Tutorial02-solutions(1)

# 3 c with 2 2 is continuous at the origin

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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 ( =...
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## This document was uploaded on 02/10/2014.

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