Part II: The Answers
204
217.
–∞
Consider what happens in the numerator and denominator as x approaches 0 from the
left. As x → 0
–
, you have (1 – x) → 1 and (e
x
– 1) → 0
–
. Therefore, as
x → 0
–
, it follows that
1
1
1
0
−
−
→
→−∞
−
x
e
x
218.
–∞
Begin by writing lim cot
lim cos
sin
x
x
x
x
x
→
→
−
−
=
0
0
. Then consider what happens in the
numerator and denominator as x approaches 0 from the left. As x → 0
–
, you have
cos x → 1 and sin x → 0
–
. Therefore, as x → 0
–
, it follows that
cos
sin
x
x
→
→−∞
−
1
0
219.
∞
Consider what happens in the numerator and denominator as x approaches 2.
As x → 2, you have 4e
x → 4e
2
and 2
0
− →
+
x
. Therefore, as x → 2, it follows that
4
2
4
0
2
e
x
e
x
−
→
→∞
+
220.
–∞
Consider what happens to each factor in the numerator and denominator as x
approaches 1
2
from the right. As x →
+
1
2
, you have x
2
1
5
4
+
(
) → and
cos
c os
π
π
x
( )→





 →
+
−
2
0 . Therefore, as x →
+
1
2
, it follows that
x
x
x
2
1
5 4
1
2
0
5 4
0
+
( )
→ ( )
→
→−∞
−
−
cos π
221.
–∞
Consider what happens in the numerator and denominator as x approaches 5. As x → 5,
you have sin x → sin 5, and because π < 5 < 2π, it follows that sin 5 < 0. And as x → 5, you
also have (5 – x)
4 → 0
+
. Therefore, as x → 5, it follows that
sin
s in
x
x
5
5
0
4
−
(
)
→
→−∞
+
Note that the limit is negative infinity because sin(5) is negative, and dividing sin(5) by
a small positive number close to zero gives you a negative number whose absolute
value is large.
Answers
201–300