sup f ( x
) D( f
) D( f
ﰲ ﺣﺎﻟﺔ ) D( fﳎﺎﻝ ﻣﻐﻠﻖ ﺗﻜﻮﻥ
ﺃﻧﻈﺮ ﻧﻈﺮﻳﺔ ﺑﻮﻟﺰﺍﻧﻮ-ﻛﻮﺷﻲ .
] Ε( f ) = [m, M
ﰲ ﺣﺎﻟﺔ ) D( fﳎﺎﻝ ﻛﻴﻔﻲ ﻳﻜﻮﻥ ) Ε( fﺃﻳﻀﺎﹰ ﳎﺎﻝ ﻛﻴﻔﻲ
ﺎﻳﺘﻪ . m, M
ﺍﳊﺎﻟﺔ [ ، Ε( f ) = ]m, Mﺃﻱ ﺃﻥ ) D( fﳎﺎﻝ ﻣﻔﺘﻮﺡ ،
ﻻﺣﻆ ﻣﻦ ﺃﺟﻞ ﻛﻞ ) y0 ∈ Ε( fﺗﻮﺟﺪ x0ﻣﻦ ) ، D( fﺣﻴﺚ ) f ( x0
= . y0
ﺣﺴﺐ ﺧﺎﺻﻴﺔ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻟـ ، ℜﻣﻦ ﺃﺟﻞ ﻛﻞ (ε > 0) , εﻳﻜﻮﻥ :
) . x0 − ε ∈ D( f ), x0 + ε ∈ D( f
ﻧﻀﻊ ) y2 = f ( x0 + ε ), y1 = f ( x0 − ε
ﲟﺎﺃﻥ fﻣﺘﺰﺍﻳﺪﺓ ،ﻓﺈﻥ . y1 < y0 < y2 :
ﻧﺄﺧﺬ ﺍﻵﻥ δ > 0ﳛﻘﻖ δ )، y1 ≤ y0 − δ , y0 + δ ≤ y2ﻣﻮﺟﻮﺩ ﺣﺴﺐ ﺧﺎﺻﻴﺔ ﺍﻻﺳﺘﻤﺮﺍﺭ ﻟـ .( ℜ
ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻣﻦ ﺃﺟﻞ ﻛﻞ yﳛﻘﻖ ، y0 − δ < y < y0 + δﻳﻜﻮﻥ y1 < y < y2 :
ﲟﺎ ﺃﻥ f −1ﻣﺘﺰﺍﻳﺪﺓ ﻓﺈﻥf −1 ( y1 ) < f −1 ( y) < f −1 ( y2 ) :
ﲟﺎ ﺃﻥ
f −1 ( y1 ) = x0 − ε = f −1 ( y0 ) − ε
f −1 ( y 0 ) + ε ,
= ، f −1 ( y2 ) = x0 + εﻓﺈﻥ :
f −1 ( y0 ) − ε < f −1 ( y) < f −1 ( y0 ) + ε
ﻭﻫﺬﺍ ﻣﻦ ﺍﺟﻞ ﻛﻞ yﲢﻘﻖ
y0 − δ < y < y0 + δ
.ﺃﻱ ﺃﻥ:
∀ε > 0 , ∃δ > 0 / (∀ ∈ Ε( f ) / y − y0 < δ ) → f −1 ( y) − f −1 ( y0 ) < ε
ﲟﺎ ﺃﻥ y0ﻛﻴﻔﻴﺔ ﻣﻦ ) ، Ε( fﻓﺈﻥ f −1ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ) . Ε( f
ﰲ ﺣﺎﻟﺔ ﻭﺟﻮﺩ ﻋﻠﻰ ﺍﻷﻗﻞ mﺃﻭ Mﰲ ) ، Ε( fﻳﻜﻔﻲ ﺃﻥ ﻧﱪﻫﻦ ﺃﻥ
ﻫﺬﺍﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ.
f −1
ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﳝﲔ ، mﻭﻋﻠﻰ ﻳﺴﺎﺭ ، M
ﺗﻌﺮﻳﻒ :ﺍﻟﺘﻄﺒﻴﻖ fﻣﻦ ﳎﺎﻝ Iﰲ ﳎﺎﻝ Jﻣﻦ ، ℜﻧﻘﻮﻝ ﺇﻧﻪ ﻫﻮﻣﻴﻮﻣﻮﺭﻓﻴﺰﻡ ،ﺇﺫﺍ ﲢﻘﻖ ﻣﺎ ﻳﻠﻲ :
ﺃ f -ﺗﻘﺎﺑﻞ
ﺏ f -ﻭ f −1ﻣﺴﺘﻤﺮﺍﻥ.
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ﻧﺘﻴﺠﺔ :
ﺃ -ﺇﺫﺍ ﻛﺎﻧﺖ fﻫﻮﻣﻴﻮﻣﻮﺭﻓﻴﺰﻡ ،ﻓﺈﻥ f −1ﻛﺬﻟﻚ.
ﺏ -ﺇﺫﺍ ﻛﺎﻧﺖ fﺭﺗﻴﺒﺔ ﲤﺎﻣﺎﹰ ﻭﻣﺴﺘﻤﺮﺓ ،ﻓﺈﻥ fﻫﻮﻣﻴﻮﻣﻮﺭﻓﻴﺰﻡ ﻣﻦ ) D( fﻋﻠﻰ ) . Ε( f
wﳎﻤﻮﻋﺘﺎﻥ ﺑﻴﻨﻬﻤﺎ ﻫﻮﻣﻴﻮﻣﻮﺭﻓﻴﺰﻡ ﻧﺴﻤﻴﻬﻤﺎ ﻫﻮﻣﻴﻮﻣﻮﺭﻓﻴﲔ.
-7.3ﺍﻟﺪﻭﺍﻝ ﺍﻟﺒﺴﻴﻄﺔ ﻭﺩﻭﺍﳍﺎ ﺍﻟﻌﻜﺴﻴﺔ :
-1.7.3ﺍﻟﺪﻭﺍﻝ ﺍﻵﺳﻴﺔ :
ﺗﻌﺮﻳﻒ :ﻣﻦ ﺃﺟﻞ ﻛﻞ (a > 0) ، aﻣﻌﻄﻰ ﺍﻟﺪﺍﻟﺔ f (x) = a xﺍﳌﻌﺮﻓﺔ ﻋﻠﻰ ﻛﻞ ℜﺗﺴﻤﻰ ﺩﺍﻟﺔ ﺁﺳﻴﺔ ﺫﺍﺕ ﺍﻷﺳﺎﺱ . a
ﰲ ﺣﺎﻟﺔ
a =1
ﺗﻜﻮﻥ . f (x) = 1
ﺧﺼﺎﺋﺺ :
ﺃ، (a < 1) ، a > 1 -ﺍﻟﺪﺍﻟﺔ fﻣﺘﺰﺍﻳﺪﺓ )،ﻣﺘﻨﺎﻗﺼﺔ(.
ﺏ، (a x )y = a x⋅ y -ﻭ . ∀x, y ∈ ℜ, a x a y = a x+ y
ﺝ f (x) = a x -ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ﻛﻞ . ℜ
ﻧﺘﺎﺋﺞ :
-1
-2
a > 0, ∀x ∈ ℜ → a x > 0
a > 0 , b > 0, ∀x ∈ ℜ → (ab ) = a xb x
x
-2.7.3ﺍﻟﺪﻭﺍﻝ ﺍﻟﻠﻮﻏﺎﺭﲤﻴﺔ :
ﺣﺴﺐ ﻧﻈﺮﻳﺔ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ
ﰲ ﺣﺎﻟﺔ a > 0ﺍﻟﺪﺍﻟﺔ ﺍﻵﺳﻴﺔ f (x) = a xﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ
ﻋﻠﻰ ﻛﻞ ℜﻭﳎﻤﻮﻋﺔ ﻗﻴﻤﻬﺎ ﻫﻲ [∞، ]0,+ﻭﺑﺎﻟﺘﺎﱄ ﻋﻠﻰ ﺍﺎﻝ
[∞ ]0,+ﺗﻌﺮﻑ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ﺍﻵﺳﻴﺔ ﻭﺗﻜﻮﻥ ﻣﺘﺰﺍﻳﺪﺓ
ﻭﻣﺴﺘﻤﺮﺓ .
ﺗﺴﻤﻰ ﻫﺬﻩ ﺍﻟﺪﺍﻟﺔ ﺑﺎﻟﺪﺍﻟﺔ ﺍﻟﻠﻮﻏﺎﺭﲤﻴﺔ ﺫﺍﺕ ﺍﻷﺳﺎﺱ ، a
ﻭﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ y = log a x
ﰲ ﺣﺎﻟﺔ ، 0 < a < 1ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ y = log a xﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ .
ﺧﺼﺎﺋﺺ :
ﺃ-
ﺏ-
x > 0 , y > 0 , log a xy = log a x + log a y
x > 0 , α ∈ ℜ , log a xα = α log a x
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ﺝ-
, a log a x = x
log b x
log b a
ﺩ-
ﺣﺎﻟﺔ ﺧﺎﺻﺔ
:ﺣﺎﻟﺔ a = e
x>0
= x > 0 , b > 0 , log a x
ﺗﺴﻤﻰ ﺍﻟﺪﺍﻟﺔ
log e x
ﺑﺎﻟﺪﺍﻟﺔ ﺍﻟﻠﻮﻏﺎﺭﻳﺘﻢ ﺍﻟﻨﻴﱪﻱ ﻭﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ . ln x
-3.7.3ﺍﻟﺪﺍﻟﺔ ﺍﻟﻘﻮﺓ :
ﺗﻌﺮﻳﻒ :ﻣﻦ ﺃﺟﻞ ﻛﻞ ﻋﺪﺩ αﺍﻟﺪﺍﻟﺔ f (x) = xαﺍﳌﻌﺮﻓﺔ ﻣﻦ ﺃﺟﻞ
x>0
ﺗﺴﻤﻰ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻘﻮﺓ ﺫﺍﺕ ﺍﻷﺱ . α
ﻣﻼﺣﻈﺔ α = 0 :ﺗﻜﻮﻥ . f (x) = 1
ﺧﺼﺎﺋﺺ :
ﺃ α < 0 -ﺍﻟﺪﺍﻟﺔ
ﺏ α > 0 -ﺍﻟﺪﺍﻟﺔ
f
f
ﻣﺘﻨﺎﻗﺼﺔ .
ﻣﺘﺰﺍﻳﺪﺓ .
ﺝ f -ﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ [∞. ]0,+
ﺩ ، x = 0 ، α > 0 -ﺗﻜﻮﻥ . f (x) = 0
-4.7.3ﺍﻟﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ ﻭﺩﻭﺍﳍﺎ ﺍﻟﻌﻜﺴﻴﺔ :
ﻟﺘﻜﻦ x 2 + y 2 = r 2ﻣﻌﺎﺩﻟﺔ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ) (0,0ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ، (r > 0 ), rﺍﺳﺘﻨﺎﺩﺍ ﺇﱃ ﺍﻟﺸﻜﻞ ﺍﺎﻭﺭ،ﺗﻌﺮﻑ
ﺍﻟﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ ﻛﺎﻟﺘﺎﱄ :
η
r
ξ
= f ( x) = cos x
r
η sin x
= = f ( x) = tgx
ξ cos x
ξ cos x
= = f ( x) = ctgx
η sin x
r
1
= = f ( x) = sec x
ξ cos x
r
1
= = f ( x) = cos ec
η sin x
= f ( x) = sin x
ﰲ ﺣﺎﻟﺔ r = 1ﺗﻜﻮﻥ :
cos x = ξ = OB , sin x = η = BC
η BC
=
= AE
ξ OB
ﺗﻨﺒﻴﻪ x :ﻻ ﺗﻔﻬﻢ ﻓﻘﻂ ﻋﻠﻰ ﺃﺎ
= tgx
ﺍﻟﺰﺍﻭﻳﺔ AOC
,
ξ OB
=
= MN
η BC
= ctg
ﺑﻞ ﻧﻔﻬﻢ ﺃﻳﻀﺎﹰ ﺿﻌﻒ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ ، OCAﺫﻟﻚ ﻷﻥ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ
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. S = 1 AC ⋅ r : ﺗﻌﻄﻰ ﺑﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ
2
. S = 1 r 2 x ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥ
2
r ⋅ x = AC
: ﻫﻲx ﻭﺍﻟﺰﺍﻭﻳﺔ
AC
S
ﻣﻌﻠﻮﻡ ﺃﻥ ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﺍﻟﻘﻮﺱ
. x = 2.S ﻳﻜﻮﻥ،()ﺩﺍﺋﺮﺓ ﺍﻟﻮﺣﺪﺓ، r = 1 ﻧﻀﻊ
: ﺃﻫﻢ ﺧﺼﺎﺋﺺ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ
sin x
π π
x ∈ − , , x ≠ 0 , cos x <
<1
x
2 2
-1
sin ( x ± y) = sin x cos y ± cos x sin y
-2
-3
cos( x ± y) = cos x cos y m sin x sin y
tgx ± tgy
ctgxctgy m 1
tg ( x ± y) =
, ctg ( x ± y) =
ctgy ± ctgx
1 m tgxtgy
-4
∀x ∈ ℜ,
sin x ≤ x
: ﻫﻨﺎﻙ ﻋﺪﺓ ﺧﺼﺎﺋﺺ ﺃﺧﺮﻯ ﳒﻤﻠﻬﺎ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ-5
=
f ( x)
ctgx
tgx
cos x
sin x
− ∞ < x < +∞
− ∞ < x < +∞
π
x ≠ + kπ
2
− ∞ < x < +∞
− ∞ < x < +∞
D( f )
− 1 ≤ f ( x) ≤ 1
Ε( f )
2π
π
x = + kπ
2
ﻧﻘﺎﻁ ﺍﻟﻘﻴﻢ ﺍﻟﻘﺼﻮﻯ
x = kπ
ﻧﻘﺎﻁ ﺍﻹﻧﻌﻄﺎﻑ
x ≠ kπ
− ∞ < f ( x) < +∞
π
x = + kπ
2
− ∞ < f ( x) < +∞
x = kπ
− 1 ≤ f ( x) ≤ 1
π
x = + kπ
2
π
π
2π
ﻻ
ﻻ
x = kπ
x=
π
+ kπ
2
x = kπ
x=
π
+ kπ
2
x = kπ
f
ﺃﺻﻔﺎﺭ
ﺍﻟﺪﻭﺭ
ﺯﻭﺟﻴﺔ
ﻓﺮﺩﻳﺔ
ﺯﻭﺟﻴﺔ
ﻓﺮﺩﻳﺔ
ﺍﻟﻨﻮﻋﻴﺔ
ﻻ
ﻻ
ﻻ
ﻻ
ﺍﻟﺮﺗﺎﺑﺔ
ﻻ
ﻻ
ﻧﻌﻢ
ﻧﻌﻢ
ﺍﶈﺪﻭﺩﻳﺔ
ﻧﻌﻢ
ﻧﻌﻢ
ﻧﻌﻢ
ﻧﻌﻢ
ﺍﻻﺳﺘﻤﺮﺍﺭﻳﺔ
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ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ :
-1ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ
ﻋﻨﺪﻧﺎ
D( f ) = ℜ
f ( x) = sin x
Ε( f ) = [− 1,1] ,
ﻻﺣﻆ ﻣﻦ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ﻳﻨﺘﺞ ﺃﻥ ﻟﻜﻞ y0ﻣﻦ ) Ε( fﻳﻜﻮﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ، f (x) = y0ﻣﺎ ﻻﺎﻳﺔ ﻣﻦ ﺍﳊﻠﻮﻝ ،ﻭﺑﺎﻟﺘﺎﱄ ﻧﻘﻮﻝ ﺇﻥ
ﻟﻠﺪﺍﻟﺔ fﻻ ﺗﻮﺟﺪ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﺗﻌﺮﻳﻔﻬﺎ ،ﺑﻞ ﻧﻘﻮﻝ ﺇﻥ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ﺃﻭ ﻛﻤﺎ ﲰﻴﻨﻬﺎ ﺳﺎﺑﻘﺎﹰ
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ﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﺗﻌﺮﻳﻔﻬﺎ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ
Arc sin y
،ﻭﻧﻜﺘﺐ ، x = f ( y) = Arc sin yﻭﻧﻘﺮﺃ xﻫﻮ
ﺍﻟﻘﻮﺱ ﺍﻟﺬﻱ ﺟﻴﺒﻪ . y
ﻻﺣﻆ ﺣﺴﺐ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ، f (x) = sin xﺍﻟﺪﺍﻟﺔ ﻣﺘﺰﺍﻳﺪﺓ ﻋﻠﻰ ﺍﺎﻻﺕ ﻣﻦ ﺍﻟﻨﻮﻉ
ﻣﺴﺘﻤﺮﺓ،ﻭﰲ ﻛﻞ ﳎﺎﻝ ﻣﻦ
π
π
ﺍﻟﻨﻮﻉ 2kπ + 2 , (2k + 1)π + 2 = J k
π
π
2kπ − 2 ,2kπ + 2 = I k
ﻭﺃﻳﻀﺎﹰ
ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺔ ﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ،ﻭﺑﺎﻟﺘﺎﱄ ﺣﺴﺐ
ﺍﻟﻨﻈﺮﻳﺔ ﻓﻘﺮﺓ ، 6.3ﻳﻜﻮﻥ ﻟﻠﺪﺍﻟﺔ f (x) = sin xﻋﻠﻰ ﺍﺎﻻﺕ I kﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ f −1ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ) ، Ε( fﻭﻋﻠﻰ
ﺍﺎﻻﺕ J kﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ f −1ﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ) ، Ε( fﻧﺮﻣﺰ ﻟﻠﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ sin xﺑﺎﻟﺮﻣﺰ ، x = arcsin y
ﻭﻧﻘﺮﺃ xﻫﻮ ﺍﻟﻘﻮﺱ ﺍﻟﺬﻱ ﺟﻴﺒﻪ ، yﻭﻧﻜﺘﺐ . x = f −1 ( y) = arcsin y
ﻣﻦ
ﺃﺟﻞ k = 0
f −1
ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻫﻲ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺪﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ . f
ﺧﺼﺎﺋﺺ :
ﺃ-
ﺏ-
ﺝ-
ﺭﺳﻢ
ﻋﻠﻰ
ﺍﺎﻝ I 0
ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ، f −1 ( y) = arcsin yﺗﺴﻤﻰ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻘﻮﺱ ﺍﳉﻴﺐ،ﺃﻭ ﻧﻘﻮﻝ ﺇﻥ
]sin (arcsin y) = y , y ∈ [− 1,1
π π
arcsin (sin x) = x , x ∈ − ,
2 2
]arcsin (− y) = − arcsin y , y ∈ [− 1,1
ﺍﻟﺪﺍﻟﺘﲔ Arc sin y
arcsin y,
:
ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ، f ( y) = Arc sin yﻫﻮ ﻧﻈﲑ
ﻟﺒﻴﺎﻥ . x ∈ D( f ) , f (x) = sin x
ﻟﻜﻦ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ، f −1 ( y) = arcsin yﻫﻮ ﻧﻈﲑ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ
π π
x ∈ − , , f ( x) = sin x
2 2
) ﺍﻟﺘﻨﺎﻇﺮ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻨﺼﻒ ﺍﻷﻭﻝ (.
ﺑﻴﺎﻥ
π π
) f −1 ( yﺣﻴﺚ x ∈ − ,
2 2
ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ . f ( y) = Arc sin y
ﻭﺍﺿﺢ ﺃﻥ:
ﻳﺴﻤىﺎﻟﻔﺮﻉ ﺍﻟﺮﺋﻴﺴﻲ ﻣﻦ
π
π
≤ ≤ arcsin y
2
2
, k = 0,±1,±2...........
∀y ∈ [− 1,1] → −
∀y ∈ [− 1,1] → Arc sin y = arcsin y + 2kπ
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]→ Arc sin y = (2k + 1)π − arcsin y , k = 0,±1,±2........... ∀y ∈ [− 1,1
ﺍﻟﺘﻔﺴﲑ ﺍﳍﻨﺪﺳﻲ :
ﺍﻧﻈﺮ ﺍﻟﺸﻜﻞ ﺍﻟﺴﺎﺑﻖ
ﻣﻦ ﺃﺟﻞ ﻛﻞ ] y0 ∈ [− 1,1ﻧﺒﺤﺚ ﻋﻦ ﻛﻞ ﺍﻷﻗﻮﺍﺱ)،ﺍﻟﺰﺍﻭﻳﺔ(،ﺍﻟﱵ ﺗﻘﺒﻞ y0ﺟﻴﺒﺎﹰ ﳍﺎ،ﻧﻼﺣﻆ ﻣﻦ ﺍﻟﺸﻜﻞ ﺃﻥ ﻛﻞ ﺍﻷﻗﻮﺍﺱ
ﺍﻟﱵ ﻣﺒﺪﺅﻫﺎ Aﻭﺎﻳﺘﻬﺎ ﰲ ﺇﺣﺪﻯ ﺍﻟﻨﻘﻄﻴﺘﲔ Mﺃﻭ M ′ﳍﺎ ﺟﻴﺐ ﻳﺴﺎﻭﻱ . y0
ﺇﺫﺍ ﻛﺎﻥ aﺃﺻﻐﺮ ﺍﻷﻗﻮﺍﺱ ﺍﻟﱵ ﺟﻴﺒﻬﺎ ، y0ﻓﺈﻧﻪ ﻳﻜﻮﻥ :
AM = 2kπ + a
AM ′ = (2k + 1)π − a
,
ﻭﻣﻨﻪ ﻳﻮﺟﺪ ﻋﺪﺩ ﻏﲑ ﻣﻨﺘﻪ ﻣﻦ ﺍﻷﻗﻮﺍﺱ ﺗﻘﺒﻞ y0ﺟﻴﺒﺎ ﳍﺎ،ﻭﺑﺎﻟﺘﺎﱄ ﻧﻜﻮﻥ ﻗﺪ ﻋﺮﻓﻨﺎ ﺩﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ﻭﻫﻲ ﺍﻟﱵ ﻋﺮﻓﺖ
ﻗﺒﻞ ﻗﻠﻴﻞ ﺃﻱ. f ( y) = Arc sin y :
ﻻﺣﻆ ﻟﻮ ﺃﺧﺬﻧﺎ ﻓﻘﻂ ﺍﻷﻗﻮﺍﺱ ﺍﻟﻮﺍﻗﻌﺔ ﰲ
ﻭﻫﻮ
ﺍﻟﻘﻮﺱ AM
.
π
ﺍﺎﻝ
2
, 2kπ +
، I k = 2kπ − πﻓﺈﻧﻪ ﻳﻮﺟﺪ ﻗﻮﺱ ﻭﺍﺣﺪ ﻳﻘﺒﻞ y0ﺟﻴﺒﺎ ﻟﻪ،
2
ﻭﺇﺫﺍ ﺃﺧﺬﻧﺎ ﻓﻘﻂ ﺍﻷﻗﻮﺍﺱ ﺍﻟﻮﺍﻗﻌﺔ ﰲ ﺍﺎﻝ ، J k = 2kπ + π , (2k + 1)π + π ﻓﺈﻧﻪ ﻳﻮﺟﺪ ﻗﻮﺱ ﻭﺍﺣﺪ ﻳﻘﺒﻞ y0ﺟﻴﺒﺎ ﻟﻪ،
2
ﻭﻫﻮ ﺍﻟﻘﻮﺱ . AM ′
ﻭﺑﺎﻟﺘﺎﱄ ﺗﻜﻮﻥ ﺩﺍﺧﻞ ﺍﺎﻻﺕ I kﺃﻭ J kﺍﻟﺪﺍﻟﺔ
2
f (x) = sin xﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻭﻫﻲ:
f −1 ( y) = arcsin y
-2ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ y = f ( x) = cos x
D( f ) = ℜ
Ε( f ) = [− 1,1],
ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﻟﺴﺎﺑﻖ ﻧﻼﺣﻆ ﺃﻧﻪ ﻣﻦ ﺃﺟﻞ
ﻛﻞ y0
ﻣﻦ ] ، [− 1,1ﻳﻜﻮﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ f (x) = y0ﻣﺎ ﻻﺎﻳﺔ ﻣﻦ ﺍﳊﻠﻮﻝ.
ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺪﺍﻟﺔ ، f (x) = sin xﳒﺪ ﻟﻠﺪﺍﻟﺔ f (x) = cos xﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ﺃﻭ ﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ﻋﻠﻰ
ﻛﻞ ﳎﺎﻝ ﺗﻌﺮﻳﻔﻬﺎ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ ، x = f ( y) = Arc cos yﻭﻧﻘﺮﺃ xﺍﻟﻘﻮﺱ ﺍﻟﺬﻱ ﺟﻴﺐ ﲤﺎﻣﻪ . y
ﻋﻠﻰ ﺍﺎﻻﺕ ] I k = [2kπ , 2kπ + πﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ ،ﻭﻋﻠﻰ ﺍﺎﻻﺕ ] J k = [2kπ + 1 , 2kπ + π
ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ .
f −1 ( y) = arccos y
ﻳﺮﻣﺰ ﻟﻠﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ، y = f (x) = cos xﺑﺎﻟﺮﻣﺰ
ﰲ ﺍﳊﺎﻟﺔ ﺍﳋﺎﺻﺔ ، k = 0ﻋﻠﻰ I 0ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ ، cos xﺃﻱ ﺍﻟﺪﺍﻟﺔ
ﺍﻟﺮﺋﻴﺴﻲ ﻟﻘﻮﺱ ﲤﺎﻡ ﺍﳉﻴﺐ ،ﺃﻱ ﻟﻠﺪﺍﻟﺔ . f ( y) = x = Arc cos y
ﺧﺼﺎﺋﺺ :
ﺃ-
]y ∈ [− 1,1
cos(arccos y) = y ,
=.x
x ∈ [0, π ] , f −1 ( y) = arccos y
ﺗﺴﻤﻰ ﺍﳉﺰﺀ
.
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ﺏ-
ﺝ-
ﺩ-
] arccos(cos x) = x , x ∈ [0, π
]arccos(− y) = π − arccos y , y ∈ [− 1,1
π
], y ∈ [− 1,1
2
ﺭﺳﻢ ﺍﻟﺪﺍﻟﺘﲔ
= arcsin y + arccos y
arccos y , Arc cos y
:
ﺭﺳﻢ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ) f ( yﻫﻮ ﻧﻈﲑ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ
) f ( xﻋﻠﻰ ﻛﻞ ) ، D( fﻭ ﺭﺳﻢ ﺑﻴﺎﻥ f −1 ( y) = arccos y
ﻫﻮ ﻧﻈﲑ ﺑﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ x ∈ [0, π ], f (x) = cos xﺑﺎﻟﻨﺴﺒﺔ
ﻟﻠﻤﻨﺼﻒ ﺍﻷﻭﻝ ،ﻭﻳﺴﻤﻰ ﺍﻟﻔﺮﻉ ﺍﻟﺮﺋﻴﺴﻴﻠﺒﻴﺎﻥ ﺍﻟﺪﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ
ﺍﻟﻘﻴﻤﺔ . f ( y) = Arc cos y
ﺍﻟﺘﻔﺴﲑ ﺍﳍﻨﺪﺳﻲ :
ﻧﺄﺧﺬ ﻋﻠﻰ ﳏﻮﺭ ﲤﺎﻡ ﺍﳉﻴﻮﺏ ﰲ ﺩﺍﺋﺮﺓ ﺍﻟﻮﺣﺪﺓ ﻧﻘﻄﺔ . y0
ﻧﺒﺤﺚ ﻋﻦ ﺍﻷﻗﻮﺍﺱ ﺍﻟﱵ ﺟﻴﺐ ﲤﺎﻣﻬﺎ . y0
ﺣﺴﺐ ﺍﻟﺸﻜﻞ،ﻧﻼﺣﻆ ﺃﻥ ﻛﻞ ﺍﻷﻗﻮﺍﺱ ﺍﻟﱵ ﻣﺒﺪﺅﻫﺎ ﻭﺎﻳﺘﻬﺎ Mﺃﻭ M ′ﻳﻜﻮﻥ ﺟﻴﺐ ﲤﺎﻣﻬﺎ . y0
ﺇﺫﺍ ﻛﺎﻥ aﺃﺻﻐﺮ ﺍﻷﻗﻮﺍﺱ ﺍﻟﱵ ﺟﻴﺐ ﲤﺎﻣﻬﺎ ، y0ﻓﺈﻥ ﻛﻞ ﺍﻷﻗﻮﺍﺱ AM = 2kπ − a , AM = 2kπ + aﺟﻴﺐ ﲤﺎﻣﻬﺎ
، y0ﻭﺑﺎﻟﺘﺎﱄ ﻧﻜﻮﻥ ﻗﺪ ﻋﺮﻓﻨﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ،ﺃﻭ ﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ
ﺍﻟﻘﻴﻤﺔ . x = Arc cos y
ﰲ ﺍﳊﺎﻟﺔ ﺍﳋﺎﺻﺔ،ﺇﺫﺍ ﺃﺧﺬﻧﺎ ﻓﻘﻂ ﺍﻷﻗﻮﺍﺱ ﺍﳌﻮﺟﻮﺩﺓ ﺩﺍﺧﻞ ﺍﺎﻻﺕ ] ، [2kπ , 2kπ + πﺃﻭ ] ، [2kπ + π , 4kπﻓﺈﻧﻪ ﻟﻜﻞ
y = cos x
y0
،ﻭﻫﻲ ﺍﻟﺪﺍﻟﺔ ﻣﺘﻌﺪﺩﺓ
ﻟﻠﺪﺍﻟﺔ y = cos xﻭﻫﻲ f −1 ( y) = arccos y
ﻳﻘﺎﺑﻠﻬﺎ ﻗﻮﺱ ﻭﺍﺣﺪ،ﻭﻣﻨﻪ ﻧﻜﻮﻥ ﻗﺪ ﻋﺮﻓﻨﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ
-3ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ y = f ( x) = tgx
=.x
π
D( f ) = ℜ − + kπ , Ε( f ) = ℜ
2
ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﻟﺴﺎﺑﻖ ﻧﻼﺣﻆ ﺃﻧﻪ ﻣﻦ ﺃﺟﻞ ﻛﻞ y0ﻣﻦ ، ℜﻟﻠﻤﻌﺎﺩﻟﺔ y0 = tgxﻣﺎ ﻻﺎﻳﺔ ﻣﻦ ﺍﳊﻠﻮﻝ .
ﻭﻣﻨﻪ ﺍﻟﺪﺍﻟﺔ y = tgxﻻ ﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻋﻠﻰ ﳎﺎﻝ ﺗﻌﺮﻳﻔﻬﺎ،ﺑﻞ ﺗﻘﺒﻞ ﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰﳍﺎ ﺑﺎﻟﺮﻣﺰ
x = f ( y) = Arctgy
ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﺪﺍﻟﺔ
y = tgx
ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﻣﻦ ﺍﻟﻨﻮﻉ ، − π + kπ , π + kπ ﺗﻜﻮﻥ ﻣﺘﺰﺍﻳﺪﺓ
ﻭﻣﺴﺘﻤﺮﺓ ،ﻭﺑﺎﻟﺘﺎﱄ ﻓﻬﻲ ﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ
ﰲ ﺍﳊﺎﻟﺔ k = 0ﻋﻠﻰ
f −1 ( y) = arctgy
2
=.x
2
π
ﺍﺎﻝ
2 2
,
− πﺍﻟﺪﺍﻟﺔ ، x = arctgyﺗﺴﻤﻰ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺼﻮﺭﺓ ﺍﻟﻌﻜﺴﻴﺔ . f ( y) = Arctgy
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ﺧﺼﺎﺋﺺ ﻭﺭﺳﻢ ﺍﻟﺒﻴﺎﻥ :
tg (arctgy) = y
ﺃ-
, y∈ℜ
ﺏ-
π π
x∈ − ,
2 2
ﺩ-
π
π
< < arctgy
2
2
ﺝ-
arctg (tgx) = x ,
arctg (− y) = −arctgy
−
ﺑﻴﺎﻥ f −1ﻳﺴﻤﻰ ﺍﻟﻔﺮﻉ ﺍﻟﺮﺋﻴﺴﻲ ﻣﻦ ﺑﻴﺎﻥ . f
-4ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ y = ctgx
}ℜ = Ε( f ) , D( f ) = ℜ−{kπ
ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﺎﻭﺭ ﻭﺍﺿﺢ ﺃﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ
y0 = ctgx , y0 ∈ ℜﻋﺪﺩ ﻻﺎﺋﻲ ﻣﻦ ﺍﳊﻠﻮﻝ،
ﻭﺑﺎﻟﺘﺎﱄ ﻟﻠﺪﺍﻟﺔ y = ctgxﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰ ﳍﺎ
ﺑﺎﻟﺮﻣﺰ . x = f ( y) = Arcctgy
ﻧﻼﺣﻆ ﺃﻥ ﻋﻠﻰ ﺍﺎﻻﺕ ﻣﻦ ﺍﻟﻨﻮﻉ [ ، ]kπ , kπ + π
ﺍﻟﺪﺍﻟﺔ y = ctgxﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ،ﻭﺑﺎﻟﺘﺎﱄ ﻓﻬﻲ
ﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ
x = f −1 ( y) = arcctgy
ﰲ ﺍﳊﺎﻟﺔ ﺍﳋﺎﺻﺔ ﻋﻠﻰ ﺍﺎﻝ [ ]0, πﺍﻟﺪﺍﻟﺔ
f −1 ( y) = arcctgyﺗﺴﻤىﺎﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺪﺍﻟﺔ
ﻣﺘﻌﺪﺩﺓ ﺍﻟﻘﻴﻤﺔ ) ، f ( yﻭﻳﻜﻮﻥ ﻋﻨﺪﻫﺎ :
Arcctgy = arcctgy + kπ
ﺑﻴﺎﻥ
f −1
ﻳﺴﻤﻰ ﺍﻟﻔﺮﻉ ﺍﻟﺮﺋﻴﺴﻲ ﻣﻦ ﺑﻴﺎﻥ . f
ﻣﻼﺣﻈﺔ :
ﻛﻞ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ ﺇﺫﺍ ﺫﻛﺮﺕ ﻣﻨﻔﺼﻠﺔ ﻋﻦ ﺩﻭﺍﳍﺎ ﺍﻷﺻﻠﻴﺔ ،ﻓﺈﻧﻨﺎ ﻧﺒﺪﻝ ﻓﻴﻬﺎ ﺍﳌﺘﻐﲑ yﺑﺎﳌﺘﻐﲑ ، xﺃﻱ
ﻧﻜﺘﺐ
y = arcsin x
y = arccos x,
y = arctgx,
y = arcctgx,
wﳒﻤﻞ ﺃﻫﻢ ﺧﺼﺎﺋﺺ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﻭﺍﻝ ﺍﳌﺜﻠﺜﻴﺔ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ:
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)f ( x
) D( f
) Ε( f
y = arcsin x
y = arccos x
y = arcctgx
y = arcctgx
−1 ≤ x ≤ 1
π
π
≤− ≤ y
2
2
−1 ≤ x ≤ 1
ℜ
ℜ
0≤ y≤π
π
π
≤≤ y
2
2
0≤ y≤π
−
ﺍﻟﺮﺗﺎﺑﺔ
ﻣﺘﺰﺍﻳﺪﺓ
ﻣﺘﻨﺎﻗﺼﺔ
ﻣﺘﺰﺍﻳﺪﺓ
ﻣﺘﻨﺎﻗﺼﺔ
ﻧﻘﻂ ﺍﻻﻧﻌﻄﺎﻑ
)(0,0
π
0,
2
)(0,0
π
0,
2
∞x → −∞ , x → +
\
\
π π
,
2 2
−
π , 0
-8.3ﺍﻟﺪﻭﺍﻝ ﺍﻟﺰﺍﺋﺪﻳﺔ)،ﺍﻟﻘﻄﻌﻴﺔ(،ﻭﺩﻭﺍﳍﺎ ﺍﻟﻌﻜﺴﻴﺔ :
ﻟﺘﻜﻦ x2 − y2 = 1ﻣﻌﺎﺩﻟﺔ ﻗﻄﻊ ﺯﺍﺋﺪ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﲔ ﻭ mﻧﻘﻄﺔ ﻣﺘﺤﺮﻛﺔ ﻋﻠﻰ ﻓﺮﻉ ﺍﻟﻘﻄﻊ ﺍﻟﺰﺍﺋﺪﺓ.
ﻧﺮﻣﺰ ﺑﺎﻟﺮﻣﺰ tﺇﱃ ﺿﻌﻒ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ ﺍﻟﺬﻱ ﳝﺴﺤﻪ ﺍﻟﺸﻌﺎﻉ
omﻋﻨﺪﻣﺎ ﺗﺘﺤﺮﻙ ﺍﻟﻨﻘﻄﺔ mﻣﻦ ﺍﻟﻮﺿﻊ Aﺇﱃ ﺍﻟﻮﺿﻊ . M
) ﻧﻌﺘﱪ ﺍﳌﺴﺎﺣﺔ ﻣﻮﺟﺒﺔ ﺇﺫﺍ ﲢﺮﻙ ﺍﻟﺸﻌﺎﻉ omﺑﻨﻔﺲ ﺍﻻﲡﺎﻩ
ﺍﳌﻮﺟﺐ ﻟﻠﺰﺍﻭﻳﺔ ﻭﺍﻟﻌﻜﺲ (.
ﺗﻌﺎﺭﻳﻒ :
ﺃ -ﻳﻌﺮﻑ ﲤﺎﻡ ﺍﳉﻴﺐ ﺍﻟﺰﺍﺋﺪﻱ ﻟﻠﻤﺘﻐﲑ ، tﺑﺄﻧﻪ ﻓﺎﺻﻠﺔ ﺍﻟﻨﻘﻄﺔ
، Mﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ، ch tﻭﻧﻜﺘﺐ OB = x = ch t
ﺏ -ﻳﻌﺮﻑ ﺍﳉﻴﺐ ﺍﻟﺰﺍﺋﺪﻱ ﻟﻠﻤﺘﻐﲑ tﺑﺄﻧﻪ ﺗﺮﺗﻴﺐ ﺍﻟﻨﻘﻄﺔ
، Mﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ، sh tﻭﻧﻜﺘﺐ BM = y = sh t
ﺝ -ﻳﻌﺮﻑ ﺍﻟﻈﻞ ﺍﻟﺰﺍﺋﺪﻱ ﻟﻠﻤﺘﻐﲑ ، tﺑﺄﻧﻪ ﺍﳌﻴﻞ ﺍﻟﺰﺍﻭﻱ ﻟﻠﺸﻌﺎﻉ ، OMﻭﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ ، th tﻭﻧﻜﺘﺐ
ﺩ -ﻳﻌﺮﻑ ﺍﻟﺘﻈﻞ ﺍﻟﺰﺍﺋﺪﻱ ﻋﻠﻰ ﺃﻧﻪ ﻣﻘﻠﻮﺏ ﺍﻟﻈﻞ ﺍﻟﺰﺍﺋﺪﻱ،ﻭ ﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ cth tﻭﻧﻜﺘﺐ:
ch t
sh t
sh t
ch t
= AC = th t
= PN = cth t
ﻗﻴﻢ ﺍﻟﺪﻭﺍﻝ ﺍﻟﺰﺍﺋﺪﻳﺔ :
ﻋﻨﺪﻧﺎ t = 2Sﺣﻴﺚ Sﻫﻲ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ OAMﺍﻟﺬﻱ ﳝﺴﺤﻪ ﺍﻟﺸﻌﺎﻉ OMﻋﻨﺪﻣﺎ ﺗﺘﺤﺮﻙ ﺍﻟﻨﻘﻄﺔ mﻣﻦ Aﺇﱃ . M
Sﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻦ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ OMBﻧﺎﻗﺺ ﻣﺴﺎﺣﺔ ﺍﳌﻘﻄﻊ . AMB
ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﺚ ﻫﻲ
1
1
OB ⋅ BM = xy
2
2
= OMB
ﺣﻴﺚ y, xﻫﻲ ﺇﺣﺪﺍﺛﻴﺎﺕ ﺍﻟﻨﻘﻄﺔ . M
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ﺎ ﻋﺒﺎﺭﺓ ﻋﻦ ﺍﳌﺴﺎﺣﺔ ﺍﻟﱵ ﳛﺼﺮﻫﺎ ﻓﺮﻉ ﺍﻟﻘﻄﻊ ﺍﻟﺰﺍﺋﺪ ﻭﺍﻟﻘﻄﻌﺔﻷ، ﳓﺴﺒﻬﺎ ﺑﻮﺍﺳﻄﺔ ﺍﻟﺘﻜﺎﻣﻞ ﺍﶈﺪﻭﺩAMB ﺍﳌﺴﺎﺣﺔ
( y = x 2 + 1 )ﻣﻌﺎﺩﻟﺔ ﺍﻟﻔﺮﻉ ﻫﻲ، ﻭﳏﻮﺭ ﺍﻟﻔﻮﺍﺻﻞBM ﺍﳌﺴﺘﻘﻴﻤﺔ
x
x
1
1
AMB = ∫ ydx = ∫ x 2 + 1
: ﻧﺴﺘﻌﻤﻞ ﺍﻟﺘﻜﺎﻣﻞ ﺑﺎﻟﺘﺠﺰﺋﺔ ﻧﻀﻊ
xdx
v = x , du =
x
x −1
x
x
∫
⇐ dv = dx , u = x 2 − 1
2
1
1
x
∫
1
x2
x2 − 1
x
dx = ∫
(x
∫
1
x2 − 1
1
)
−1 +1
x
x
dx = ∫ x − 1dx + ∫
2
1
1
x
x
1
x
dx
2 ∫ x 2 − 1dx = xy − ∫
1
x2 − 1
1
x
∫
x2 − 1
1
x 2 − 1dx = xy − ∫ x 2 − 1dx − ∫
x
x2
dx = xy − ∫
dx
ﺍﻟﺘﻜﺎﻣﻞ
x2 − 1
1
x
2
x
x2
x 2 − 1dx = x x 2 − 1 − ∫
(
1
dx
x2 − 1
dx
x2 − 1
= xy − ln x + x 2 − 1
)
x
1
xy 1
− ln ( x + y)
2 2
x 2 − 1dx =
1
(1)
(2)
x2 − y 2 = 1
x + y = et
: ﳒﺪ،
x − y = e −t
,
:ﺃﻱ ﺃﻥ
ﰲ ﺍﻟﻨﻬﺎﻳﺔ ﳒﺪ
ﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ ﻋﻨﺪﻧﺎ
( ﳓﺼﻞ ﻋﻠﻰ1) ( ﻋﻠﻰ2) ﺑﻘﺴﻤﺔ
x − y = e −t
e t + e −t
x=
2
:ﻭﻣﻨﻪ
:ﻭﺑﺎﻟﺘﺎﱄ
1
1
1
t = 2 S = 2 xy − xy + ln ( x + y) = ln( x + y)
2
2
2
x + y = et
: ﺃﻱ
ﲝﻞ ﺍﳉﻤﻠﺔ
e t − e −t
y=
2
: ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥ
sh t =
e t − e −t
e t + e −t
, ch t =
,
2
2
th t =
e t − e −t
e t + e −t
,
cth
t
=
e t + e −t
e t − e −t
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ﻣﻼﺣﻈﺔ :ﺇﺫﺍ ﺍﻋﺘﱪﻧﺎ ﺍﻟﺪﻭﺍﻝ ﺍﻟﺰﺍﺋﺪﻳﺔ ﺩﻭﺍﻝ ﰲ ﺍﳌﺘﻐﲑ ، xﻓﺈﺎ ﺗﻜﺘﺐ ﻛﺎﻟﺘﺎﱄ :
y = ch x
y = th x ,
y = sh x ,
y = cth x ,
ﺧﻮﺍﺹ ﺍﻟﺪﻭﺍﻝ ﺍﻟﺰﺍﺋﺪﻳﺔ :ﳒﻤﻞ ﺃﻫﻢ ﺍﳋﻮﺍﺹ ﰲ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ :
)f ( x
y = sh x
y = ch x
y = th x
) D( f
∞− ∞ < x < +
∞− ∞ < x < +
∞− ∞ < x < +
) Ε( f
∞− ∞ < y < +
∞1 ≤ y < +
−1 < y < 1
ﺃﺻﻔﺎﺭ
x=0
ﻻ
x=0
ﺍﳋﻄﻮﻁ ﺍﳌﻘﺎﺭﺑﺔ
ﻻ
ﻻ
y = ±1
∞x → ±
∞±
∞+
±1
ﺍﻟﺮﺗﺎﺑﺔ
ﻣﺘﺰﺍﻳﺪﺓ
ﻧﻘﺎﻁ ﺍﻟﻘﻴﻢ
ﺍﻟﻘﺼﻮﻯ
ﻻ
[ ]− ∞,0ﻣﺘﻨﺎﻗﺼﺔ
[∞ ]0,+ﻣﺘﺰﺍﻳﺪﺓ
ﺻﻔﺮ
x=0
ﻧﻘﻂ ﺍﻻﻧﻌﻄﺎﻑ
x=0
ﻻ
ﺍﻟﻔﺮﺩﻱ ﻭﺍﻟﺰﻭﺟﻲ
ﻓﺮﺩﻳﺔ
ﺯﻭﺟﻴﺔ
y = cth x
x≠0
∞− ∞ < x < +
− ∞ < y < −1
∞+ 1 < y < +
ﻻ
x=0
ﺃﻭ
y = ±1
±1
ﻣﺘﺰﺍﻳﺪﺓ
[ ]− ∞,0ﻣﺘﻨﺎﻗﺼﺔ
[∞ ]0,+ﻣﺘﻨﺎﻗﺼﺔ
ﻻ
ﻻ
x=0
ﻻ
ﻓﺮﺩﻳﺔ
ﺯﻭﺟﻴﺔ
ch ( x ± y) = ch x ch y ± sh x sh y
sh ( x ± y) = sh x ch y ± ch x sh y
ﺍﻟﺮﺳﻢ :
ch 2 x − sh 2 x = 1
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ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﻭﺍﻝ ﺍﻟﺰﺍﺋﺪﻳﺔ :
-1ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ y = f ( x) = sh x
D ( f ) = ℜ , Ε( f ) = ℜ
ﺍﻟﺪﺍﻟﺔ fﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ ) ، D( fﻭﻣﻨﻪ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﻦ ) Ε( fﰲ ) ، D( fﺗﻜﻮﻥ ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ،ﻳﺮﻣﺰ ﳍﺎ
ﺑﺎﻟﺮﻣﺰ . x = f −1 ( y) = arsh y
"، " arﻫﻲ ﺍﳊﺮﻭﻑ ﺍﻷﻭﱃ ﻣﻦ ﺍﻟﻜﻠﻤﺔ ﺍﻹﳒﻠﻴﺰﻳﺔ " " areaﺗﻌﲏ ﻣﺴﺎﺣﺔ ﻭﻧﻘﺮﺃ xﻫﻲ ﺃﺭﻳﺎ ﺍﳉﻴﺐ ﺍﻟﺰﺍﺋﺪﻱ ﻟـ ، yﺃﻭ
xﻫﻲ ﺍﳌﺴﺎﺣﺔ ﺍﻟﱵ ﺟﻴﺒﻬﺎ ﺍﻟﺰﺍﺋﺪﻱ . y
ﻋﻨﺪﻧﺎ :
e x − e− x
2
= y = sh x
1
1
ﺑﻮﺿﻊ z = e xﻳﻜﻮﻥ ، = e − xﻭﻣﻨﻪ
z
z
ﺣﻠﻮﻝ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺑﻘﺔ
ﲟﺎ ﺃﻥ e x > 0ﻓﺈﻥ
ﻫﻲ y 2 + 1
، 2 y = z −ﺃﻱ . z 2 − 2 yz − 1 = 0 :
z= y±
e x = y + y2 + 1
،ﺃﻱ:
y2 + 1
،ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
)
.ex = y ±
(
arsh y = x = ln y + 1 + y 2
-2ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ y = f ( x) = ch x
Ε( f ) = [1,+∞[ ,
D( f ) = ℜ
ﻣﻦ ﺍﻟﺸﻜﻞ ﺍﻟﺴﺎﺑﻖ ﻭﺍﺿﺢ ﺃﻥ
ﻟﻠﻤﻌﺎﺩﻟﺔ y0 ∈ Ε( f ) , y0 = ch x
ﺃﻛﺜﺮ ﻣﻦ ﺣﻞ .
ﻭﺑﺎﻟﺘﺎﱄ ﻟﻠﺪﺍﻟﺔ fﺻﻮﺭﺓ ﻋﻜﺴﻴﺔ ﻋﻠﻰ ﻛﻞ ﳎﺎﻝ ﺗﻌﺮﻳﻔﻬﺎ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ . x = f ( y) = Arch y
ﺍﻟﺪﺍﻟﺔ fﻋﻠﻰ ﺍﺎﻝ [∞ [0,+ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ،ﻭﺑﺎﻟﺘﺎﱄ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ . x = arch + y
ﻭﻋﻠﻰ ﺍﺎﻝ [ ]− ∞,0ﻣﺘﻨﺎﻗﺼﺔ ﻭﻣﺴﺘﻤﺮﺓ ،ﻭﺑﺎﻟﺘﺎﱄ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ،ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ x = arch − y
ﺍﻟﺪﺍﻟﺔ
ﺍﻟﻌﻜﺴﻴﺔ x = f −1 ( y) = arch + y
ﻋﻨﺪﻧﺎ :
e x + e−x
2
،ﺗﺴﻤﻰ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺼﻮﺭﺓ ﺍﻟﻌﻜﺴﻴﺔ . f
= y = ch x
ﺑﻮﺿﻊ ، z = e xﳓﺼﻞ ﻋﻠﻰ ﺍﳌﻌﺎﺩﻟﺔ
z 2 − 2 yz + 1 = 0
ﲝﻠﻬﺎ ﳒﺪ ، z = e = y ± y − 1ﺃﻱ ﺃﻥ) :
ﻭﺑﺎﻟﺘﺎﱄ ﻳﻜﻮﻥx = ln (y + y − 1 ) , x = ln (y − y − 1 ) :
1
) = − ln (y + y − 1
x = ln
ﻻﺣﻆ ﺃﻥ:
y + y −1
)x = arch y = ln (y + y − 1
, x≥0
ﻭﻣﻨﻪ
2
y2 − 1
x
2
. x = ln (y ±
2
1
2
2
2
2
2
+
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(
x = arch − y = − ln y −
y2 − 1
)
, x≤0
f ( x) = y = th x ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ-3
Ε( f ) = ]− 1,1[ ,
D( f ) = ℜ
ﻧﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ، Ε( f ) ﻭﺑﺎﻟﺘﺎﱄ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰ، D( f ) ﻣﺘﺰﺍﻳﺪﺓ ﻭﻣﺴﺘﻤﺮﺓ ﻋﻠﻰf ﺍﻟﺪﺍﻟﺔ
x = f −1 ( y) = arth y
y = th x =
(
e x − e−x
e x + e−x
)
y e x + e − x = e x − e − x ⇒ e x (1 − y) = e − x (1 + y) ⇒ e 2 x =
ex =
1+ y
1 1+ y
⇒ x = arth y = ln
1− y
2 1− y
:ﻋﻨﺪﻧﺎ
y <1
,
:ﻭﻣﻨﻪ ﻳﻜﻮﻥ
1+ y
⇒
1− y
y < 1
,
f ( x) = y = cth x ﺍﻟﺪﺍﻟﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺪﺍﻟﺔ-4
Ε( f ) = ]− ∞,−1[ U ]1,+∞[ ,
D( f ) = ℜ − {0}
ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺮﻣﺰ، Ε( f ) ﻭﺑﺎﻟﺘﺎﱄ ﺗﻮﺟﺪ ﳍﺎ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ، D( f ) ﻣﺘﻨﺎﻗﺼﺔ ﻋﻠﻰf ﺍﻟﺪﺍﻟﺔ
x = f −1 ( y) = arcth y
y = cthx =
e x + e −x
e x − e −x
e x ( y − 1) = e − x ( y + 1) ⇒ e 2 x =
y +1
⇒ ex =
y −1
1 y + 1
x = arcth y = ln
2 y − 1
,
(
)
y = arsh x = ln x + 1 + x 2
1 1+ x
, x <1
arth x = ln
2 1− x
:ﻋﻨﺪﻧﺎ
y >1
,
:ﻭﻣﻨﻪ ﻳﻜﻮﻥ
y +1
⇒
y −1
y > 1
: ﰲ ﻋﺒﺎﺭﺓ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻌﻜﺴﻴﺔ ﻳﻜﻮﻥx ﺑـy ﻧﺴﺘﺒﺪﻝ: ﻣﻼﺣﻈﺔ
,
(
)
Arch x = ± ln x + x 2 − 1
1 x + 1
,
arcth x = ln
2 x − 1
(
arch x ± arch y = arch (xy ±
x ≥1
,
x >1
,
)
(x − 1)(y − 1))
arsh x ± arsh y = arsh x 1 + y2 ± y 1 + x2
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2
2
: ﺧﺼﺎﺋﺺ
-1
-2
. arch − x ﺃﻭ، arch + x ﺗﻌﲏ
arch x
ﺣﻴﺚ
x± y
arth x ± arth y = arth
1 ± xy
1 ± xy
arcth x ± arcth y = arcth
x± y
y = arcth x
− ∞ < x < −1
+ 1 < x < +∞
− ∞ < y < +∞
y≠0
x=0
y = 0, x = ±1
x = ±1
x → ±∞, f
ﻻ
ﻓﺮﺩﻳﺔ
→0
1 ≤ x < +∞
1 ≤ x < +∞
0 ≤ y < +∞
−∞ < y≤ 0
x =1
x =1
− ∞ < y < +∞
ﻻ
− ∞ < x < +∞
D( f )
− ∞ < y < +∞
Ε( f )
x=0
ﻻ
ﻻ
x → ±∞, f −1 → +∞
lim f −1 ( x) = ±∞
x → ±∞, f −1 → −∞
x→ ±∞
x=0
ﻻ
x=0
ﻓﺮﺩﻳﺔ
ﻻ
ﻓﺮﺩﻳﺔ
x = arcth y,
x = arth y,
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-4
y = arsh x
y = arch − x
− 1 < x < +1
ﻻ
−1
y = arch + x
y = arth x
-3
x = arsh y,
x = arch y
f
ﺃﺻﻔﺎﺭ
ﺧﻂ ﻣﻘﺎﺭﺏ
∞
ﺍﻟﺴﻠﻮﻙ ﰲ
ﻧﻘﺎﻁ
ﺍﻻﻧﻌﻄﺎﻑ
ﺯﻭﺟﻲ ﻓﺮﺩﻱ
: ﺭﺳﻢ ﺍﻟﺪﻭﺍﻝ
-9.3ﻣﻘﺎﺭﻧﺔ ﺍﻟﺪﻭﺍﻝ ﰲ ﺟﻮﺍﺭ ﻧﻘﻄﺔ ﺭﻣﺰ ﺍﻻﻧﺪﻭ ""0" ، "º
-1.9.3ﺍﻟﺪﻭﺍﻝ ﺍﳌﺘﻜﺎﻓﺌﺔ :
ﺗﻌﺮﻳﻒ :ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺘﲔ
g, f
ﻣﺘﻜﺎﻓﺌﺘﺎﻥ ﻋﻨﺪﻣﺎ ، x → aﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻞ ﻣﻦ
g, f
ﻣﻌﺮﻓﺘﲔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ
ﻟﻠﻨﻘﻄﺔ ، Vδ (a ) , aﻭﺇﺫﺍ ﻭﺟﺪﺕ ﺩﺍﻟﺔ hﻣﻌﺮﻓﺔ ﰲ ) ، Vδ (aﲝﻴﺚ :
•
•
h( x) = 1
lim
x→ a
)f ( x) = g ( x)h( x
,
ﻭﻧﻜﺘﺐ ) f (x) ≈ g (xﻋﻨﺪﻣﺎ x → aﺃﻭ f ≈ gﻋﻨﺪﻣﺎ x → a
ﻣﺜﺎﻝ :
ﺃ-
f ( x) = sin x
g, f
ﻣﺘﻜﺎﻓﺌﺘﺎﻥ ﻋﻨﺪﻣﺎ ، x → 0ﻷﻥ
ﻣﻊ ﺍﻟﻌﻠﻢ ﺃﻥ
ﺏ-
g ( x) = x ,
sin x
)= g ( x)h(x
x
f ( x) = sin x = x
lim h( x) = lim
sin x
=1
x→0
x
x4
x2 + x
= )f (x
,
x→0
g ( x) = x2
x2
f ( x) = x 2
)= g ( x)h( x
x +1
ﻣﺘﻜﺎﻓﺌﺘﺎﻥ ﻋﻨﺪﻣﺎ ∞ → ، xﻷﻥ
2
ﺧﺼﺎﺋﺺ :
،ﻭ
x2
=1
lim h( x) = lim 2
∞→x
x→∞ x + 1
ﺃ -ﺍﻟﻌﻼﻗﺔ "≈" ﻋﻼﻗﺔ ﺗﻜﺎﻓﺆ .
ﺏ f1 ≈ g1 , f ≈ g ) -ﻋﻨﺪﻣﺎ ff1 ≈ gg1 ) ⇐ (x → aﻋﻨﺪﻣﺎ (x → a
ﻧﻈﺮﻳﺔ :ﺇﺫﺍ ﻛﺎﻧﺖ
f ≈ f1
g ≈ g 1ﻋﻨﺪﻣﺎ ، x → 0ﻓﺈﻧﻪ ﻣﻦ ﻭﺟﻮﺩ ﺎﻳﺔ
,
ﻭﺟﻮﺩ ﺎﻳﺔ ﺍﻟﻨﺴﺒﺔ ) f (xﻋﻨﺪﻣﺎ
)g ( x
ﺍﻟﱪﻫﺎﻥ :
x→0
)
)lim h ( x) = 1
lim h1 ( x) = 1
x→ a
2
ﲟﺎ ﺃﻥ
)f1 ( x
)g1 ( x
ﻛﻞ ﻣﻦ
lim
x→ a
g1 , f1 , h1
x→ a
/
ﻋﻨﺪﻣﺎ ، x → 0ﻧﺴﺘﻨﺘﺞ
.ﻭﻳﻜﻮﻥ ﻋﻨﺪﻫﺎ :
)f ( x
)f ( x
= lim 1
x
→
a
)g ( x
)g1 ( x
/
)f1 (x
ﺍﻟﻨﺴﺒﺔ
)g1 ( x
،
lim
x→ a
(
)(g ≈ g ) ⇔ (g (x) = g (x)h (x
)≈ f1 ) ⇔ f ( x) = f1 ( x)h1 ( x
2
1
(f
1
، limﻓﺈﻧﻪ ﻣﻦ ﺧﺼﺎﺋﺺ ﺎﻳﺎﺕ ﺍﻟﺪﻭﺍﻝ،ﻧﺴﺘﻨﺘﺞ ﻭﺟﻮﺩ (δ > 0), δﲝﻴﺚ ﺗﻜﻮﻥ
ﻣﻮﺟﻮﺩﺓ ﻭ h1 ( x) = 1
x→ a
ﻣﻌﺮﻓﺔ ،ﻭ
g 1 ( x) ≠ 0 , h1 ( x) ≠ 0
•
ﰲ ﺍﳉﻮﺍﺭ ) . Vδ (a
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•
ﻫﺬﺍ ﻳﻌﲏ ﺃﻥ ﺍﻟﺪﺍﻟﺔ ) g (xﺣﻴﺚ )، g (x) = g1 (x)h1 (xﺗﻜﻮﻥ ﻣﻌﺮﻓﺔ ﰲ ﺍﳉﻮﺍﺭ ) Vδ (aﻭﲢﻘﻖ . g (x) ≠ 0
)f ( x) h1 ( x) f1 ( x
=
ﻭﺑﺎﻟﺘﺎﱄ ﳝﻜﻦ ﺃﻥ ﻧﻜﺘﺐ:
)g ( x) h2 ( x) f1 ( x
ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
)f ( x
)f ( x
lim
= lim 1
)x→ a g ( x
)x→ a g ( x
1
ﻣﺜﺎﻝ :ﺣﺴﺎﺏ )
(
arcsin x e x − 1
lim
x→0 cos x − cos 3 x
ﻋﻨﺪﻧﺎ ، sin x ≈ xﻭ
ﻫﺬﺍ ﻳﻌﲏ ﺣﺴﺐ ﺍﳋﺎﺻﻴﺔ –ﺏ -ﺃﻥ :
،ﻋﻨﺪﻣﺎ ). (x → 0
e x − 1 ≈ x , arcsin x ≈ x
)
(
arcsin x e x − 1 ≈ x2
x→0
sin 2 x ≈ 2 x
ﻋﻨﺪﻧﺎ
ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
cos x − cos 3 x = 2 sin x sin 2 x
x→0
ﻭﺑﺎﻟﺘﺎﱄ ﺗﻜﻮﻥ:
cos x − cos 3 x ≈ 4 x2
)
(
arcsin x e x − 1
x2
1
lim
= = lim 2
x→0 cos x − cos 3 x
x→0 4 x
4
-2.9.3ﻣﻔﻬﻮﻡ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻼﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ ﺑﺎﻟﻨﺴﺒﺔ ﻟﺪﺍﻟﺔ ﺃﺧﺮﻯ :
ﺭﻣﺰ ﻻﻧﺪﻭ ") "°ﺍﻟﺼﻔﺮ ﺍﻟﺼﻐﲑ( :
ﺗﻌﺮﻳﻒ :ﺇﺫﺍ ﻋﺮﻓﺖ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ ﻟﻠﻨﻘﻄﺔ aﺍﻟﺪﻭﺍﻝ f , g , αﲝﻴﺚ
f ( x) = g ( x)α ( x) ,
lim α ( x) = 0
x→a
ﻓﺈﻧﻨﺎ ﻧﻘﻮﻝ ﺇﻥ ﺍﻟﺪﺍﻟﺔ fﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺪﺍﻟﺔ ، gﺃﻭ ﺍﻟﺪﺍﻟﺔ fﻗﺎﺑﻠﺔ ﻟﻺﳘﺎﻝ ﺃﻣﺎﻡ ﺍﻟﺪﺍﻟﺔ gﻋﻨﺪﻣﺎ ، x → 0
ﻭﻧﻜﺘﺐ :
x→a
f ( x) = o( g ( x)) ,
ﺃﻭ
) f = o( g
ﻣﻼﺣﻈﺔ :
ﺃ -ﰲ ﺣﺎﻟﺔ ، g ≡ 1ﺃﻱ ) ، x → 0 , f = o(1ﺗﻜﻮﻥ fﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ.
ﺏ -ﰲ ﺣﺎﻟﺔ g ≡ 0ﺍﻟﻌﻼﻗﺔ ) ، f = o(gﺗﻌﻦ. lim o (g ) = 0 :
g
ﺝ-
g, f
x→ 0
ﻟﻴﺲ ﺣﺘﻤﺎ ﻻ ﻣﺘﻨﺎﻫﻴﺘﲔ ﰲ ﺍﻟﺼﻐﺮ.
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ﻣﺜﺎﻝ
:1ﻟﺘﻜﻦ f ( x) = x 2
g ( x) = x 4 ,
ﻭﺍﺿﺢ ﺃﻥ ) f = o(gﻋﻨﺪﻣﺎ ∞ → . x
ﻟﻜﻦ g, fﻻ ﻣﺘﻨﺎﻫﻴﺘﺎﻥ ﰲ ﺍﻟﻜﱪ ﻋﻨﺪﻣﺎ ∞ → . x
ﺗﻨﺒﻴﻪ :
ﺃ -ﺇﺫﺍ ﻛﺎﻧﺖ gﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ،ﻳﻘﺎﻝ ﺇﻥ fﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ ﻣﻦ ﺭﺗﺒﺔ ﺃﻛﱪ ﻣﻦ ﺭﺗﺒﺔ ﺍﻟﻼ ﺗﻨﺎﻫﻲ ﰲ ﺍﻟﺼﻐﺮ . g
k ، limﺛﺎﺑﺖ ،ﻧﻘﻮﻝ ﺇﻥ f , gﻻ ﻣﺘﻨﺎﻫﻴﺘﺎﻥ ﰲ ﺍﻟﺼﻐﺮ
ﺏ -ﺇﺫﺍ ﻛﺎﻧﺖ ﻛﻞ ﻣﻦ g, fﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ ﻭ α (x) = k
x→ a
ﻣﻦ ﻧﻔﺲ ﺍﻟﺪﺭﺟﺔ.
ﻣﺜﺎﻝ:2
g ( x) = x
f ( x) = x 2 ,
ﺃ -ﻟﺘﻜﻦ
ﻭﺍﺿﺢ ﺃﻥ ) ، x → 0 , f = o(gﺃﻱ fﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ ﻣﻦ ﺩﺭﺟﺔ ﺃﻛﱪ ﻣﻦ ﺩﺭﺟﺔ ﺍﻟﻼ ﺗﻨﺎﻫﻲ ﰲ ﺍﻟﺼﻐﺮ . g
ﺏ -ﻟﺘﻜﻦ g ( x) = x , f ( x) = 4 + x − 2
ﻻﺣﻆ ﺃﻥ:
4+ x−2 1
=
x
2
ﻣﻼﺣﻈﺔ :ﺍﻟﻜﺘﺎﺑﺔ
lim
x→ 0
) , f = o( g
،ﺃﻱ ﺛﺎﺑﺖ،ﻭﻣﻨﻪ g, fﳍﻤﺎ ﻧﻔﺲ ﺩﺭﺟﺔ ﺍﻟﻼ ﺗﻨﺎﻫﻲ ﰲ ﺍﻟﺼﻐﺮ ﻋﻨﺪﻣﺎ . x → 0
x→a
ﻻ ﺗﻌﲏ ﺍﳌﺴﺎﻭﺍﺓ ﲟﻌﻨﺎﻫﺎ ﺍﳊﻘﻴﻘﻲ ،ﻷﻥ ) o ( gﻳﻌﲎ ﻗﺴﻢ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻼﻣﺘﻨﺎﻫﻴﺔ ﰲ
ﺍﻟﺼﻐﺮ ﻣﻦ ﺩﺭﺟﺔ ﺃﻛﱪ ﻣﻦ ﺩﺭﺟﺔ ﺍﻟﻼ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ )، gﻫﺬﺍ ﰲ ﺣﺎﻟﺔ gﻻ ﻣﺘﻨﺎﻫﻴﺔ ﰲ ﺍﻟﺼﻐﺮ (.
ﺃﻣﺎ ﰲ ﺣﺎﻟﺔ gﻛﻴﻔﻴﺔ ﻓﻬﻮ ﻳﻌﲏ ﻗﺴﻢ ﺍﻟﺪﻭﺍﻝ ﺍﻟﻘﺎﺑﻠﺔ ﻟﻺﳘﺎﻝ ﺃﻣﺎﻡ ، gﻭﺑﺎﻟﺘﺎﱄ ﺃﺻﻞ ﺍﻟﻜﺘﺎﺑﺔ ﻫﻮ, f ∈ o( g ) :
ﻟﻜﻦ ﻟﺘﺴﻬﻴﻞ ﺍﳊﺴﺎﺑﺎﺕ ﻧﻜﺘﺐ
) , f = o( g
x→a
.x→a
ﺧﺼﺎﺋﺺ ﺍﻟﺼﻔﺮ ﺍﻟﺼﻐﲑ)ﻋﻨﺪﻣﺎ :( x → a
-1
-2
-3
-4
-5
-6
-7
-8
) o (cg ) = o(g
) o ( g ) + o( g ) = o( g
) o ( g + o( g )) = o( g
) (
g n −1 o ( g ) = o g n
n ∈ Ν,
) C o ( g ) = o( g
) o (o ( g )) = o( g
( ) () (
)
) (o (g )) = o(g
) o (g
= o(g ) -9
g
n, m ∈ Ν, o g n o g m = o g n + m
n
n −1
n
n
n ∈ Ν,
•
, x ∈ Vδ (a ) , n ∈ Ν ،
.g ≠ 0
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ﺭﻣﺰ ﻻﻧﺪﻭ " "0ﺍﻟﺼﻔﺮ ﺍﻟﻜﺒﲑ :
ﺗﻌﺮﻳﻒ :ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﻭﺍﻝ ϕ , g, fﻣﻌﺮﻓﺔ ﰲ ﺟﻮﺍﺭ ﻣﺎ ﻣﺜﻘﻮﺏ
•
ﻟﻠﻨﻘﻄﺔ ، Vδ (a ) , aﲝﻴﺚ ) f ( x) = g ( x)ϕ (xﻭ )ϕ (x
•
ﳏﺪﻭﺩﺓ ﻋﻠﻰ ) ، Vδ (aﺃﻱ :
•
∃c > 1 / ∀x ∈ Vδ (a ) → ϕ ( x) ≤ c
ﻓﺈﻧﻨﺎ ﻧﻘﻮﻝ ﺇﻥ fﺻﻔﺮ ﻛﺒﲑ ﻣﻦ gﻋﻨﺪﻣﺎ ، x → aﻭﻧﻜﺘﺐ :
f (x) = 0(g (x)) , x → aﺃﻭ
ﻣﺜﺎﻝ :
ﺃ-
, f ( x) = x 2 + 2 x 3
) , f = 0( g
x→a
x → 0 , g ( x) = x 2
ﻭﺍﺿﺢ ﺃﻥ:
)/ ϕ ( x) = (1 + 2 x
ﻻﺣﻆ
)f ( x) = x 2 + 2 x 3 = x 2 (1 + 2 x) = g ( x)ϕ ( x
lim ϕ ( x) = 1
x→ 0
•
)Vδ (0
ﻫﺬﺍ ﻳﻌﲏ ﻭﺟﻮﺩ ﺟﻮﺍﺭ
ﺗﻜﻮﻥ ﻓﻴﻪ ) ϕ (xﻣﻌﺮﻓﺔ ﻭﳏﺪﻭﺩﺓ ﻭ
))f ( x) = 0( g ( x
, f ( x) = x 2 + 2 x3
ﺏ-
ﻭﺍﺿﺢ ﺃﻥ:
ﻻﺣﻆ
g, f
ﻣﻌﺮﻓﺘﲔ،ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
x → 0,
x → 0 , g ( x) = x 3
1
/ ϕ ( x) = 2 +
x
lim ϕ ( x) = 2
1
)f ( x) = x 2 + 2 x 3 = x 3 2 + = g ( x)ϕ ( x
x
∞ →x
•
ﺃﻱ ﺃﻥ ) ϕ (xﳏﺪﻭﺩﺓ ﰲ ﺍﳉﻮﺍﺭ ) ∞( Vδﻭ g, fﻣﻌﺮﻓﺘﲔ ﻓﻴﻪ ،ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
))x → ∞ , f ( x) = 0( g ( x
ﻣﻼﺣﻈﺔ :
ﺃ -ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺪﻭﺍﻝ ϕ , g, fﻣﻌﺮﻓﺔ ﻋﻠﻰ ﺍﺎﻝ ، Iﻭ ϕﳏﺪﻭﺩﺓ ﻋﻠﻴﻪ ،ﻓﺈﻥ
ﺏ -ﺍﻟﻜﺘﺎﺑﺔ ) x ∈ I , f (x) = 0(1ﺗﻌﲏ ﺃﻥ ﺍﻟﺪﺍﻟﺔ fﳏﺪﻭﺩﺓ ﻋﻠﻰ . I
) x ∈ I , f = 0( g
-3.9.3ﻣﻌﻴﺎﺭ ﺍﻟﺪﻭﺍﻝ ﺍﳌﺘﻜﺎﻓﺌﺔ :
ﻧﻈﺮﻳﺔ :ﺗﻜﻮﻥ ﺍﻟﺪﺍﻟﺘﺎﻥ
g, f
ﻣﺘﻜﺎﻓﺌﺘﲔ ﻋﻨﺪﻣﺎ ، x → aﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ
)), f ( x) = g ( x) + o( g ( x
x → x0
ﺃﻭ
))g ( x) = f (x) + o( f ( x
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( x → a ﻋﻨﺪﻣﺎf ﺗﺴﻤﻰ ﺍﳉﺰﺀ ﺍﻟﺮﺋﻴﺴﻲ ﻟﻠﺪﺍﻟﺔg )ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ
x→a
: ﺃﻱ ﺃﻥ، x → a , f ≈ g ]⇐[ ﻧﻔﺮﺽ ﺃﻥ: ﺍﻟﱪﻫﺎﻥ
, f ( x) = g ( x)h( x)
/ lim h(x) = 1
x→ a
:ﻋﻨﺪﻧﺎ
f ( x) − g ( x) = g ( x)h( x) − g ( x) = g ( x)(h( x) − 1) = g ( x)α ( x) / α ( x) = h( x) − 1 x→
0
→1
f ( x) − g ( x) = o( g ( x))
x→a
, f ( x) = g ( x) + o( g ( x))
x→a
, f ( x) = g ( x) + o( g ( x))
:ﻭﻣﻨﻪ ﻳﻜﻮﻥ
:ﺃﻱ ﺃﻥ
: ]⇒[ ﻧﻔﺮﺽ ﺃﻥ
: ﻋﻨﺪﻧﺎ
o ( g ( x)) = g ( x)α (x) / α ( x)
→ 0
x→a
:ﻭﻣﻨﻪ ﻳﻜﻮﻥ
f ( x) = g ( x) + g ( x)α ( x) = g ( x)(1 + α ( x)) = g ( x)h( x) /
h( x) = 1 + α ( x) x
→1
→a
x→a
, f ≈ g :ﺃﻥ
ﺃﻱ
(x → 0) : ﺃﻣﺜﻠﺔ
sin x = x + o( x) ⇔ sin x ≈ x
tg x = x + o( x) ⇔ tg x ≈ x
arcsin x = x + o( x) ⇔ arcsin x ≈ x
arctg x = x + o( x) ⇔ arctg x ≈ x
e x − 1 = x + o( x ) ⇔ e x − 1 ≈ x
sh x = x + o( x) ⇔ sh x ≈ x
ln(1 + x) = x + o( x) ⇔ ln(1 + x) ≈ x
(1 + x)
α
− 1 = x + o( x) ⇔ (1 + x) − 1 ≈ x
α
-1
-2
-3
-4
-5
-6
-7
-8
:1ﻣﺜﺎﻝ
ex − 3 1+ x
x→0 2 arctg x − arcsin x
lim
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: ﺣﺴﺎﺏ-ﺃ
arctg x = x + o( x)
arcsin x = x + o(x)
،
،
ex − 3 1 − x =
3
x → 0 ﻋﻨﺪﻧﺎ ﳌﺎ
e x −1 = x + o( x)
1+ x −1 =
1
x + o( x)
3
2
x + o( x)
3
:ﻭﻣﻨﻪ ﻳﻜﻮﻥ
2 arctg x − arcsin x = x + o( x)
: ﻭﺑﺎﻟﺘﺎﱄ
2 o ( x)
2
x + o( x)
+
ex − 3 1+ x
x =2
lim
= lim 3
= lim 3
x→0 x + o( x)
x→0
x→0 2 arctg x − arcsin x
o ( x) 3
1+
x
1
: ﺣﺴﺎﺏ-ﺏ
2
lim cos x 2 x
x→0
(
) ﻭ، cos
ln cos 2 x 1
ln 1 − sin 2 2 x
lim ln cos 2 x = lim
=
lim
x→0
x→0
x2
2 x→0
x2
1
x2
(
1
2
1
x2
2x = e
ln cos x 2 x
ﻋﻨﺪﻧﺎ
: ﻣﻦ ﻧﺎﺣﻴﺔ ﺛﺎﻧﻴﺔ ﺑﺴﻬﻮﻟﺔ ﳒﺪ
)
ln 1 − sin 2 2 x ≈ − sin 2 2 x ≈ −(2 x) = −4 x2
(
2
:ﻭﺑﺎﻟﺘﺎﱄ ﺗﻜﻮﻥ
)
1
ln 1 − sin 2 2 x
1
4 x2
lim
=
−
lim
= −2
2 x→0
x2
2 x→0 x2
1
lim cos x 2 x = −2
2
x→0
1
lim cos x 2 x = e x→0
2
1
2
lim ln cos x 2 x
x→0
=
1
e2
ﺃﻱ ﺃﻥ
: ﻓﺈﻥ، ﻣﺴﺘﻤﺮﺓe x ﲟﺎ ﺃﻥ ﺍﻟﺪﺍﻟﺔ
1
cos x x ln (1+ x)
lim
x→0 ch x
: ﺣﺴﺎﺏ-ﺝ
(cos x − 1 + 1) xln (1+ x) −1
cos x x ln (1+ x) lim
lim
= x→0
=e
1
x→0 ch x
lim(ch x − 1 + 1) xln (1+ x)
1
1
x→0
.
x→0 ,
t = ϕ −1 ( x) ﻋﻜﺴﻴﺔ
( ) ﻭ، ch x − 1 = x2 + o(x )ﻭ، cos x − 1 = − x2
2
x ln(1 + x) = x2 + o x2
2
2
( ) ﺫﺍﻟﻚ ﻷﻥ
+ o x2
: ﺍﻟﺪﻭﺍﻝ ﺍﳌﻌﻄﺎﺓ ﺑﺸﻜﻞ ﻭﺳﻴﻄﻲ-10.3
ﳍﺎ ﺩﺍﻟﺔϕ ﻧﻔﺮﺽ ﺃﻥ ﺍﻟﺪﺍﻟﺔ، ℜ ﻣﻦI ﺩﺍﻟﺘﲔ ﻣﻌﺮﻓﺘﲔ ﻋﻠﻰ ﳎﻤﻮﻋﺔy = ψ (t ),
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x = ϕ (t )
ﻟﺘﻜﻦ
ﻋﻠﻰ ﺍﻤﻮﻋﺔ ) ، ϕ (Iﻭﻣﻨﻪ ﻋﻠﻰ ﺍﻤﻮﻋﺔ ) ϕ (Iﺗﻌﺮﻑ ﺍﻟﺪﺍﻟﺔ ﺍﳌﺮﻛﺒﺔ )، y = ψ (ϕ −1 (x)) = f (xﺍﻟﱵ ﺗﺴﻤﻰ ﺍﻟﺪﺍﻟﺔ
ﺍﳌﻌﻄﺎﺓ ﺑﺎﻟﺸﻜﻞ ﺍﻟﻮﺳﻴﻄﻲ ﺍﻟﺘﺎﱄ:
ﻣﺜﺎﻝ :
) x = ϕ (t
x = ϕ (t ) = R cos t
ﺃ-
. y = ψ (t ),
y = ψ (t ) = R sin t ,
ﺍﻟﺪﺍﻟﺔ ϕﻋﻠﻰ ﺍﺎﻝ ] [0, πﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ
x
R
0≤t ≤π ،
t = arccos
ﻭﻣﻨﻪ ﻳﻜﻮﻥ:
x2
x
x
)y = sin arccos = R 1 − cos 2 arccos = R 1 − 2 = R2 − x2 = f ( x
R
R
R
ﺃﻱ ﺃﻥ ﺍﻟﺪﺍﻟﺔ
y = f ( x) = R2 − x2
ﲤﺜﻞ ﻭﺳﻴﻄﻴﺎ ﺑﺎﳌﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ:
0≤t ≤π
ﺏ-
x = ϕ (t ) = R cos t
y = R sin t
y = ψ (t ) = R sin t ,
x = R cos t ,
،
π ≤ t ≤ 2π
ﺍﻟﺪﺍﻟﺔ ϕﻋﻠﻰ ﺍﺎﻝ ] [π ,2πﺗﻘﺒﻞ ﺩﺍﻟﺔ ﻋﻜﺴﻴﺔ ﻣﻌﺮﻓﺔ ﻛﺎﻟﺘﺎﱄ . y = − R2 − x2
ﻭﺑﺎﻟﺘﺎﱄ ﻧﻘﻮﻝ ﺇﺫﺍ ﺗﻐﲑ tﰲ ﺍﺎﻝ ] ، [0,2πﻓﺈﻥ ﺍﳌﻌﺎﺩﻻﺕ y = R sin t , x = R cos tﺗﻌﺮﻑ ﺩﺍﻟﺘﲔ ﺑﻴﺎﻧﻴﻬﻤﺎ ﻫﻮ
ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﻌﺎﺩﻟﺘﻬﺎ . x2 + y2 = R2
ﺃﻭ ﺃﻥ ﺍﻟﺪﺍﻟﺔ x2 + y2 = R2ﲤﺜﻞ ﻭﺳﻴﻄﻴﺎ ﺑﺎﳌﻌﺎﺩﻻﺕ
y = ψ (t ) = b sin t , x = ϕ (t ) = a cos t ، 0 ≤ t ≤ 2π
ﺝ-
x = R cos t
ﺑﲔ ﺃﻥ ﻫﺬﻩ ﺍﳌﻌﺪﻻﺕ ﲤﺜﻞ ﻭﺳﻴﻄﻴﺎ ﺍﻟﻘﻄﻊ ﺍﻟﻨﺎﻗﺺ ﺍﻟﺬﻱ ﻣﻌﺎﺩﻟﺘﻪ
y = R sin t ,
x2 y 2
+
=1
a b2
t ∈ [0,2π ],
.
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