Thomas Calculus 13th [Solutions]

110 Chapter 2 Limits and Continuity
Step 2: X 1 x 1 1 x 1
4 4 4 4 .
Then
1 1 1 1
4 4 2 4 4 2 4(2 ) , or 1 1 1 1
4 4 2 4 2 4 4(2 ) .
Choose
4(2 ) , the smaller of the two values. Then 0 x
1 1
2 and
4 2x
By the continuity test, g( x ) is continuous at x 1
9. Show lim h( x) lim 2x 3 1 h(2).
x 2 x 2
Step 1: 2x
3 1 2x
3 1
Step 2: | x 2| x 2 or 2 x 2.
Then
4 .
(1
2
) 3 (1
2
) 3 1 (1
2
)
2 2 2
2 2
2
2 . Choose 2
2
so lim 2 3 1.
x
2
1
4
1
2x
lim 2.
x
2 2
(1 ) 3 (1 ) 3
x
2 2
1 2x
3 1 .
2 2
(1 ) 3
, or 2
2 2
, the smaller of the two values. Then, 0 | x 2| 2 x 3 1 ,
x By the continuity test, h( x ) is continuous at x 2.
2 2
(1 ) 3 (1 ) 1
2
2 2
10. Show lim F( x) lim 9 x 2 F(5).
x 5 x 5
2 2
Step 1: 9 x 2 9 x 2 9 (2 ) x 9 (2 ) .
Step 2: 0 | x 5| x 5 5 x 5.
2 2 2
2 2 2
Then 5 9 (2 ) (2 ) 4 2 , or 5 9 (2 ) 4 (2 ) 2 .
2
Choose 2 , the smaller of the two values. Then, 0 | x 5| 9 x 2 , so lim 9 x 2.
x 5
By the continuity test, F( x ) is continuous at x 5.
11. Suppose L 1 and L 2 are two different limits. Without loss of generality assume 2 1 . 1
3 ( 2 L 1 lim
x x0
f ( ) L 1 1 0 such that 0 | x x0 | 1 | f ( x) L1
|
f ( x)
L1
1 ( L 1
2 L1 ) L1 f ( x) ( L
3 3 2 L ) L
1 1 4L1 L2 3 f ( x) 2 L1 L 2.
Likewise, lim
x x0
f ( ) L 2 there is a 2
such that 0 | x x0 | 2 | f ( x) L2 | f ( x)
L 1 1
2 ( L2 L1 ) L
3 2 f ( x) ( L
3 2 L1 ) L2
2L2 L1 3 f ( x) 4L2 L1 L1 4L2 3 f ( x) 2 L2 L 1.
If min{ 1, 2} both inequalities must
4L1 L2 3 f ( x) 2L1 L2
hold for 0 | x x0
| :
5( L1 L2 ) 0 L1 L2.
That is, L
L1 4L2 3 f ( x) 2L2 L
1 L 2 0 and
1
L1 L 2 0, a contradiction.
12. Suppose lim f ( x) L . If k 0, then lim k f ( x ) lim 0 0 0 lim f ( x ) and we are done. If k 0, then given
x c x c x c
x c
any 0, there is a 0 so that 0 | x c| | f ( x) L| | k || f ( x) L | | k( f ( x) L)|
| k |
|( kf ( x)) ( kL )| . Thus lim k f ( x) kL k lim f ( x) .
x c x c
3 3 3
3
13. (a) Since x 0 , 0 x x 1 ( x x) 0 lim f ( x x)
lim f ( y) B where y x x.
x 0
y 0
3 3 3
3
(b) Since x 0 , 1 x x 0 ( x x) 0 lim f ( x x)
lim f ( y) A where y x x.
x 0
y 0
4 2 2 4 2 4
2 4
(c) Since x 0 , 0 x x 1 ( x x ) 0 lim f ( x x ) lim f ( y) A where y x x .
x 0
y 0
4 2 2 4
2 4
(d) Since x 0 , 1 x 0 0 x x 1 ( x x ) 0 lim f ( x x ) A as in part (c).
x 0
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