Part II: The Answers
214
247.
− 5
2
Notice that if you consider the limit immediately, you have the indeterminate
form –∞ + ∞.
Create a fraction and multiply the numerator and denominator by the conjugate of the
expression x
x
x
+
+
(
)
2
5 . The conjugate is x
x
x
−
+
(
)
2
5 , so you have
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→−∞
→−∞
+
+
(
)
=
+
+
(
) − +
−
+






=
2
2
2
2
5
5
1
5
5
x x
x
x
x
x
→−∞
−
−
+
5
5
2
Next, multiply the numerator and denominator by 1
x
so that you can simplify the
expression underneath the square root:
lim
lim
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
→−∞
→−∞
→−∞
−
−
+
=
−
( )
−
+
(
)
=
−
( )
5
5
1 5
1
5
5
1
2
2
− −
+
=
−
−
+
→−∞
1
5
5
1 1
5
2
2
x
x
x
x
x
x
x
lim
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
x
x
x
x
x
x
x
x
x
x
→−∞
→−∞
→−∞
−
−
+
=
−
− −
+






=
−
+
5
1 1
5
5
1
1
5
5
1
2
2
2
2 2
2
2
5
5
1 1 5
x
x
x
x
x
+
=
−
+ +
→−∞
lim
Now apply the limit:
=
−
+ +
= −
5
1 1 0
5
2
Answers
201–300