Quiz 6
Math 1220–7
October 26, 2012
Key
Directions: Show all work for full credit. Clearly indicate all answers. Simplify all mathematical expressions completely. No calculators are allowed on this quiz. Each part of each
question is worth 15 points.
1. Find each limit:
3 sec x + 5
(#5 from 8.2)
(a) lim
x→π/2
tan x
This has the indeterminate form
∞
,
∞
so we can use L’Hôpital’s rule:
3 sec x tan x
3 sec x + 5
= lim
x→π/2
x→π/2
tan x
sec2 x
3 sin xx
3 tan x
= lim
= lim cos
= lim 3 sin x = 3
1
x→π/2 sec x
x→π/2
x→π/2
cos x
lim
You could have also started by multiplying the numerator and denominator of the
cos x
original fraction by cos x, leaving you with 3+5
, which you can evaluate without
sin x
L’Hôpital’s rule.
(b) lim x1/x (#23 from 8.2)
x→∞
This has the indeterminate form ∞0 , so we need to take the natural logarithm
before we can apply L’Hôpital’s rule:
ln x
x→∞ x
1/x
= lim
=0
x→∞ 1
lim ln x1/x = lim
x→∞
To find the limit of limx→∞ x1/x , we exponentiate our result to get e0 = 1.
2. Evaluate each improper integral or show that it diverges:
Z 1
dx
(#15 from 8.3)
(a)
3
−∞ (2x − 3)
Z 1
Z 1
dx
(2x − 3)−3 dx
= lim
3
a→−∞
(2x
−
3)
−∞
a
1
1
= lim − (2x − 3)−2 a→−∞
4
a
1
−2
= lim − (−1) − (2a − 3)−2
a→−∞
4
1
=−
4
Z
(b)
0
4
dx
(#11 from 8.4)
(2 − 3x)1/3
Since this function is discontinuous at x = 2/3, we need to rewrite the function as:
Z 4
Z 4
Z 2/3
dx
−1/3
(2 − 3x)−1/3 dx
(2 − 3x)
dx +
=
1/3
(2
−
3x)
2/3
0
0
Z t
Z 4
−1/3
= lim −
(2 − 3x)
dx + lim +
(2 − 3x)−1/3 dx
t→2/3
t→2/3
0
t
t
4
−1
−1
2/3 2/3 = lim −
(2 − 3x) + lim +
(2 − 3x) t→2/3
t→2/3
2
2
0
t
1
1
= − 0 − 22/3 − (−10)2/3 − 0
2
2
1 2/3
= 2 − 102/3
2