Answers and Explanations 207
Therefore, as x → 25
+
, you have
5
25
10
0
+
−
→
→∞
+
x
x
Because the left-hand limit doesn’t equal the right-hand limit, the limit doesn’t exist.
229.
–∞
Consider what happens to each factor in the numerator and denominator as x
approaches 0. As x → 0, you have (x
2
+ 4) → 4, x
2 → 0
+,
and (x – 1) → –1. Therefore, as
x → 0, it follows that
x
x x
2
2
4
1
4
0
1
4
0
+
−
(
)
→
−
( )
→
→−∞
+
−
230.
–∞
Consider what happens to each factor in the numerator and denominator as x
approaches 0 from the left. You have (x – 1) → –1, x
2 → 0
+
, and (x + 2) → 2. Therefore, as
x → 0
–
, you get the following:
x
x x
−
+
(
)
→ −
( )
→ − → −∞
+
+
1
2
1
0 2
1
0
2
231.
∞
Consider what happens in the numerator and denominator as x approaches 3. As
x → 3, you have (3 + x) → 6 and 3
0
− →
+
x
. Therefore, as x → 3, it follows that
3
3
6
0
+
−
→
→∞
+
x
x
232.
− π
2
As the x values approach –∞, the y values approach − π
2
so that lim
x
f x
→−∞
( ) = − π
2
.
Answers
201–300