Review of Limits

Chapter 1, Preliminaries: Section 1.
1, Review of
CalculusReview of Limits And Absolute Value
C. Byron Harding
December 19, 2015
Hello,
At times, life throws a curve ball. So, I am taking time to document a review path that I use
to review crucial topics for learning numerical methods[1;2]. It will help me later if I must return
after some time away, and I will learn LaTex as well.
Limits and Continuity
Denition 1.1.
Assume that f(x) is dened on a set S of real numbers. Then f is said
to have the limit L at x =x0 and we write
lim f (x) = L
(1)
xx0
if, given any
implies that
> 0, there exists

|f (x) L| < [1].
>0
such that, whenever x
S,
0 < |x x0 | <
After many years of not doing math, the above abstract description of limits seemed daunting. I
felt that I understood it, but I also felt that I was mistaken. So, I decided to review my math books.
From [3;3a], I discovered that the above description of limits is the Rigorous Approach [3]. There
are some major details overlooked in the numerical method textbook[1], but, hey, a person is suppose to know calculus or review topics of calculus before embarking.
As an example, my calculus textbook shows
lim f (x) = L
(2)
xa
[Page 130, 3]
actually describes, intuitively[3], that the value of f(x) approaches L as x approaches a from
either side.
So, this is a two-sided limit problem and that is important.
The limit in (sic 2) is intended to describe the behavior of f when x is near but dierent
from a. The value of f at a is irrelevant to the limit.

In contrast, it is expected that f(x) approaches L means that f(x) can take on the value of
L.
Note, the use of a graphic really helps me understand because it is an example, but I need to
get back to more abstract thinking.
Also, I need to emphasize that x can
approach
a but
not take on
the value it approaches.
As an example, I provide an equation that makes the point[3]
lim
x0
sin(x)
=1
x
(3)
If x became 0, we would have an undened resultdivision by zero is innity. Still, the limit can
become 1.
I should remind myself that the mathematics of absolute values really means that a distance is
measured from two dierent directions to the center point.
L < f (x) < L +
is equivalent to
|f (x) L| <
Rigorous Approach Limit Example

First, I would like to quote a paragraph from [3].
As Illustrated in Example (sic below), a limit proof proceeds as follows: We assume
that an arbitrary positive number
is given to us.
Then we try to nd a positive number
such that the conditions in the denitions are fullled. The idea is to nd a formula
for
in terms of
so
that no matter what
formula the required
is
chosen, we automatically obtain from the
.[3]
lim (3x 5) = 1
(4)
x2
Solution. We must show that given any positive number , we can nd a positive number
that
f (x) = 3x 5
such
satises[3]
|f (x) 1| < = |(3x 5) 1| <
(5)
0 < |x a| < 0 < |x 2| <
(6)
whenever x satises
To nd
we can rewrite (5) as
|3(x) 6| < 3|x 2| < |x 2| <

The trick is to choose

3
(7)
so that (7) is satised whenever (6) is satised(1).
=
To prove that this choice for

3
(8)
works, suppose that x satises (6). Because

3 , it follows
from (6) that x satises
0 < |x 2| <

3
(9)

Absolute Value Point: Remember, |x 2| < 3 is the same as 3 < x 2 < 3 and has a solution

set of 2
3 < x < 2 + 3 . The latter is a result of Theorem for distance formula between points on,
for example, a number line[3].
In fact, I should review a little more about absolute values and provide a diagram to help explain
the above results.
Denitions of Absolute Value[4]

Absolute values measure distance!
|| = if 0 and || = if 0
As an example,
|5| = 5
and
(10)
| 5| = (5) = 5.
The above denition is quite important and must be remembered technically when working with
complicated problems.
Two Properties of Absolute Value[4]

if
> 0
|m| < is equivalent to < m <
(11)
|m| is equivalent to m
(12)
With the above denitions (10) and properties (11-12), and the below diagram, I can better
understand and derive the alternative form of the inequality and the solution set of (9) given above.
In fact, I will use a general diagram.
Must remember that the absolute value measures distance. In the case of the above diagram,
|x a|
has a value
(x a) > 0
so the denition is satised that
|x a| = (x a)(10).
Also, the
property(11) tells us
|m| < where m = (x a) so that < (x a) <

4
(13)
Finally, the solution set is
a <x<+a
(14)
Conclusion: Limits and Continuity
Denition 1.1.
Assume that f(x) is dened on a set S of real numbers. Then f is said
to have the limit L at x =x0 and we write
lim f (x) = L
(15)
xx0
if, given any
implies that
> 0, there exists

|f (x) L| < [1].
>0
such that, whenever x
For clarity and connection to above derivation,
x0 = a
S,
0 < |x x0 | <
It can be seen that the
0 < |x x0 | <
satises the denitions of absolute values(10) and the property(11).

Remember that absolute values measure distance, and the properties means that the above is a
two sided limit problem.
Also, the MIT video on the Rigorous Approach[5] of the Limit is quite valuableespecially when
seeing how absolute values are used and the reason for the rigorous approach.
As such, I need
to supply some more properties of absolute values and one particular inequality property that is
common sense but I easily overlook, abstract sense, unless I plug in the numbers. Yep, it is really easy. To be most eective, a review of properties and examples in my math books is advised too.
If
|a| 0
for every real number and
| a| = |a|
|ab| = |a||b|
|a + b| |a| + |b|
|a| = 0
if and only if
a = 0.
Triangle inequality
Taking reciprocals reverses an inequality of numbers with the same sign:
a < b,
then
1
a
>
1
b
Yes, I know that the a larger denominator means a smaller outcome and vice versa. I just forget
it when doing variable math.
The MIT professor begins the very important discussion from video time: 35:12 to end at 46
minutes[5].
To compliment his discussion, and a nice review of Theorem 2.5.1 (Page 119; 3), I provide the
important theorem 2.5.1
Theorem 2.5.1.
limxto .
If
Let lim stand for one of the limits
L1 = lim f (x)
and
L2 = lim g(x)
limxa , limxa , limxa+ , limx+ ,
or
both exists, then
(a) lim [f (x) + g(x)] = lim f (x) + lim g(x) = L1 + L2
(16a)
(b) lim [f (x) g(x)] = lim f (x) lim g(x) = L1 L2
(16b)
(c) lim [f (x)g(x)] = lim f (x) lim g(x) = L1 L2
(16c)
f (x)
lim f (x)
L1
=
=
g(x)
lim g(x)
L2
p
p
p
(e) lim n f (x) = n lim f (x) = n L1
(d) lim
(16d)
(16e)
(a) The limit of a sum is the sum of the limits.

(b) The limit of the dierence is the dierence of the limits.
(c) The limit of a product is the product of the limits.
(d) The limit of the quotient is the quotient of the limits provided the limit of the denominator is nonzero.
(e) The limit of an nth root is the nth root of the limit.
Closing Quote
In truth, the above examples are simplied. Real world problems will not likely work out so
easily. Here is what Dr. Anton had to say:
Example 1 is about as easy a limit proof can get; most limit proofs require a little more
algebraic and logical ingenuity. (sic: read basic knowledge should be well understood.)
The reader who nds
00
00
discussions hard going should not become discouraged;
the concepts and techniques are intrinsically dicult. In fact, a precise understanding
of limits evaded the nest mathematical minds for centuries.[3]
References:
[1] Mathews, John H.; Fink, Kurtis, D. (1999).
Numerical Methods Using Matlab.
(Third
Edition). Prentice Hall

[2] Quarteroni, Alo; Saleri, Fausto; Gervasio, Paola. Scientic Computing with MATLAB and
Octave. Fourth Edition. Retrieved (2015, Dec. 18). Shannon[online]
[3] Anton, Howard. Calculus with Analytic Geometry. Fifth Edition.
[3a] Barker, William H.; Ward, James E. (1995). The Calculus Companion to Accompany Calculus with Analytic Geometry. Fifth Edition by Howard Anton. John Wiley
[4] Grossman, Stanley I. (1992) Algebra and Trigonometry, Second Edition. Saunders College
[5] MIT OpenCourseWare.
(2011, May 05).
Unit I: Lec 5 | MIT Calculus Revisted: Single
Variable Calculus, Unit !: Lecture 5: A More Rigorous Approach to Limits. Retrieved (2015, Dec.
28). youtube[online].

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Chapter 1, Preliminaries: Section 1.

Limits and Continuity

Assume that f(x) is dened on a set S of real numbers. Then f is said

to have the limit L at x =x0 and we write

 > 0, there exists

such that, whenever x

So, this is a two-sided limit problem and that is important.

from a. The value of f at a is irrelevant to the limit.

the value it approaches.

As an example, I provide an equation that makes the point[3]

L  < f (x) < L + 

Rigorous Approach Limit Example

Then we try to nd a positive number

that no matter what

formula the required

chosen, we automatically obtain from the

|f (x) 1| <  = |(3x 5) 1| < 

0 < |x a| < 0 < |x 2| <

we can rewrite (5) as

|3(x) 6| <  3|x 2| <  |x 2| <

so that (7) is satised whenever (6) is satised(1).

works, suppose that x satises (6). Because

from (6) that x satises

Denitions of Absolute Value[4]

Two Properties of Absolute Value[4]

|m| < is equivalent to < m <

so the denition is satised that

|m| < where m = (x a) so that < (x a) <

Finally, the solution set is

Conclusion: Limits and Continuity

Assume that f(x) is dened on a set S of real numbers. Then f is said

to have the limit L at x =x0 and we write

 > 0, there exists

such that, whenever x

For clarity and connection to above derivation,

It can be seen that the

satises the denitions of absolute values(10) and the property(11).

for every real number and

Taking reciprocals reverses an inequality of numbers with the same sign:

Let lim stand for one of the limits

limxa , limxa , limxa+ , limx+ ,

both exists, then

(a) lim [f (x) + g(x)] = lim f (x) + lim g(x) = L1 + L2

(b) lim [f (x) g(x)] = lim f (x) lim g(x) = L1 L2

(c) lim [f (x)g(x)] = lim f (x) lim g(x) = L1 L2

(a) The limit of a sum is the sum of the limits.

discussions hard going should not become discouraged;

Numerical Methods Using Matlab.

Edition). Prentice Hall

(2011, May 05).

Unit I: Lec 5 | MIT Calculus Revisted: Single

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Assume that f(x) is dened on a set S of real numbers. Then f is said

> 0, there exists

from a. The value of f at a is irrelevant to the limit.

L < f (x) < L +

Then we try to nd a positive number

|f (x) 1| < = |(3x 5) 1| <

|3(x) 6| < 3|x 2| < |x 2| <

so that (7) is satised whenever (6) is satised(1).

works, suppose that x satises (6). Because

from (6) that x satises

Denitions of Absolute Value[4]

so the denition is satised that

Assume that f(x) is dened on a set S of real numbers. Then f is said

> 0, there exists

satises the denitions of absolute values(10) and the property(11).