Since M 0 there exists a positive number d 1 such that for all x x c d 1 g x M

# Since m 0 there exists a positive number d 1 such

• Notes
• 56

This preview shows page 12 - 15 out of 56 pages.

Since M 0, there exists a positive number d 1 such that for all x 0 x c d 1 g x M M 2 . (3) continued
574 Appendices For any numbers A and B it can be shown that A B A B and B A A B , from which it follows that A B A B . With A g x and B M , this becomes g x M g x M , which can be combined with the inequality on the right in (3) to get, in turn, g x M M 2 M 2 g x M M 2 M 2 g x 3 2 M g 1 x M 2 g 3 x . Multiply by 2 M g x . (4) Therefore, 0 x c d implies that g 1 x M 1 M Mg g x x M 1 g 1 x M g x M 1 M 2 M g x . Inequality (4) (5) Since 1 2 M 2 e 0, there exists a number d 2 0 such that for all x in D 0 x c d 2 M g x 2 e M 2 . (6) If we take d to be the smaller of d 1 and d 2 , the conclusions in (5) and (6) both hold for all x such that 0 x c d . Combining these conclusions gives 0 x c d 2 g 1 x M 1 e . This completes the proof of the Limit Quotient Rule. The last proof we give is of the Sandwich Theorem (Theorem 4) of Section 2.1. THEOREM 4 The Sandwich Theorem If g x f x h x for all x c in some interval about c , and lim x c g x lim x c h x L , then lim x c f x L. Proof for Right-hand Limits Suppose that lim x c g x lim x c h x L. Then for any e 0 there exists a d 0 such that for all x the inequality c x c d implies L e g x L e and L e h x L e . continued
Section A3 Using the Limit Definition 575 These inequalities combine with the inequality g x f x h x to give L e g x f x h x L e , L e f x L e , e f x L e . Thus, for all x , the inequality c x c d implies f x L e . Therefore, lim x c f x L. Proof for Left-hand Limits Suppose that lim x c g x lim x c h x L. Then for any e 0 there exists a d 0 such that for all x the inequality c d x c implies L e g x L e and L e h x L e . We conclude as before that for all x , c d x c implies f x L e . Therefore, lim x c f x L. Proof for Two-sided Limits If lim x c g x lim x c h x L , then g x and h x both approach L as x c and x c ; so lim x c f x L and lim x c f x L . Hence lim x c f x exists and equals L. Section A3 Exercises In Exercises 1 and 2, sketch the interval a , b on the x -axis with the point c inside. Then find a value of d 0 such that for all x , 0 x c d a x b. 1. a 4 9, b 4 7, c 1 2 2. a 2.7591, b 3.2391, c 3 In Exercises 3 and 4, use the graph to find a d 0 such that for all x 0 x c d f x L e . 3. 4. Exercises 5–8 give a function f x and numbers L , c , and e . Find an open interval about c on which the inequality f x L e holds. Then give a value for d 0 such that for all x satisfying 0 x c d the inequality f x L e holds. Use algebra to find your answers. 5. f x 2 x 2, L 6, c 2, e 0.02 6. f x x 1, L 1, c 0, e 0.1 7. f x 19 x , L 3, c 10, e 1 8. f x x 2 , L 4, c 2, e 0.5 Exercises 9–12 give a function f x , a point c , and a positive number e . (a) Find L lim x c f x . Then (b) find a number d 0 such that for all x 0 x c d f x L e .