Part II: The Answers
210
Next, divide the numerator and denominator by the highest power of x that appears in
the denominator, x
5
, and simplify:
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→∞
→∞
+
−
+
−
=
+
−
+
−
5
5
2
2
3
3
5
5
2
2
3
4
5
3
2
4
5
5
5
5
3
5
2
5
3 3
5
5
2 2
3
3
5
5
2
3
5
x
x x
x
x
x
x
=
+
−
+
−
→∞
lim
Then apply the limit:
=
+
− + −
=
0 0
2 0 0 0
0
241.
−1
Begin by multiplying the numerator and denominator by 1
x
so that you can simplify the
expression underneath the square root:
lim
lim
x
x
x
x
x
x
x
x
→−∞
→−∞
+
=
( )
+
2
2
1
1
1
1
Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to
use the substitution 1
1
2
x
x
= −
in the denominator. Therefore, you have
lim
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
→−∞
→−∞
→−∞
( )
+
=
−
+
=
−
+
=
1
1
1
1
1
1
1
1
2
2
2
2
2
2
→ →−∞ − +
1
1 1
2
x
Answers
201–300