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Basic Calculus Learner’s Material First Edition 2016
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Development Team of the Basic Calculus Learner’s Material Jose Maria P. Balmaceda, PhD
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Published by the Department of Education Secretary: Leonor M. Briones, PhD Undersecretary: Dina S. Ocampo, PhD
LESSON 1: The Limit of a Function: Theorems and Examples
LEARNING OUTCOMES: At the end of the lesson, the learner shall be able to:
TOPIC 1.1: The Limit of a Function
LIMITS
Consider a function f of a single variable x. Consider a constant c which the variable x will approach (c may or may not be in the domain of f ). The limit, to be denoted by L, is the unique real value that f (x) will approach as x approaches c. In symbols, we write this process as
lim x→c f (x) = L.
This is read, ‘ ‘The limit of f (x) as x approaches c is L.”
To illustrate, let us consider
xlim→ 2 (1 + 3x).
Here, f (x) = 1 + 3x and the constant c, which x will approach, is 2. To evaluate the given limit, we will make use of a table to help us keep track of the effect that the approach of x toward 2 will have on f (x). Of course, on the number line, x may approach 2 in two ways: through values on its left and through values on its right. We first consider approaching 2 from its left or through values less than 2. Remember that the values to be chosen should be close to 2.
x f (x) 1 4
Now we consider approaching 2 from its right or through values greater than but close to 2.
EXAMPLE 2: Investigate lim x→ 0 |x| through a table of values.
Approaching 0 from the left and from the right, we get the following tables:
x |x| − 0. 3 0. 3 − 0. 01 0. 01 − 0. 00009 0. 00009 − 0. 00000001 0. 00000001
x |x|
Hence,
xlim→ 0 |x|^ = 0.
EXAMPLE 3: Investigate
xlim→ 1 x
(^2) − 5 x + 4 x − 1
by constructing tables of values. Here, c = 1 and f (x) = x
(^2) − 5 x + 4 x − 1
Take note that 1 is not in the domain of f , but this is not a problem. In evaluating a limit, remember that we only need to go very close to 1; we will not go to 1 itself.
We now approach 1 from the left.
x f (x)
Approach 1 from the right.
x f (x)
5
The tables show that as x approaches 1, f (x) approaches −3. In symbols,
xlim→ 1 x
(^2) − 5 x + 4 x − 1
EXAMPLE 4: Investigate through a table of values
xlim→ 4 f^ (x),
if
f (x) =
x + 1 if x < 4 (x − 4)^2 + 3 if x ≥ 4.
This looks a bit different, but the logic and procedure are exactly the same. We still approach the constant 4 from the left and from the right, but note that we should evaluate the appropriate corresponding functional expression. In this case, when x approaches 4 from the left, the values taken should be substituted in f (x) = x + 1. Indeed, this is the part of the function which accepts values less than 4. So,
x f (x)
On the other hand, when x approaches 4 from the right, the values taken should be substituted in f (x) = (x − 4)^2 + 3. So,
x f (x)
6
x^2 − 5 x + 4 x − 1 =^ −3 and lim x→ 1 +
x^2 − 5 x + 4 x − 1
LOOKING AT THE GRAPH OF y = f (x)
If one knows the graph of f (x), it will be easier to determine its limits as x approaches given values of c.
Consider again f (x) = 1 + 3x. Its graph is the straight line with slope 3 and intercepts (0, 1) and (− 1 / 3 , 0). Look at the graph in the vicinity of x = 2.
You can easily see the points (from the table of values in page 4) (1, 4), (1. 4 , 5 .2), (1. 7 , 6 .1), and so on, approaching the level where y = 7. The same can be seen from the right (from the table of values in page 4). Hence, the graph clearly confirms that
xlim→ 2 (1 + 3x) = 7. x
y
y = 1 + 3x
− 1 0 1 2 3 4
1
2
3
4
5
6
7
8
Let us look at the examples again, one by one.
Recall Example 1 where f (x) = x^2 + 1. Its graph is given by
− 3 − 2 − 1 0 1 2 3
1
2
3
4
5
6
7
8
x
y
y = x^2 + 1
It can be seen from the graph that as values of x approach −1, the values of f (x) approach 2.
Recall Example 2 where f (x) = |x|.
x
y y = |x|
It is clear that lim x→ 0 |x| = 0, that is, the two sides of the graph both move downward to the origin (0, 0) as x approaches 0.
Recall Example 3 where f (x) = x
(^2) − 5 x + 4 x − 1
So, in general, if we have the graph of a function, such as below, determining limits can be done much faster and easier by inspection.
− 3 − 2 − 1 0 1 2 3 4 5 6
1
2
3
4
5
6
x
y
For instance, it can be seen from the graph of y = f (x) that:
a. (^) xlim→− 2 f (x) = 1.
b. (^) xlim→ 0 f (x) = 3. Here, it does not matter that f (0) does not exist (that is, it is undefined, or x = 0 is not in the domain of f ). Always remember that what matters is the behavior of the function close to c = 0 and not precisely at c = 0. In fact, even if f (0) were defined and equal to any other constant (not equal to 3), like 100 or −5000, this would still have no bearing on the limit. In cases like this, lim x→ 0 f (x) = 3 prevails regardless of the value of f (0), if any. c. (^) xlim→ 3 f (x) DNE. As can be seen in the figure, the two parts of the graph near c = 3 do not move toward a common y-level as x approaches c = 3.
Solved Examples
LOOKING AT TABLES OF VALUES
EXAMPLE 1: Determine lim x→ 1
x^3 − 1
Solution. Approaching x = 1 from the left,
x f (x)
Now, taking values from the right of x = 1,
x f (x)
Thus, lim x→ 1
x^3 − 1
EXAMPLE 2: Determine lim x→ 0 |x + 2|.
Solution. Taking values from the left of 0,
x f (x) − 0. 1 1. 9 − 0. 05 1. 95 − 0. 01 1. 99 − 0. 005 1. 995 − 0. 001 1. 999
Approaching 0 from the right,
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