ABSTRACT: No text of College Algebra and Pre-calculus provides a comprehensive rigorous explanation of unbounded functions. In Calculus, Advanced Calculus, and Real Analysis, unbounded functions are usually defined abstractly without using tangible computational methods. In this presentation, using a simple example, the behavior of a rational function near a vertical asymptote is explained in a comprehensive manner that is understandable to students in
College Algebra, Calculus, Advanced Calculus, and Real Analysis.
(1) A Typical Example of a Rational Function Used in College Algebra or Applied
Calculus Texts f ( x )

( x

1 )( x x

5

2 )( x

4 )
Problems:
(a) No rigorous definition of a vertical asymptote is given.
(b) No Prototype that shows the core idea of a rational function with a vertical asymptote is shown.
(c) No mechanism of a blow-up or blow-down near a vertical asymptote is shown .
(2) Definition: A Function Bounded Above:
A function f(x) is bounded above when there is a constant M that serves as a ceiling of the graph of f, i.e., f(x) < M, for all x in Dom(f).
Definition: A Strictly Increasing Function:
A function f(x) is strictly increasing on an interval I when for any a, b in I, with a < b, f(a)
< f(b) holds.
(3) Definition: A Function Unbounded Above is defined by Negating the Definition in (2).
A function f(x) is unbounded above when no matter how large a constant M is used for a ceiling of function f, the ceiling of height y = M is surpassed by Graph(f).
Ex. Prove that f(x) = x 2 , as x →∞, is unbounded above.

Illustration using M = 10 6 .
(i) Take a > 0. Then for any x > 0, (x + a) 2 = x 2 + 2ax + a 2 > x 2 . So, f(x) is strictly increasing for x > 0.
(ii) Solve x
2
= 10
6
to find the point of intersection of y = x
2
with ceiling y = 10 6 .
X = 10 3 .
Since f is strictly increasing for x > 0, x > 10
3  x
2
> 10
6
. The ceiling has been surpassed by Graph(f).
(iii) Repeat the argument with M = 10
9
, M = 10
12
.
(iv) Replace 10 12 with M. Find the point of intersection of y = x 2 with y = M, to get x = M
1/2
. So, x > M
1/2
implies x
2
> M. As M →∞, x
2 →∞.
(4) Exponential Functions are Unbounded Above
Consider f(x) = 10 x
. Set a ceiling of height y = M (as large as we want).
Solve 10 x
= M for x by taking log with base 10. x
 log
10
M f ( x
 a )

10 x
 a 
. f is a strictly increasing function since
10 x 
10 a  f ( x )

10 a  f ( x ) .
 a

R,
So, x
 log
10
M
 f ( x )

M .
(5) Prototype Rational Function with a Vertical Asymptote
Consider f ( x )

1 x having a vertical asymptote at x = 0.
(i) Investigate the Behavior of f(x) near x = 0.
(ii) Evaluate this function at x

10

6 for x > 0 near 0, and at x
 
10

6 for x < 0 near 0. f ( 10

6
)

1
10

6

10
0
10

6

10
0

(

6 ) 
10
6 
1 , 000 , 000 .
(Blow-Up) f (

10

6
)

1

10

6

10
0

10

6
 
10
0

(

6 )  
10
6  
1 , 000 , 000 .
(Blow-Down)
(iii) Replace x

10

6 in (ii) with x = 10 -n for x > 0 near 0, and at x
 
10

6 for x < 0 with x =
−10 -n
near 0. Then, by (4), f ( 10
 n
)

10 n  
as n →∞ (Blow-Up). f (

10
 n )
 
10 n  
as n →∞ (Blow-Down).

Example 1 (Minor Variation of Prototype):
(i) Let f
2
( x )
 x
1

2
, and w = x – 2 . Then, g ( w )

1 w has the same graph on the xw-plane as f(x). w = 0 corresponds to x = 2 . So, f
2
( x ) is the right translation of f(x) by 2.
(ii) Numerically, evaluate this function at 2 values very close to the vertical asymptote x = 2, x

2

10
 n
, x

2

10
 n to get f
2
( 2

10
 n
)

2
1

10
 n 
2
 
10 n   as n →∞, and f
2
( 2

10
 n
)

2
1

10
 n 
2

10 n   as n →∞,
Example 2:
Consider f ( x )

1
( x

1 )( x

2 )
having 2 vertical asymptotes, and evaluate this function at 2 values very close to the vertical asymptote x = 2, x

2

10

6
, x

2

10

6 , and at 2 values very close to the vertical asymptote x = −1, x
 
1

10

6
, x
 
1

10

6
.
(i) I expect that students will get stuck in (ii). They always get stuck. I let them struggle for about 2 minutes. For example, f (

1

10

6
)

(

1

10

6
1

1 )(

1

10

6 
2 )

10

6
(

3
1

10

6
)
. They get stuck here.
(ii) I tell them that what matters here is not the exact value of the function at x
 
1

10

6
, but the magnitude of the most dominant term in the expression. Adding 10

6 to

3 does not change

3 significantly. Introducing the approximate equality symbol

, I show

f (

1

10

6
)

1
10

6 (

3

10

6 )

1
10

6 (

3 )
 
1
3

10
6
(Blow-Down).
(iv) I let the students handle the remaining 3 cases f (

1

10

6
), f ( 2

10

6
), f ( 2

10

6
) .
(v) Replace 10
-6 in (iv) with 10
-n
.
Example 3:
(i) I write f ( x )

( x
1

1 )( x

2 )( x

4 )
.
(ii) I tell the students to repeat the procedure of Example 2.
Example 4:
(i) I write f ( x )

( x

1 )( x x

5

2 )( x

4 )
.
(ii) I tell the students to repeat the procedure of Example 3.