A6F08Sol-138

A6F08Sol-138 - Math 138 Assignment 6 Solutions Fall 2008 1....

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: Math 138 Assignment 6 Solutions Fall 2008 1. a) Use the- definition of lim x a f ( x ) = L to prove the limit: lim x - 5 4- 3 x 5 = 7 . Solution Given any > 0, we must show that there is a such that 4- 3 x 5- 7 < for all x such that < | x + 5 | < . 4- 3 x 5- 7 =- 3- 3 x 5 = | - 3 | 1 + x 5 = 3 5 | 5 + x | . Thus 4- 3 x 5- 7 = 3 5 | 5 + x | < if | x + 5 | < 5 3 . Choosing = 5 3 for the given , we thus have 4- 3 x 5- 7 = 3 5 | x + 5 | < 3 5 5 3 = if | x + 5 | < = 5 3 , as required. b) The definition for the limit of f ( x ) as x tends to infinity is: lim x f ( x ) = b means that for all > 0 there exists R > such that for all x , if x > R then | f ( x )- b | < . Use this definition to prove that lim x x 2 + x + 1 x 2 + 1 = 1. Solution For all x > 0 we have | f ( x )- 1 | = x 2 + x + 1 x 2 + 1- 1 = x x 2 + 1 = x x 2 + 1 < x x 2 = 1 x and so given > 0 we can choose R = 1 , and then we have x > R = | f ( x )- 1 | < 1 x < 1 R = , as required....
View Full Document

Page1 / 4

A6F08Sol-138 - Math 138 Assignment 6 Solutions Fall 2008 1....

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