Answers and Explanations 189
Answers
101–200
Then substitute 0 for x:
lim
x
x
→
+
= +
=
0
1
1
1
0 1
1
178.
0
The given function is continuous everywhere, so you can simply substitute in the limiting value:
lim
x
x
→
− = − =
4
4
4 4 0
179.
3
Note that substituting the limiting value, –1, into the function x
x
3
1
1
+
+
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 x
→−
→−
→−
+
+
=
+
− +
(
)
+
=
− +
(
)
1
3
1
2
1
2
1
1
1
1
1
1
Then substitute –1 for x:
lim
( )
x
x x
→−
− +
(
) = −
( ) − − + =
1
2
2
1
1
1 1 3
180.
1
4
Note that substituting the limiting value, 0, into the function 4
2
+ −
h
h
gives you the
indeterminate form 0
0
.
To find the limit, first multiply the numerator and denominator by the conjugate of the
numerator and then simplify:
lim
lim
lim
h
h
h
h
h
h
h
h
h
h
h
h
→
→
→
+ −
=
+ −
(
) + +
(
)
+ +
(
)
=
+ −
+ +
0
0
0
4
2
4
2
4
2
4
2
4
4
4
2 2
4
2
1
4
2
0
0
(
)
=
+ +
(
)
=
+ +
→
→
lim
lim
h
h
h
h
h
h
Then substitute 0 for h:
lim h
h
→
+ +
=
+ +
=
0
1
4
2
1
4 0 2
1
4