131a3 - Homework 3 NOTE: This was a half-length assignment,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 3 NOTE: This was a half-length assignment, so only two problems were graded: 8.8 (worth 10 points) and 9.6 (worth 15 points). The other problems were graded for completion (worth 5 points each excpt for 9.1, which was worth 10) to bring the total score to 50 points. Scores were than doubled to make this assignment worth 100 points like the others. 8.8. Before beginning, we prove the following lemma. Lemma: Suppose A and B are strictly positive numbers and { c n } is a se- quence of strictly positive numbers which converges to C . Then A B + c n converges to A B + C . Proof: We note that | A B + c n- A B + C | = | A ( B + C )- A ( B + c n ) ( B + c n )( B + C ) | A | C- c n | B ( B + C ) ; it suffices to show that this last expression converges to zero. Fixing > we can pick N such that for n > N , | C- c n | < B ( B + C ) A (as c n C ); this tells us that A | C- c n | B ( B + C ) < A B ( B + C ) B ( B + C ) A = so our sequence is within of its desired limit; letting go to zero proves the lemma. Armed with this lemma, we now establish our limits. (a) | n 2 + 1- n | = 1 | n 2 +1+ n | (multiplying numerator and denominator by n 2 + 1+ n ) 1 n + n (as is an increasing function and 1 /x < 1 /y if x > y > 0) = 1 2 n ; picking > 0 we therefore have that if n > 1 2 then | ( n 2 + 1- n )- | < showing that the limit of...
View Full Document

Page1 / 3

131a3 - Homework 3 NOTE: This was a half-length assignment,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online