Homework 7 Solutions
1.) Give a geometric interpretation of the Cartesian product [2, 4]×[−3, 3]×[0, 1], include
dimensions where appropriate.
This Cartesian product is a solid rectangular prism in R3 with corners (2, −3, 0), (2, −3, 1), (2, 3, 0),
(2, 3, 1), (4, −3, 0), (4, −3, 1), (4, 3, 0), and (4, 3, 1). It has width 4−2 = 2, length 3−(−3) = 6,
and height 1 − 0 = 1.
2.) Prove that the set N3 is countable.
We will show that the function f (x, y, z) = 2x ∗ 3y ∗ 5z is an injective function from N3 to
N, from which it follows that N3 is countable.
For (x, y, z) ∈ N3 , 2x ∗ 3y ∗ 5z is a positive integer, so the image of f is a subset of N.
Now assume f (x1 , y1 , z1 ) = f (x2 , y2 , z2 ).
Then 2x1 ∗ 3y1 ∗ 5z1 = 2x2 ∗ 3y2 ∗ 5z2 .
Since each number in the products is prime, the corresponding exponents must be equal,
i.e. x1 = x2 , y1 = y2 , z1 = z2 .
Thus (x1 , y1 , z1 ) = (x2 , y2 , z2 ), so f is injective.
3.) Prove that | [2, 5] | = | [15, 249] | by constructing a bijective function from one set
to the other, finding its inverse, and showing that the composition of the function with its
inverse equals the identity function.
Let f (x) = 2x3 − 1, the image of [2, 5] under f is [15, 249].
p
Here f −1 (x) = 3 (x + 1)/2.
√
p
p
3
Thus f −1 (x) ◦ f (x) = 3 (f (x) + 1)/2 = 3 (2x3 + 1 − 1)/2 = x3 = x.
4.) Use the -δ definition of a limit to show that limx→3 x3 = 27.
Here |f (x) − L| = |x3 − 27| = |(x2 + 3x + 9)(x − 3)|.
The function x2 + 3x + 9 is increasing when x > −3/2, so on the interval (3 − r, 3 + r) it
is bounded above by (3 + r)2 + 3(3 + r) + 9 = r2 + 9r + 27.
Thus |(x2 + 3x + 9)(x − 3)| < (r2 + 9r + 27)|x − 3|, so choose δ = /(r2 + 9r + 27).
1
Then |x − 3| < δ =⇒ |x3 − 27| < , so limx→3 x3 = 27.
5.) Use the -δ definition of continuity to show that f (x) =
2
(x−3)2
is continuous at x = 5.
First, f (5) = 1/2.
2
4−(x−3)
−(x−5)(x−1)
(x−1)
1
−x +6x−5
2
| = 12 | (x−3)
Then |f (x)−f (5)| = | (x−3)
2 − 2 | = | 2(x−3)2 | = | 2(x−3)2 | = |
2 (x−5)|
2(x−3)2
Now
x−1
(x−3)2
2
is decreasing when x > 3, so it is bounded above on (5 − r, 5 + r) by
So, choose δ = ∗
at x = 5.
(2−r)2
,
4−r
then |x − 5| < δ =⇒ |f (x) − f (5)| < , and so f is continuous
6.) Use the -δ definition of an infinite limit to show that limx→∞
ax+b
cx+d
bc−ad
Here |f (x) − L| = | ax+b
− ac | = | (ax+b)c−a(cx+d)
| = | (cx+d)c
| = | c2bc−ad
|.
cx+d
(cx+d)c
(x+d/c)
This is less than if and only if
So choose M =
|bc−ad|
∗c2
Therefore limx→∞
ax+b
cx+d
4−r
.
(2−r)2
|bc−ad|
∗c2
< (x + d/c).
− d/c, then x > M =⇒ | ax+b
− ac | < .
cx+d
= ac .
2
= ac .