Education.com
Try
Brainzy
Try
Plus

Calculus and the Idea of Limits Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Calculus and the Idea of Limits

The single most important idea in calculus is the idea of limit. More than 2000 years ago, the ancient Greeks wrestled with the limit concept, and they did not succeed . It is only in the past 200 years that we have finally come up with a firm understanding of limits. Here we give a brief sketch of the essential parts of the limit notion.

Suppose that f is a function whose domain contains two neighboring intervals: f : ( a , c ) ∪ ( c , b ) → Foundations of Calculus 2.1 Limits . We wish to consider the behavior of f as the variable x approaches c . If f ( x ) approaches a particular finite value ℓ as x approaches c , then we say that the function f has the limit ℓ as x approaches c . We write

Foundations of Calculus 2.1 Limits

The rigorous mathematical definition of limit is this:

Definition: Let a < c < b and let f be a function whose domain contains ( a , c ) ∪ ( c , b ). We say that f has limit ℓ at c , and we write lim xc f ( x ) = ℓ when this condition holds: For each Foundations of Calculus 2.1 Limits > 0 there is a δ > 0 such that

f ( x ) − ℓ∣ < Foundations of Calculus 2.1 Limits

whenever 0 < ∣ xc ∣ < δ .

It is important to know that there is a rigorous definition of the limit concept, and any development of mathematical theory relies in an essential way on this rigorous definition. However, in the present book we may make good use of an intuitive understanding of limit. We now develop that understanding with some carefully chosen examples.

Examples

Example 1

Define

Foundations of Calculus 2.1 Limits

See Fig. 2.1 . Calculate lim x → 1 f ( x )

Foundations of Calculus 2.1 Limits

Fig. 2.1

Solution 1

Observe that, when x is to the left of 1 and very near to 1 then f ( x ) = 3 − x is very near to 2. Likewise, when x is to the right of 1 and very near to 1 then f ( x ) = x 2 + 1 is very near to 2. We conclude that

Foundations of Calculus 2.1 Limits

We have successfully calculated our first limit. Figure 2.1 confirms the conclusion that our calculations derived.

Example 2

Define

Foundations of Calculus 2.1 Limits

Calculate lim x → 2 g ( x ).

Solution 2

We observe that both the numerator and the denominator of the fraction defining g tend to 0 as x → 2 (i.e., as x tends to 2). Thus the question seems to be indeterminate.

However, we may factor the numerator as x 2 − 4 = ( x − 2)( x + 2). As long as x ≠ 2 (and these are the only x that we examine when we calculate lim x → 2 ), we can then divide the denominator of the expression defining g into the numerator. Thus

g ( x ) = x + 2 for x ≠ 2.

Now

Foundations of Calculus 2.1 Limits

Foundations of Calculus 2.1 Limits

Fig. 2.2

The graph of the function g is shown in Fig. 2.2 . We encourage the reader to use a pocket calculator to calculate values of g for x near 2 but unequal to 2 to check the validity of our answer. For example,

x

g ( x ) = [ x 2 − 4]/[x − 2]

1.8

3.8

1.9

3.9

1.99

3.99

1.999

3.999

2.001

4.001

2.01

4.01

2.1

4.1

2.2

4.2

We see that, when x is close to 2 (but unequal to 2), then g ( x ) is close (indeed, as close as we please) to 4.

You Try It: Calculate the limit Foundations of Calculus 2.1 Limits .

Math Note: It must be stressed that, when we calculate lim xc f ( x ), we do not evaluate f at c . In the last example it would have been impossible to do so. We want to determine what we anticipate f will do as x approaches c, not what value (if any) f actually takes at c . The next example illustrates this point rather dramatically.

Example

Define

Foundations of Calculus 2.1 Limits

Calculate lim x → 7 h ( x ).

 

Foundations of Calculus 2.1 Limits

Fig. 2.3

Solution

It would be incorrect to simply plug the value 7 into the function h and thereby to conclude that the limit is 1. In fact when x is near to 7 but unequal to 7, we see that h takes the value 3. This statement is true no matter how close x is to 7. We conclude that lim x → 7 h ( x ) = 3.

You Try It: Calculate lim x → 4 [ x 2x − 12]/[ x − 4].

View Full Article
Add your own comment