Part II: The Answers
186
Answers
101–200
167.
1
As x approaches 3 from the left, the y values approach 1 so that lim ( )
x
f x
→
−
=
3
1.
168.
3
As x approaches 3 from the right, the y values approach 3 so that lim ( )
x
f x
→
+
=
3
3.
169.
2
As x approaches –3 both from the left and from the right, the y values approach 2 so
that lim ( )
x
f x
→−
=
3
2.
Note that the actual value of f
(–3) doesn’t matter when you’re finding the limit.
170.
3
As x approaches 1 from the left, the y values approach 3 so that lim ( )
x
f x
→
−
=
1
3.
171.
3
As x approaches 1 from the right, the y values approach 3 so that lim ( )
x
f x
→
+
=
1
3.
172.
5
As x approaches –2 both from the left and from the right, the y values approach 5 so
that lim ( )
x
f x
→−
=
2
5.
173.
4
Note that substituting the limiting value, 3, into the function x
x
x
2
2
3
3
−
−
−
gives you the
indeterminate form 0
0
.
To find the limit, first factor the numerator and simplify:
lim
lim
(
)(
)
(
)
lim(
)
x
x
x
x
x
x
x
x
x
x
→
→
→
− −
−
=
−
+
−
=
+
3
2
3
3
2
3
3
3
1
3
1
Then substitute 3 for x:
lim(
)
x
x
→
+ = + =
3
1 3 1 4
186
Answers
101–200
167.
1
As x approaches 3 from the left, the y values approach 1 so that lim ( )
x
f x
→
−
=
3
1.
168.
3
As x approaches 3 from the right, the y values approach 3 so that lim ( )
x
f x
→
+
=
3
3.
169.
2
As x approaches –3 both from the left and from the right, the y values approach 2 so
that lim ( )
x
f x
→−
=
3
2.
Note that the actual value of f
(–3) doesn’t matter when you’re finding the limit.
170.
3
As x approaches 1 from the left, the y values approach 3 so that lim ( )
x
f x
→
−
=
1
3.
171.
3
As x approaches 1 from the right, the y values approach 3 so that lim ( )
x
f x
→
+
=
1
3.
172.
5
As x approaches –2 both from the left and from the right, the y values approach 5 so
that lim ( )
x
f x
→−
=
2
5.
173.
4
Note that substituting the limiting value, 3, into the function x
x
x
2
2
3
3
−
−
−
gives you the
indeterminate form 0
0
.
To find the limit, first factor the numerator and simplify:
lim
lim
(
)(
)
(
)
lim(
)
x
x
x
x
x
x
x
x
x
x
→
→
→
− −
−
=
−
+
−
=
+
3
2
3
3
2
3
3
3
1
3
1
Then substitute 3 for x:
lim(
)
x
x
→
+ = + =
3
1 3 1 4
