Calculus Of One Real Variable 
1.1.2 
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 
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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.
Remark 2.1
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3.
Properties Of Limits 
Theorem 3.1

Proof
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.
i.
We have:
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.
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4.
Direct Substitution 
As an example we have:
We stepbystep apply the above
theorems on properties of limits to evaluate the limit. The line preceding the last
line in the
above calculation, 4(2^{3})  10(2^{2}) + 3(2) + 5, can be obtained by
substituting x = 2 directly into the
function of the limit, 4x^{3} 
10x^{2} + 3x +
10. So in practice the evaluation isn't done in such a stepbystep 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, (3^{2}  1)/(3 
1), can be obtained by substituting x
= 3 directly into the function of the
limit, (x^{2}  1)/(x  1). So again in practice the
evaluation isn't done in such a stepbystep detail. Instead it's done by
substituting the value x = 3 directly into the given function, as follows.
Evaluate the following limit if it exists:
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 casedefined functions) at points of formula change, as demonstrated in Section 1.1.4 Part 6.
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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 
Theorem 5.1 – The Squeeze Theorem

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
3.
Let the function f be defined by:
Note
A graph, as shown below, will guide us in finding the limits.

y = f(x) 
where f (x) = x/x and x = x for x sufficiently close to 2/5 such that x > 0 > –2.
Solution

y = x^{2} (red), 
Let g_{1} and g_{2} be defined by:
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