This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Notes on Homework #7 Exercise 4.2.1 (c) Most of you had the right idea, writing  x 3 8  =  x 2  x 2 + 2 x + 4  . Now you need to produce an upper bound for the second factor. If we commit ourselves to choosing 1 then  x 2  < implies that  x 2  < 1, so that 1 < x 2 < 1 1 < x < 3 7 = 1 2 + 2 1 + 4 < x 2 + 2 x + 4 < 3 2 + 2 3 + 4 = 19  x 2 + 2 x + 4  < 19 . (It is important that we produce these bounds outside the absolute value. That is because a < b does not imply  a  <  b  without a lower bound an a .) Now continuing in the usual way, you will come to the choice of = min(1 , / 19). Exercise 4.2.8 The most efficient way to do this is using Theorem 4.2.3 to relate the functional limits to sequence limits, and then appealing to the Order Limit Theorem (Theorem 2.3.4) for sequences. Some of you tried to work directly from the definition of limit. That is possible, but more difficult than using sequences. Here is how it would go. Let L f = lim x...
View
Full
Document
 '08
 KBHANNSGEN
 Calculus

Click to edit the document details