Part II: The Answers
206
226.
limit does not exist
Begin by examining both the left-hand limit and the right-hand limit to determine
whether they’re equal.
To find the limits, consider what happens in the numerator and denominator as x
approaches e
2
. For the left-hand limit, as x → e
2–
, you have –x → –e
2
and (ln x – 2) → 0
–
.
Therefore, as x → e
2–
, it follows that
−
−
→ − → ∞
−
x
x
e
ln
2
0
2
Note that the limit is positive infinity because dividing –e
2
by a small negative number
close to zero gives you a large positive number.
For the right-hand limit, as x → e
2+
, you have –x → –e
2
and (ln x – 2) → 0
+
. Therefore, as
x → e
2+
, it follows that
−
−
→ − → −∞
+
x
x
e
ln
2
0
2
Because the left-hand limit doesn’t equal the right-hand limit, the limit doesn’t exist.
227.
limit does not exist
Begin by examining the left-hand limit and the right-hand limit to determine whether
they’re equal.
To find the limits, consider what happens in the numerator and denominator as x
approaches 2. For the left-hand limit, as x → 2
–
, you have (x + 2) → 4 and
(x
2
– 4) → 0
–
. Therefore, as x → 2
–
, it follows that
x
x
+
−
→
→−∞
−
2
4
4
0
2
For the right-hand limit, as x → 2
+
, you have (x + 2) → 4 and (x
2
– 4) → 0
+
. Therefore, as
x → 2
+
, it follows that
x
x
+
−
→
→∞
+
2
4
4
0
2
Because the left-hand limit doesn’t equal the right-hand limit, the limit doesn’t exist.
228.
limit does not exist
Begin by examining the left-hand limit and the right-hand limit to determine whether
they’re equal.
To find the limits, consider what happens in the numerator and denominator as x
approaches 25. For the left-hand limit, as x → 25
–
, you have 5
5
25 10
+
(
) → + =
x
and
(x – 25) → 0
–
. Therefore, as x → 25
–
, it follows that
5
25
10
0
+
−
→
→−∞
−
x
x
For the right-hand limit, as x → 25
+
, you have 5
5
25 10
+
(
) → + =
x
and (x – 25) → 0
+
.
Answers
201–300