3 c with 2 2 is continuous at the origin

Info iconThis preview shows page 1. Sign up to view the full content.

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: = 2 + 3 · 12 2 7 4 (b) We see that for ( )= 4 +3 4 ( if we approach the origin along the -axis, we have 0) = 0 → 0 as →0 as →0 while if we approach along the -axis (0 )= 1 1 → 3 3 Since these limits are different, lim ( )→(0 0) does not exist. 3 ( ) )= √ (c) With ( 2+ +2 , is continuous at the origin so lim )→(0 0) ( 2+ (d) With ( ) = ++ 2 (the -axis), we have ( ) = (0 0) = 0 2 . Approaching the origin along the line ( 2+ 4 ) = (0 0 ) = 0 → 0 ( as →0 On the other hand, approaching the origin along the line ( 2 ( )= ( 0) = 2 2 1 1 → 2 2 = ) = (0 0 ) as )=( 0) gives 0 Since these limits are different, lim )→(0 0 0) ( ( ) does not exist. (e) We have for ( 0≤ 3 + 2+ ) = (0 0) that | 3 2 3 3 + = ( ) | ≤ 2 ( | |3 + | |3 ≤ ( )2 ) 3+ ( ) ( )2 3 =2 ( ) →0 as ( ) → (0 0) :) so we can conclude (by the Squeeze theorem) that 3 + 2+ lim ( )→(0 0) 3 2 =0 3+ 3 An alternative solution is to change to polar coordinates: With ( have in polar coordinates = cos θ , = sin θ , 3 3 + 2+ Since ( 2 3 cos3 θ + = lim )→(0 0) sin3 θ 2 ) → (0 0) is equivalent to ( 3 ( =...
View Full Document

This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online