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Pre-Calc Homework Solutions 55

# Pre-Calc Homework Solutions 55 - Section 2.3(b lim f(x x2...

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59. lim x } l l o n g x x } 5 ln (10), since } l l o n g x x } 5 } (ln x ) ln /(l x n 10) } 5 ln 10. 60. lim x } ln l ( n x 1 x 1) } 5 1 Since ln ( x 1 1) 5 ln 3 x 1 1 1 } 1 x } 24 5 ln x 1 ln 1 1 1 } 1 x } 2 , } ln ( l x n 1 x 1) } 5 5 1 1 But as x , 1 1 1/ x approaches 1, so ln(1 1 1/ x ) approaches ln (1) 5 0. Also, as x , ln x approaches infinity. This means the second term above approaches 0 and the limit is 1. Section 2.3 Continuity (pp. 73–81) Exploration 1 Removing a Discontinuity 1. x 2 2 9 5 ( x 2 3)( x 1 3). The domain of f is ( 2‘ , 2 3) < ( 2 3, 3) < (3, ) or all x 6 3. 2. It appears that the limit of f as x 3 exists and is a little more than 3. [ 2 3, 6] by [ 2 2, 8] 3. f (3) should be defined as } 1 3 0 } . 4. x 3 2 7 x 2 6 5 ( x 2 3)( x 1 1)( x 1 2), x 2 2 9 5 ( x 2 3)( x 1 3), so f ( x ) 5 } ( x 1 x 1 1 )( x 3 1 2) } for x 3. Thus, lim x 3 } ( x 1 x 1 1 )( x 3 1 2) } 5 } 2 6 0 } 5 } 1 3 0 } . 5. lim x 3 g ( x ) 5 } 1 3 0 } 5 g (3), so g is continuous at x 5 3. Quick Review 2.3 1. lim x –1 } 3 x 2 x 2 3 1 2 x 4 1 1 } 5 5 } 6 3 } 5 2 2. (a) lim x 2 1 2 f ( x ) 5 lim x 2 1 2 int ( x ) 5 2 2 (b) lim x 2 1 1 f ( x ) 5
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