hw4sol - Math 125A Fall 2009 Solutions for Homework...

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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 125A, Fall 2009 Solutions for Homework 4 (Prepared by Matt Low) 1. Regardless of the argument of the cosine function, we have | cos u | ≤ 1, and so: lim x → x cos ( 3 x 2 + e sin x- ln | x | ) ≤ lim x → | x | = 0 and consequently: lim x → x cos ( 3 x 2 + e sin x- ln | x | ) = 0 2. (a) False. For example, the function f ( x ) = sin( πx ) is continuous and satisfies: lim n →∞ a n = lim n →∞ sin( nπ ) = lim n →∞ 0 = 0 But lim x →∞ f ( x ) does not even exist. (b) False. For example, for the functions f ( x ) = x and g ( x ) =- x , we have f ( x ) → ∞ and g ( x ) → -∞ as x → ∞ , and: lim x →∞ [ f ( x ) + g ( x )] = lim x →∞ 0 = 0 (c) False. Likewise, we can take f ( x ) = x and g ( x ) = 1 /x . Of course, we have f ( x ) → ∞ and g ( x ) → 0 as x → ∞ , and: lim x →∞ f ( x ) g ( x ) = lim x →∞ 1 = 1 3. (a) Let ε > 0 be given. In accordance with statement (i), there is a δ > 0 so that if | h | < δ , then | f ( x + h )- f...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

hw4sol - Math 125A Fall 2009 Solutions for Homework...

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