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.
 Spring '13

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