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Pre-Calc Homework Solutions 456

Pre-Calc Homework Solutions 456 - 456 Appendix A5.1 L and 0...

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18. If L , M , and c are real numbers and lim x c f ( x ) 5 L and lim x c g ( x ) 5 M , show that for any e . 0, there is a d . 0 such that 0 , ) x 2 c ) , d ) f ( x ) 2 g ( x ) 2 ( L 2 M ) ) , e . Proof: ) f ( x ) 2 g ( x ) 2 ( L 2 M ) ) 5 ) f ( x ) 2 L 1 M 2 g ( x ) ) # ) f ( x ) 2 L ) 1 ) M 2 g ( x ) ) 5 ) f ( x ) 2 L ) 1 ) g ( x ) 2 M ) by the triangle inequality. Since lim x c f ( x ) 5 L , lim x c g ( x ) 5 M , for any e there exist d 1 , d 2 such that 0 , ) x 2 c ) , d 1 ) f ( x ) 2 L ) , } 2 e } and 0 , ) x 2 c ) , d 2 ) g ( x ) 2 M ) , } 2 e } . Choose d 5 min { d 1 , d 2 }. Then 0 , ) x 2 c ) , d ) f ( x ) 2 L ) 1 ) g ( x ) 2 M ) , } 2 e } 1 } 2 e } 5 e ) f ( x ) 2 g ( x ) 2 ( L 2 M ) ) , e . 19. lim x c [ f 1 ( x ) 1 f 2 ( x ) 1 f 3 ( x )] 5 lim x c [ f 1 ( x ) 1 f 2 ( x )] 1 L 3 5 L 1 1 L 2 1 L 3 , by two applications of the Sum Rule. To generalize: Step 1 ( n 5 1): lim x c f 1 ( x ) 5 L 1 as given. Step 2: Suppose lim x c [ f 1 ( x ) 1 f 2 ( x ) 1 1 f k ( x )] 5 L 1 1 L 2 1 1 L k . Then lim x c [ f 1 ( x ) 1 f 2 ( x ) 1 1 f k 1 1 ( x )] 5 lim[ f 1 ( x ) 1 f 2 ( x ) 1 1 f k ( x )] 1 L k 1 1 5 L 1 1 L 2 1 1 L k 1 1 , by the Sum Rule.
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