pbïšbî‹ÜaZò†b¾a
ò‹Ñïå‚@óibïä
‘ì‹«@‹Ð
p@Ë@l@‘2@óåÜaZõín¾a
ðäbîÜa@âbÕÜa@ðic@óîíäbq
2HZŒb−fia@ò‡à
Zßìÿa@æî‹ánÜa 3
1− 3 x Arc tan ( x − 2 x + 2 ) = 0 (iii 3 + 125 = 0 (ii x 36 + 27 = 0 (i: ﻤﺎﻴﻠﻲℝ ﺤل ﻓﻲ 1+ x 2 . Arc tan ( x − 2 x ) ≺ 0 (iv 2
ZðäbrÜa@æî‹ánÜa :ﺃﺤﺴﺏ ﻤﺎﻴﻠﻲ .C =
3 31π ، A = Arc tan − (1 3 3 2 tan ( 3 x ) x . lim ( x sin ( x ) + 2 x ) ﻭlim ، lim (2 x →0 1 − cos ( 2 x ) x →+∞ x →0 7x
1 2
1 3
Arc tan + Arc tan ﻭB = Arc tan tan −
3
. lim
x →+∞
3 x +2 x +7 −2 ، lim (3 x − 1 x →1 x − 1
ZsÜbrÜa@æî‹ánÜa
؟ﺇﺫﺍ ﻜﺎﻥx 0 = 0 ﺘﻘﺒل ﺘﻤﺩﻴﺩﺍ ﺒﺎﻹﺘﺼﺎل ﻓﻲ
1 * f ﻫل ﺍﻝﺩﺍﻝﺔ. f ( x ) = x sin 3 : ﺒﻤﺎﻴﻠﻲℝ + ﺍﻝﺩﺍﻝﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰf ﻝﺘﻜﻥ 4 x − x .ﺍﻝﺠﻭﺍﺏ ﺒﻨﻌﻡ ﻓﺤﺩﺩ ﻫﺫﺍ ﺍﻝﺘﻤﺩﻴﺩ ZÊia‹Üa@æî‹ánÜa
.f
(c ) =
1
c
] ﺒﺤﻴﺙ0,1[ ﻤﻥ
c [ ﺒﻴﻥ ﺃﻨﻪ ﻴﻭﺠﺩ1, 2 ] [ ﻨﺤﻭ0,1] ﺩﺍﻝﺔ ﻤﺘﺼﻠﺔ ﻭﻤﻌﺭﻓﺔ ﻤﻥf ﻝﺘﻜﻥ Z÷àb©a@æî‹ánÜa
.f
2 x 1+ x
( x ) = 2Arc tan
: ﺒﻤﺎﻴﻠﻲℝ ﺍﻝﺩﺍﻝﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰf ﻝﺘﻜﻥ
Arc tan (t ) ﻭlim f ( x ) : ﻭﺃﺤﺴﺏ ﺍﻝﻨﻬﺎﻴﺎﺕ ﺍﻝﺘﺎﻝﻴﺔD f ﺤﺩﺩ:(ﺃ1 t →0 x →+∞ t f (x ) . lim = +∞ :ﺒﻴﻥ ﺃﻥ:ﺏ x →0 x . I = [1, +∞[ ﻤﻊI ﻋﻠﻰf ﻗﺼﻭﺭ ﺍﻝﺩﺍﻝﺔg (ﻝﺘﻜﻥ2 . ﻴﻨﺒﻐﻲ ﺘﺤﺩﻴﺩﻩJ ﻨﺤﻭ ﻤﺠﺎلI ﺘﻘﺎﺒل ﻤﻥg ﺒﻴﻥ ﺃﻥ:ﺃ −1 . J ﻤﻥx ﻝﻜلg ( x ) ﺤﺩﺩ:ﺏ π . f ( x ) ≻ ﻴﺤﻘﻕ ﺍﻝﻤﺘﺭﺍﺠﺤﺔ2 ( ﺒﻴﻥ ﺃﻥ ﺍﻝﻌﺩﺩ ﺍﻝﺤﻘﻴﻘﻲ3
. lim
+
3
Bonne chance
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