Answers and Explanations 191
Answers
101–200
lim
lim
lim
x
x
x
x
x
x
x
x
x
→
→
→
−
−
−
−
=
− −
=
+
0
2
0
2
0
2
2
2
2
2
2
2
=
+
→
−
lim
x
x
x
0
2
2 2
Now consider the numerator and denominator as x → 0
–
. As x → 0
–
, you have
(2 + 2x) → 2 and x
2 → 0
+
so that 2 2
2
0
2
+
→
→∞
+
x
x
. The limit is positive infinity
because dividing 2 by a very small positive number close to zero gives you a very large
positive number.
184.
limit does not exist
Note that substituting in the limiting value gives you the indeterminate form 0
0
.
Examine both the left-hand limit and right-hand limit to determine whether the limits
are equal. To find the left-hand limit, consider values that are slightly smaller than 4
3
and
substitute into the limit. Notice that to simplify the absolute value in the denominator of
the fraction, you replace the absolute values bars with parentheses and add a negative
sign, because substituting in a value less than 4
3
will make the number in the parentheses
negative; the extra negative sign will make the value positive again.
lim
lim
lim
x
x
x
x
x
x
x x
x
x
→
→
→
−
−
−
−
−
=
−
(
)
−
−
(
)
=
−
= −
4
3
2
4
3
4
3
3
4
3
4
3
4
3
4
1
4 3
1
= = − 4
3
You deal with the right-hand limit similarly. Here, when removing the absolute value
bars, you simply replace them with parentheses; you don’t need the negative sign
because the value in the parentheses is positive when you’re substituting in a value
larger than 4
3
.
lim
lim
(
)
(
)
lim
x
x
x
x
x
x
x x
x
x
→
→
→
+
+
+
−
−
=
−
−
=
=
4
3
2
4
3
4
3
3
4
3
4
3
4
3
4
1
4
3
Because the right-hand limit doesn’t equal the left-hand limit, the limit doesn’t exist.
Answers
101–200
lim
lim
lim
x
x
x
x
x
x
x
x
x
→
→
→
−
−
−
−
=
− −
=
+
0
2
0
2
0
2
2
2
2
2
2
2
=
+
→
−
lim
x
x
x
0
2
2 2
Now consider the numerator and denominator as x → 0
–
. As x → 0
–
, you have
(2 + 2x) → 2 and x
2 → 0
+
so that 2 2
2
0
2
+
→
→∞
+
x
x
. The limit is positive infinity
because dividing 2 by a very small positive number close to zero gives you a very large
positive number.
184.
limit does not exist
Note that substituting in the limiting value gives you the indeterminate form 0
0
.
Examine both the left-hand limit and right-hand limit to determine whether the limits
are equal. To find the left-hand limit, consider values that are slightly smaller than 4
3
and
substitute into the limit. Notice that to simplify the absolute value in the denominator of
the fraction, you replace the absolute values bars with parentheses and add a negative
sign, because substituting in a value less than 4
3
will make the number in the parentheses
negative; the extra negative sign will make the value positive again.
lim
lim
lim
x
x
x
x
x
x
x x
x
x
→
→
→
−
−
−
−
−
=
−
(
)
−
−
(
)
=
−
= −
4
3
2
4
3
4
3
3
4
3
4
3
4
3
4
1
4 3
1
= = − 4
3
You deal with the right-hand limit similarly. Here, when removing the absolute value
bars, you simply replace them with parentheses; you don’t need the negative sign
because the value in the parentheses is positive when you’re substituting in a value
larger than 4
3
.
lim
lim
(
)
(
)
lim
x
x
x
x
x
x
x x
x
x
→
→
→
+
+
+
−
−
=
−
−
=
=
4
3
2
4
3
4
3
3
4
3
4
3
4
3
4
1
4
3
Because the right-hand limit doesn’t equal the left-hand limit, the limit doesn’t exist.