Calc 1 WA2 - Calculus I Thomas Edison WA2 Section 2.2 x - 3...

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Unformatted text preview: Calculus I Thomas Edison WA2 Section 2.2 x - 3 lim ( 2 x + 5 ) = -1 given > 0 ( 2 x + 5) + 1 < 2 x+3 < = 40) let = if 0 < x + 3 < = x+3 < ( 2 x + 5) + 1 < f ( x) + L < Section 2.3 x 2 - 3x h( x ) = x a ) lim h( x ) x g -2 xg 0 b) lim h( x ) x 2 - 3x -22 - (3)(-2) 4 + 6 = = = -5 x -2 -2 42) x 2 - 3 x 02 - (3)(0) b) = x 0 2 x - 3x x ( x - 3) = = x - 3 = 0 - 3 = -3 x x lim h( x ) = -5 a) x g -2 xg 0 lim h( x) = -3 x g -1 lim g ( x ) = lim f ( x) = 2 x 2 - x - 3 (2)(-1) 2 - (-1) - 3 0 = = =0 x +1 (-1) + 1 0 2 x 2 - x - 3 ( x + 1) (2 x - 3) = = 2 x + 3 = (2)( -1) + 3 = 1 x +1 ( x + 1) 46) x g -1 L =1 lim g ( x ) = lim f ( x ) = 1 x v -1 x -1 g ( x) = f ( x ) except at x = -1 3- x 3-3 = =0 xg 3 x2 - 9 9 - 9 52) (3 - x) 3- x 1 1 1 = = = = 2 x - 9 ( - x - 3) ( - x + 3) (- x - 3) -3 + (-3) -6 lim f ( x) = lim f ( x) = xg 3 58) ( x - 3) x +1 - 2 x +1 + 2 = = x - 3 x +1 + 2 x - 3 ) x +1 + 2 ( x +1 - 2 0 = x-3 0 ( ) 1 1 1 = = x +1 + 2 4+2 4 1 1 - 4 ( x + 4) 0 lim = = xg 0 x 0 60) 1 1 - 4 -( x ) ( -1 -1 1 x + 4 ) 4 - ( x + 4) = = = = =- x (4) x( x + 4) (4) ( x) ( x + 4) (4)( x + 4) (4)(0 + 4) 16 62) Vx g 0 lim = ( x +Vx) 2 - x 2 ( x +Vx ) 2 - x 2 = Vx 0 2 ( x +Vx) 2 - x 2 ( x +Vx )( x +Vx ) + ( x 2 ) x 2 + 2( x )(Vx ) + (Vx ) 2 - x 2 2( x) (Vx) + (Vx) = = = = 2x Vx Vx Vx Vx lim = 0 (Cos )(Tan ) 0 = 0 72) (Cos )(Tan ) ( Sin ) (Cos ) Sin = = =1 (Cos ) ( ) 1 - Tan( ) 1 - Tan( x) 4 = = -1.000 78) lim = Sin( x) - Cos ( x) Sin( ) - Cos ( ) x 4 4 4 sin 2 x sin 2 2 2 x 1 2 3 x sin 2 x 3x = - = = sin 3 x 3 1 sin 3 x 2 3 x 2x 3 sin 3x 82) lim = xg 0 s(t ) = -16 ( t 2 ) + 1000 -16 ( t 2 ) = -1000 - 1000 t2 = -16 - 1000 t= = 7.9sec -16 108) s(a ) - s (t ) s (7.905) - s (t ) lim = = tg a a-t 7.905 - t 0 + (16(t 2 ) - 1000) 16(t 2 - 62.05) 16(t - 7.905)(t - 7.905) = = 7.905 - t -(t - 7.905) -(t - 7.905) = 16 (t - 7.905) (t - 7.905) -1(t - 7.905) ft sec y = -16 ( t 2 ) + 1000 = 16(7.905 + 7.905) ft = -252.96 -1 sec final speed = 252.96 Section 2.4 lim = -2 + c c x 4) x lim = -2 - x c lim = 2 not continuous when c a -2 8) lim = + x 2 ( 2 - x) 2- x 2- x 1 1 1 = = = = =- 2 2 x - 4 - ( 4 - x ) - ( 2 - x ) ( 2 + x ) - ( 2 + x ) - ( 2 + 2) 4 10) lim = - x 4 x -2 = x-4 ( ( x -2 x -2 )( ) x +2 ) = 1 1 1 = = x +2 4+2 4 2 - 4 x + 6, x 2 x lim f ( x) = 2 xg 2 - x + 4 x - 2, x 2 16) f ( x ) = x 2 - 4 x + 6 = 22 - (4)(2) + 6 = 4 - 8 + 6 = 2 f ( x ) = - x 2 + 4 x - 2 = -(22 ) + (4)(2) - 2 = 2 L=2 30) f ( x ) = x2 -1 is continuous to the left and right, but not at (-1) x +1 50) x 1 x 1 - 2 x + 3, x 1 f ( x) = 2 x , x V1 lim f ( x ) = -2 x + 3 = (-2)(1) + 3 = 1 - lim f ( x) = x 2 = 12 = 1 + f(x) is continuous 66) 2 - a2 x ,x g ( x) = x - a x=a 8, 2 - a2 ( x - a ) ( x + a ) x lim = = = 8 + 8 = 16 xg 8 ( x - a) x - a 16 = x + a a = 8 Should be a + a = 8 or 2a = 8 or a = 4 Section 2.5 2+ x x (1 - x) 2 +1 3 = 2 1 (1 - 1) 0 2 g ( x) = 12) x =1 Vertical asymptotes @1 Also at x = 0 22) f ( x) = 4 ( x - 2 ) ( x + 3) ( x + 3) 4 ( x2 + x - 6) 4 x 2 + 4 x - 24 4 = = = 4 3 2 x - 2 x - 9 x + 18 x x ( x 3 - 2 x 2 - 9 x + 18 ) x ( x 2 - 9 ) ( x - 2 ) ( x - 3) ( x + 3) x = 4 1 4 = 2 x x x - 3 - 3x x=3 x=3 4 4 = 9 -9 0 ( 4 ) ( 32 ) + ( 4 ) ( 3) - 24 4 x 2 + 4 x - 24 96 = = 4 4 3 2 3 2 x 0 - 2 x - 9 x + 18 x ( 3 ) - ( 2 ) ( 3 ) - ( 9 ) ( 3 ) + ( 18 ) ( 3) Vertical asymptotes @ 3 x = -1 f ( x) = 34) f ( x) = x 2 - 6 x - 7 (-1) 2 - (6)(-1) - 7 7 - 7 = = x +1 ( -1) + 1 0 x 2 - 6 x - 7 ( x + 1) ( x - 7 ) -1 - 7 = = = -8 x +1 1 ( x + 1) Removable discountinuity @ -1 x(2 + x) 1(2 + 1) 3 = = 1- x 1-1 0 40) x 1+ lim = lim = x2 42 16 1 lim = 2 = 2 = = - x 4 x + 16 4 + 16 32 2 42) 1 lim = 2 2 lim = 2 - x 0- 48) x lim = + x 50) xV 2 lim + = -2 = cos x -2 -2 = = -2.000 .999 cos 2 = Actually cos 0 so the limit is a 2 lim = e -0.5 x sin x 52) x 0+ lim = - ...
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This note was uploaded on 06/17/2008 for the course MATH 251 taught by Professor All during the Spring '08 term at Thomas Edison State.

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