Calculus Of One Real Variable By Pheng Kim Ving Chapter 1: Limits And Continuity Section 1.1.2: Properties Of Limits 1.1.2 Properties Of Limits

 1. Absolute Values

The definition of the limit involves absolute values; see Section 1.1.1 Definition 2.1. Not surprisingly, the proofs of many
theorems about limits require the use of properties of absolute values. First let's recall the definition of absolute values.
We have
|2| = 2 and |–2| = 2 = –(–2). To get the absolute value 2 of –2 we remove the minus sign, by multiplying –2 by –1.
The absolute value of 0 is 0:
|0| = 0. So, the absolute value of any real number x is formally defined as follows: The properties of absolute values often used are: where a and b are arbitrary real numbers and k is an arbitrary positive constant. Properties i and ii are obvious. Let's
consider property iii. For example:

|(2)(–3)| = | 6| = 6,
|2||–3| = (2)(3) = 6,
so that
|(2)(–3)| = |2||–3|. Next let's examine property iv. Two examples are:

|2 + 5| = |7| = 7,
|2| + |5| = 2 + 5 = 7,
so that
|2 + 5| = |2| + |5|;

|2 + (– 5)| = |–3| = 3,
|2| + | 5| = 2 + 5 = 7,
so that
|2 + (– 5)| < |2| + | 5|.

The property can be proved as follows: ABC, as seen in Fig. 1.1. For any two points P and Q, denote the vector joining and directed from P to Q and its length the reason for its name. # Fig. 1.1   Fig. 1.2 2. Uniqueness Of Limits Theorem 2.1

 If a function f (x) has a limit L at a point x = a, then the limit L is unique.

Proof which is a contradiction. Consequently, we must have M = L. That is, the limit of f (x) at x = a is unique.

### EOP

Remark 2.1 3. Properties Of Limits

Theorem 3.1 Proof  ### EOP

Theorem 3.2 Note
Part i in the above theorem expresses the limit of the sum of 2 functions, which is a new function, in terms of the limits of the
original 2 functions. Similarly for the remaining parts.

###### Proof

i.   We have:    ### EOP

Remark 3.1 theorem 2.1, , so that you can see in later proofs why we can bypass it. Here and in the remainder of this tutorial we
bypass it and the like, as they do in textbooks. 4. Direct Substitution

As an example we have: We step-by-step apply the above theorems on properties of limits to evaluate the limit. The line preceding the last line in the
above calculation, 4(23) - 10(22) + 3(2) + 5, can be obtained by substituting x = 2 directly into the function of the limit, 4x3 -
10x2 + 3x + 10. So in practice the evaluation isn't done in such a step-by-step detail. Instead it's done by substituting the value
x = 2 directly into the given function, as follows.

Example 4.1

Find this limit: EOS

In this practical solution, we merely substitute the value x = 2 directly into the given function to get the value of the limit.

As another example we have: The last fraction in the above calculation, (32 - 1)/(3 - 1), can be obtained by substituting x = 3 directly into the function of the
limit, (x2 - 1)/(x - 1). So again in practice the evaluation isn't done in such a step-by-step detail. Instead it's done by
substituting the value x = 3 directly into the given function, as follows.

## Example 4.2

Evaluate the following limit if it exists: ###### Solution ###### EOS

In this practical solution, we merely substitute the value x = 3 directly into the given function to get the value of the limit.

Now consider this limit: In this section we deal only with limits that can be evaluated by direct substitution. This is the case where the limit of a function f (x) as x approaches a is obtained by substituting x = a directly into the given expression of  f (x), which is
possible by the properties of limits as stated in the above theorems.

The form 0/0 is called an indeterminate form. Limits for which direct substitution isn't applicable because it would lead to the form 0/0 or other indeterminate forms will be handled in later sections. Direct substitution isn't applicable either to piecewise functions (also called case-defined functions) at points of formula change, as demonstrated in Section 1.1.4 Part 6. 5. The Squeeze Theorem

Refer to Fig. 5.1. There's an open interval containing x = a such that on it the graph of g is squeezed between those of f and
h. If the graphs of f and h touch each other or almost do so at a, then of course that of g touches them or almost does so
there too. In Fig. 5.1, the 3 graphs touch each other at the point (
a, L), which belongs to them all. If the point (a, L) is a hole,
so that it doesn't belong to any of them, then the 3 graphs almost touch each other there. Fig. 5.1   There's an open interval that contains a and where graph of g is squeezed between those of f and h.

Theorem 5.1 – The Squeeze Theorem # Proof

## EOP

Remark 5.1

Consider the phrase “an open interval containing a except possibly at a itself” in the above theorem. That means that the
values of
f, g, and h at a don't matter. Each may or may not exist, and if it exists, it can be any number. None affects the
conclusion of the theorem. Even if all the 3 values
f (a), g(a), and h(a) exist, they don't have to satisfy the inequalities in the
hypothesis of the theorem. In Fig. 5.1, whether the point (
a, L) is solid or a hole, the conclusion of the theorem is always valid.

 Problems & Solutions ###### Solution   Solution  3.  Let the function f be defined by: Note

A graph, as shown below, will guide us in finding the limits. y = f(x)

###### Solution where f (x) = |x|/x and |x| = x for x sufficiently close to 2/5 such that x > 0 > –2.  Solution y = x2 (red), y = x4 (gray), y = f(x) ( black).

Let g1 and g2 be defined by:   ###### Solution  