251sol5 - Page 1 Section 14.1 Page 2 Section 14.1 Page 1...

Info iconThis preview shows pages 1–8. 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

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight 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: Page 1 Section 14.1 Page 2 Section 14.1 Page 1 Section 14.2 Page 2 Section 14.2 Page 1 Section 14.3 Page 2 Section 14.3 Assignment 5 Multivariate Calculus Math 251 (Fall 2010) A1. For what values of p is ths function f ( x, y ) = braceleftBigg ( x + y ) p x 2 + y 2 ( x, y ) negationslash = (0 , 0) otherwise continuous on all of R 2 ? Hints: For what values of p is g ( x ) = x p continuous on R ? Try to bound the denominator by a term that looks like radius. Solution This problem is particularly challenging to solve completely. We claim that f ( x, y ) is continuous on R 2 if and only if p > 2. We only need to check the continuity of f at (0 , 0) since elsewhere, it is a rational function. We are given that f (0 , 0) = 0 so we need to see whether or not lim ( x,y ) → (0 , 0) f ( x, y ) = 0. That is, we need to check whether ( x + y ) p x 2 + y 2 goes to 0 as radicalbig ( x − 0) 2 + ( y − 0) 2 goes to zero....
View Full Document

This note was uploaded on 12/13/2010 for the course MATH 251 taught by Professor Unknown during the Spring '08 term at Simon Fraser.

Page1 / 8

251sol5 - Page 1 Section 14.1 Page 2 Section 14.1 Page 1...

This preview shows document pages 1 - 8. Sign up to view the full document.

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