Pre-Calc Homework Solutions 55

Pre-Calc Homework Solutions 55 - Section 2.3 (b) lim f (x)...

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Unformatted text preview: Section 2.3 (b) lim f (x) x2 55 59. lim ln x x log x ln (10), since ln x log x ln x (ln x)/(ln 10) lim (4 x2 x) 4 2 2 ln 10. ln (x 1) ln x x (c) lim f (x) does not exist, because the left- and x2 right-hand limits are not equal. 1 (d) f (2) 1) ln x 60. lim 4 2 2 2 Since ln (x ln (x 1) ln x ln x 1 ln 1 ln x 1 x 1 x ln x ln 1 ln 1 1 x 1 , x 1 ln x 1 4. ( f g)(x) f (g(x)) f 1 x 2(1 x) x x 2 ,x 0 (1 x) 5x 6x 1 1 x 1 x 1 1 1 5 But as x , 1 1/x approaches 1, so ln(1 1/x) approaches ln (1) 0. Also, as x , ln x approaches infinity. This means the second term above approaches 0 and the limit is 1. (g f )(x) x 2x 5 1 g( f (x)) 2x 2x 1 1 g 3x 2x 2x x 1 5 4 ,x 1 1 2x x 1 5 1 5 s Section 2.3 Continuity (pp. 7381) Exploration 1 1. x 9 ( 3, 3) 2 5. Note that sin x 2 (g f )(x) g( f (x)) g(x 2). Therefore: g(x) sin x, x 0 ( f g)(x) f (g(x)) = f (sin x) (sin x)2 or sin2 x, x 6. Note that , 3) 1 x 0 Removing a Discontinuity (g f )(x) f(x) 1 1 g( f (x)) 1 for x x f(x) 1. (x 3)(x 3). The domain of f is ( (3, ) or all x 3. Therefore, gives f (x) ( f g)(x) 1 x x 1 1 0. Squaring both sides 1 x2 1 x 1 2. It appears that the limit of f as x 3 exists and is a little more than 3. f (g(x)) x x 1 . Therefore, f (x) x2 1 ( 1 x 1) 2 1, x 1 0. 1 ,x 1 7. 2x 2 [ 3, 6] by [ 2, 8] 9x 1)(x 5 5) 0 0 5 (2x 10 . 3 3. f (3) should be defined as 4. x 3 (x 7x 6 3)(x (x 10 3 x3 Solutions: x 2), x 2 8. 9 3. 1 ,x 2 (x 3)(x 1)(x (x 20 6 3), so f (x) 1)(x 2) x 3 Thus, lim 5. lim g(x) x3 1)(x 2) for x x 3 10 . 3 [ 5, 5] by [ 10, 10] g(3), so g is continuous at x 3. Solution: x 0.453 1. Quick Review 2.3 1. lim x1 3x 2 x3 2x 1 4 3( 1) 2( 1) ( 1)3 4 2 1 6 3 9. For x 3, f (x) 4 when 5 x 4, which gives x (Note that this value is, in fact, 3.) 2 2. (a) lim x 1 f (x) f (x) lim int (x) x 1 2 1 For x 3, f (x) 4 when x 2 6x 8 4, which gives x 2 6x 12 0. The discriminant of this equation is b 2 4ac ( 6)2 4(1)(12) 12. Since the discriminant is negative, the quadratic equation has no solution. The only solution to the original equation is x 1. (b) lim x 1 lim x 1 f (x) (c) lim f (x) does not exist, because the left- and x 1 right-hand limits are not equal. (d) f( 1) int ( 1) lim (x 2 x2 1 4x 5) 22 4(2) 5 1 3. (a) lim f (x) x2 ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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