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Calculus Volume 1

2.2 The Limit of a Function

Calculus Volume 12.2 The Limit of a Function
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 2.2.1. Using correct notation, describe the limit of a function.
  • 2.2.2. Use a table of values to estimate the limit of a function or to identify when the limit does not exist.
  • 2.2.3. Use a graph to estimate the limit of a function or to identify when the limit does not exist.
  • 2.2.4. Define one-sided limits and provide examples.
  • 2.2.5. Explain the relationship between one-sided and two-sided limits.
  • 2.2.6. Using correct notation, describe an infinite limit.
  • 2.2.7. Define a vertical asymptote.

The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years. In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the formal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We therefore begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. At the end of this chapter, armed with a conceptual understanding of limits, we examine the formal definition of a limit.

We begin our exploration of limits by taking a look at the graphs of the functions

f(x)=x24x2,g(x)=|x2|x2,andh(x)=1(x2)2,f(x)=x24x2,g(x)=|x2|x2,andh(x)=1(x2)2,

which are shown in Figure 2.12. In particular, let’s focus our attention on the behavior of each graph at and around x=2.x=2.

Three graphs of functions.  The first is f(s) = (x^2 – 4) / (x-2), which is a line of slope, x intercept (-2,0), and open circle at (2,4). The second is g(x) = |x – 2 | / (x-2), which contains two lines: x=1 for x>2 and x= -1 for x < 2. There are open circles at both endpoints (2, 1) and (-2, 1). The third is h(x) = 1 / (x-2)^2, in which the function curves asymptotically towards y=0 and x=2 in quadrants one and two.
Figure 2.12 These graphs show the behavior of three different functions around x=2.x=2.

Each of the three functions is undefined at x=2,x=2, but if we make this statement and no other, we give a very incomplete picture of how each function behaves in the vicinity of x=2.x=2. To express the behavior of each graph in the vicinity of 2 more completely, we need to introduce the concept of a limit.

Intuitive Definition of a Limit

Let’s first take a closer look at how the function f(x)=(x24)/(x2)f(x)=(x24)/(x2) behaves around x=2x=2 in Figure 2.12. As the values of x approach 2 from either side of 2, the values of y=f(x)y=f(x) approach 4. Mathematically, we say that the limit of f(x)f(x) as x approaches 2 is 4. Symbolically, we express this limit as

limx2f(x)=4.limx2f(x)=4.

From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of the limit of a function at a number a as being the one real number L that the functional values approach as the x-values approach a, provided such a real number L exists. Stated more carefully, we have the following definition:

Definition

Let f(x)f(x) be a function defined at all values in an open interval containing a, with the possible exception of a itself, and let L be a real number. If all values of the function f(x)f(x) approach the real number L as the values of x(a)x(a) approach the number a, then we say that the limit of f(x)f(x) as x approaches a is L. (More succinct, as x gets closer to a, f(x)f(x) gets closer and stays close to L.) Symbolically, we express this idea as

limxaf(x)=L.limxaf(x)=L.
2.3

We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is described in the following Problem-Solving Strategy.

Problem-Solving Strategy: Evaluating a Limit Using a Table of Functional Values
  1. To evaluate limxaf(x),limxaf(x), we begin by completing a table of functional values. We should choose two sets of x-values—one set of values approaching a and less than a, and another set of values approaching a and greater than a. Table 2.1 demonstrates what your tables might look like.
    x f(x)f(x) x f(x)f(x)
    a0.1a0.1 f(a0.1)f(a0.1) a+0.1a+0.1 f(a+0.1)f(a+0.1)
    a0.01a0.01 f(a0.01)f(a0.01) a+0.01a+0.01 f(a+0.01)f(a+0.01)
    a0.001a0.001 f(a0.001)f(a0.001) a+0.001a+0.001 f(a+0.001)f(a+0.001)
    a0.0001a0.0001 f(a0.0001)f(a0.0001) a+0.0001a+0.0001 f(a+0.0001)f(a+0.0001)
    Use additional values as necessary. Use additional values as necessary.
    Table 2.1 Table of Functional Values for limxaf(x)limxaf(x)
  2. Next, let’s look at the values in each of the f(x)f(x) columns and determine whether the values seem to be approaching a single value as we move down each column. In our columns, we look at the sequence f(a0.1),f(a0.01),f(a0.001).,f(a0.0001),f(a0.1),f(a0.01),f(a0.001).,f(a0.0001), and so on, and f(a+0.1),f(a+0.01),f(a+0.001),f(a+0.0001),f(a+0.1),f(a+0.01),f(a+0.001),f(a+0.0001), and so on. (Note: Although we have chosen the x-values a±0.1,a±0.01,a±0.001,a±0.0001,a±0.1,a±0.01,a±0.001,a±0.0001, and so forth, and these values will probably work nearly every time, on very rare occasions we may need to modify our choices.)
  3. If both columns approach a common y-value L, we state limxaf(x)=L.limxaf(x)=L. We can use the following strategy to confirm the result obtained from the table or as an alternative method for estimating a limit.
  4. Using a graphing calculator or computer software that allows us graph functions, we can plot the function f(x),f(x), making sure the functional values of f(x)f(x) for x-values near a are in our window. We can use the trace feature to move along the graph of the function and watch the y-value readout as the x-values approach a. If the y-values approach L as our x-values approach a from both directions, then limxaf(x)=L.limxaf(x)=L. We may need to zoom in on our graph and repeat this process several times.

We apply this Problem-Solving Strategy to compute a limit in Example 2.4.

Example 2.4

Evaluating a Limit Using a Table of Functional Values 1

Evaluate limx0sinxxlimx0sinxx using a table of functional values.

Solution

We have calculated the values of f(x)=(sinx)/xf(x)=(sinx)/x for the values of x listed in Table 2.2.

x sinxxsinxx x sinxxsinxx
−0.1 0.998334166468 0.1 0.998334166468
−0.01 0.999983333417 0.01 0.999983333417
−0.001 0.999999833333 0.001 0.999999833333
−0.0001 0.999999998333 0.0001 0.999999998333
Table 2.2 Table of Functional Values for limx0sinxxlimx0sinxx

Note: The values in this table were obtained using a calculator and using all the places given in the calculator output.

As we read down each (sinx)x(sinx)x column, we see that the values in each column appear to be approaching one. Thus, it is fairly reasonable to conclude that limx0sinxx=1.limx0sinxx=1. A calculator-or computer-generated graph of f(x)=(sinx)xf(x)=(sinx)x would be similar to that shown in Figure 2.13, and it confirms our estimate.

A graph of f(x) = sin(x)/x over the interval [-6, 6]. The curving function has a y intercept at x=0 and x intercepts at y=pi and y=-pi.
Figure 2.13 The graph of f(x)=(sinx)/xf(x)=(sinx)/x confirms the estimate from Table 2.2.

Example 2.5

Evaluating a Limit Using a Table of Functional Values 2

Evaluate limx4x2x4limx4x2x4 using a table of functional values.

Solution

As before, we use a table—in this case, Table 2.3—to list the values of the function for the given values of x.

x x2x4x2x4 x x2x4x2x4
3.9 0.251582341869 4.1 0.248456731317
3.99 0.25015644562 4.01 0.24984394501
3.999 0.250015627 4.001 0.249984377
3.9999 0.250001563 4.0001 0.249998438
3.99999 0.25000016 4.00001 0.24999984
Table 2.3 Table of Functional Values for limx4x2x4limx4x2x4

After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward 0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that limx4x2x4=0.25.limx4x2x4=0.25. We confirm this estimate using the graph of f(x)=x2x4f(x)=x2x4 shown in Figure 2.14.

A graph of the function f(x) = (sqrt(x) – 2 ) / (x-4) over the interval [0,8]. There is an open circle on the function at x=4. The function curves asymptotically towards the x axis and y axis in quadrant one.
Figure 2.14 The graph of f(x)=x2x4f(x)=x2x4 confirms the estimate from Table 2.3.
Checkpoint 2.4

Estimate limx11x1x1limx11x1x1 using a table of functional values. Use a graph to confirm your estimate.

At this point, we see from Example 2.4 and Example 2.5 that it may be just as easy, if not easier, to estimate a limit of a function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example 2.6, we evaluate a limit exclusively by looking at a graph rather than by using a table of functional values.

Example 2.6

Evaluating a Limit Using a Graph

For g(x)g(x) shown in Figure 2.15, evaluate limx−1g(x).limx−1g(x).

The graph of a generic curving function g(x). In quadrant two, there is an open circle on the function at (-1,3) and a closed circle one unit up at (-1, 4).
Figure 2.15 The graph of g(x)g(x) includes one value not on a smooth curve.

Solution

Despite the fact that g(−1)=4,g(−1)=4, as the x-values approach −1 from either side, the g(x)g(x) values approach 3. Therefore, limx−1g(x)=3.limx−1g(x)=3. Note that we can determine this limit without even knowing the algebraic expression of the function.

Based on Example 2.6, we make the following observation: It is possible for the limit of a function to exist at a point, and for the function to be defined at this point, but the limit of the function and the value of the function at the point may be different.

Checkpoint 2.5

Use the graph of h(x)h(x) in Figure 2.16 to evaluate limx2h(x),limx2h(x), if possible.

A graph of the function h(x), which is a parabola graphed over [-2.5, 5]. There is an open circle where the vertex should be at the point (2,-1).
Figure 2.16

Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the value of the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need to develop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them in the next section; however, at this point we introduce two special limits that are foundational to the techniques to come.

Theorem 2.1

Two Important Limits

Let a be a real number and c be a constant.

  1. limxax=alimxax=a
    2.4
  2. limxac=climxac=c
    2.5

We can make the following observations about these two limits.

  1. For the first limit, observe that as x approaches a, so does f(x),f(x), because f(x)=x.f(x)=x. Consequently, limxax=a.limxax=a.
  2. For the second limit, consider Table 2.4.
x f(x)=cf(x)=c x f(x)=cf(x)=c
a0.1a0.1 c a+0.1a+0.1 c
a0.01a0.01 c a+0.01a+0.01 c
a0.001a0.001 c a+0.001a+0.001 c
a0.0001a0.0001 c a+0.0001a+0.0001 c
Table 2.4 Table of Functional Values for limxac=climxac=c

Observe that for all values of x (regardless of whether they are approaching a), the values f(x)f(x) remain constant at c. We have no choice but to conclude limxac=c.limxac=c.

The Existence of a Limit

As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional values must approach a single real-number value at that point. If the functional values do not approach a single value, then the limit does not exist.

Example 2.7

Evaluating a Limit That Fails to Exist

Evaluate limx0sin(1/x)limx0sin(1/x) using a table of values.

Solution

Table 2.5 lists values for the function sin(1/x)sin(1/x) for the given values of x.

x sin(1x)sin(1x) x sin(1x)sin(1x)
−0.1 0.544021110889 0.1 −0.544021110889
−0.01 0.50636564111 0.01 −0.50636564111
−0.001 −0.8268795405312 0.001 0.826879540532
−0.0001 0.305614388888 0.0001 −0.305614388888
−0.00001 −0.035748797987 0.00001 0.035748797987
−0.000001 0.349993504187 0.000001 −0.349993504187
Table 2.5 Table of Functional Values for limx0sin(1x)limx0sin(1x)

After examining the table of functional values, we can see that the y-values do not seem to approach any one single value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematic approach. Take the following sequence of x-values approaching 0:

2π,23π,25π,27π,29π,211π,….2π,23π,25π,27π,29π,211π,….

The corresponding y-values are

1,−1,1,−1,1,−1,….1,−1,1,−1,1,−1,….

At this point we can indeed conclude that limx0sin(1/x)limx0sin(1/x) does not exist. (Mathematicians frequently abbreviate “does not exist” as DNE. Thus, we would write limx0sin(1/x)limx0sin(1/x) DNE.) The graph of f(x)=sin(1/x)f(x)=sin(1/x) is shown in Figure 2.17 and it gives a clearer picture of the behavior of sin(1/x)sin(1/x) as x approaches 0. You can see that sin(1/x)sin(1/x) oscillates ever more wildly between −1 and 1 as x approaches 0.

The graph of the function f(x) = sin(1/x), which oscillates rapidly between -1 and 1 as x approaches 0. The oscillations are less frequent as the function moves away from 0 on the x axis.
Figure 2.17 The graph of f(x)=sin(1/x)f(x)=sin(1/x) oscillates rapidly between −1 and 1 as x approaches 0.
Checkpoint 2.6

Use a table of functional values to evaluate limx2|x24|x2,limx2|x24|x2, if possible.

One-Sided Limits

Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information about the behavior of the function at that particular point. To see this, we now revisit the function g(x)=|x2|/(x2)g(x)=|x2|/(x2) introduced at the beginning of the section (see Figure 2.12(b)). As we pick values of x close to 2, g(x)g(x) does not approach a single value, so the limit as x approaches 2 does not exist—that is, limx2g(x)limx2g(x) DNE. However, this statement alone does not give us a complete picture of the behavior of the function around the x-value 2. To provide a more accurate description, we introduce the idea of a one-sided limit. For all values to the left of 2 (or the negative side of 2), g(x)=−1.g(x)=−1. Thus, as x approaches 2 from the left, g(x)g(x) approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1. Symbolically, we express this idea as

limx2g(x)=−1.limx2g(x)=−1.

Similarly, as x approaches 2 from the right (or from the positive side), g(x)g(x) approaches 1. Symbolically, we express this idea as

limx2+g(x)=1.limx2+g(x)=1.

We can now present an informal definition of one-sided limits.

Definition

We define two types of one-sided limits.

Limit from the left: Let f(x)f(x) be a function defined at all values in an open interval of the form (c, a), and let L be a real number. If the values of the function f(x)f(x) approach the real number L as the values of x (where x<a)x<a) approach the number a, then we say that L is the limit of f(x)f(x) as x approaches a from the left. Symbolically, we express this idea as

limxaf(x)=L.limxaf(x)=L.
2.6

Limit from the right: Let f(x)f(x) be a function defined at all values in an open interval of the form (a,c),(a,c), and let L be a real number. If the values of the function f(x)f(x) approach the real number L as the values of x (where x>a)x>a) approach the number a, then we say that L is the limit of f(x)f(x) as x approaches a from the right. Symbolically, we express this idea as

limxa+f(x)=L.limxa+f(x)=L.
2.7

Example 2.8

Evaluating One-Sided Limits

For the function f(x)={x+1ifx<2x24ifx2,f(x)={x+1ifx<2x24ifx2, evaluate each of the following limits.

  1. limx2f(x)limx2f(x)
  2. limx2+f(x)limx2+f(x)

Solution

We can use tables of functional values again Table 2.6. Observe that for values of x less than 2, we use f(x)=x+1f(x)=x+1 and for values of x greater than 2, we use f(x)=x24.f(x)=x24.

x f(x)=x+1f(x)=x+1 x f(x)=x2−4f(x)=x2−4
1.9 2.9 2.1 0.41
1.99 2.99 2.01 0.0401
1.999 2.999 2.001 0.004001
1.9999 2.9999 2.0001 0.00040001
1.99999 2.99999 2.00001 0.0000400001
Table 2.6 Table of Functional Values for f(x)={x+1ifx<2x24ifx2f(x)={x+1ifx<2x24ifx2

Based on this table, we can conclude that a. limx2f(x)=3limx2f(x)=3 and b. limx2+f(x)=0.limx2+f(x)=0. Therefore, the (two-sided) limit of f(x)f(x) does not exist at x=2.x=2. Figure 2.18 shows a graph of f(x)f(x) and reinforces our conclusion about these limits.

The graph of the given piecewise function. The first piece is f(x) = x+1 if x < 2. The second piece is x^2 – 4 if x >= 2. The first piece is a line with x intercept at (-1, 0) and y intercept at (0,1). There is an open circle at (2,3), where the endpoint would be. The second piece is the right half of a parabola opening upward. The vertex at (2,0) is a solid circle.
Figure 2.18 The graph of f(x)={x+1ifx<2x24ifx2f(x)={x+1ifx<2x24ifx2 has a break at x=2.x=2.
Checkpoint 2.7

Use a table of functional values to estimate the following limits, if possible.

  1. limx2|x24|x2limx2|x24|x2
  2. limx2+|x24|x2limx2+|x24|x2

Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at that point. It seems clear that if the limit from the right and the limit from the left have a common value, then that common value is the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on different values, the limit of the function does not exist. These conclusions are summarized in Relating One-Sided and Two-Sided Limits.

Theorem 2.2

Relating One-Sided and Two-Sided Limits

Let f(x)f(x) be a function defined at all values in an open interval containing a, with the possible exception of a itself, and let L be a real number. Then,

limxaf(x)=L.if and only iflimxaf(x)=Landlimxa+f(x)=L.limxaf(x)=L.if and only iflimxaf(x)=Landlimxa+f(x)=L.

Infinite Limits

Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to characterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functions that do not have finite limits.

We now turn our attention to h(x)=1/(x2)2,h(x)=1/(x2)2, the third and final function introduced at the beginning of this section (see Figure 2.12(c)). From its graph we see that as the values of x approach 2, the values of h(x)=1/(x2)2h(x)=1/(x2)2 become larger and larger and, in fact, become infinite. Mathematically, we say that the limit of h(x)h(x) as x approaches 2 is positive infinity. Symbolically, we express this idea as

limx2h(x)=+.limx2h(x)=+.

More generally, we define infinite limits as follows:

Definition

We define three types of infinite limits.

Infinite limits from the left: Let f(x)f(x) be a function defined at all values in an open interval of the form (b,a).(b,a).

  1. If the values of f(x)f(x) increase without bound as the values of x (where x<a)x<a) approach the number a, then we say that the limit as x approaches a from the left is positive infinity and we write
    limxaf(x)=+.limxaf(x)=+.
    2.8
  2. If the values of f(x)f(x) decrease without bound as the values of x (where x<a)x<a) approach the number a, then we say that the limit as x approaches a from the left is negative infinity and we write
    limxaf(x)=.limxaf(x)=.
    2.9

Infinite limits from the right: Let f(x)f(x) be a function defined at all values in an open interval of the form (a,c).(a,c).

  1. If the values of f(x)f(x) increase without bound as the values of x (where x>a)x>a) approach the number a, then we say that the limit as x approaches a from the right is positive infinity and we write
    limxa+f(x)=+.limxa+f(x)=+.
    2.10
  2. If the values of f(x)f(x) decrease without bound as the values of x (where x>a)x>a) approach the number a, then we say that the limit as x approaches a from the right is negative infinity and we write
    limxa+f(x)=.limxa+f(x)=.
    2.11

Two-sided infinite limit: Let f(x)f(x) be defined for all xaxa in an open interval containing a.

  1. If the values of f(x)f(x) increase without bound as the values of x (where xa)xa) approach the number a, then we say that the limit as x approaches a is positive infinity and we write
    limxaf(x)=+.limxaf(x)=+.
    2.12
  2. If the values of f(x)f(x) decrease without bound as the values of x (where xa)xa) approach the number a, then we say that the limit as x approaches a is negative infinity and we write
    limxaf(x)=.limxaf(x)=.
    2.13

It is important to understand that when we write statements such as limxaf(x)=+limxaf(x)=+ or limxaf(x)=limxaf(x)= we are describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For the limit of a function f(x)f(x) to exist at a, it must approach a real number L as x approaches a. That said, if, for example, limxaf(x)=+,limxaf(x)=+, we always write limxaf(x)=+limxaf(x)=+ rather than limxaf(x)limxaf(x) DNE.

Example 2.9

Recognizing an Infinite Limit

Evaluate each of the following limits, if possible. Use a table of functional values and graph f(x)=1/xf(x)=1/x to confirm your conclusion.

  1. limx01xlimx01x
  2. limx0+1xlimx0+1x
  3. limx01xlimx01x

Solution

Begin by constructing a table of functional values.

x 1x1x x 1x1x
−0.1 −10 0.1 10
−0.01 −100 0.01 100
−0.001 −1000 0.001 1000
−0.0001 −10,000 0.0001 10,000
−0.00001 −100,000 0.00001 100,000
−0.000001 −1,000,000 0.000001 1,000,000
Table 2.7 Table of Functional Values for f(x)=1xf(x)=1x
  1. The values of 1/x1/x decrease without bound as x approaches 0 from the left. We conclude that
    limx01x=.limx01x=.
  2. The values of 1/x1/x increase without bound as x approaches 0 from the right. We conclude that
    limx0+1x=+.limx0+1x=+.
  3. Since limx01x=limx01x= and limx0+1x=+limx0+1x=+ have different values, we conclude that
    limx01xDNE.limx01xDNE.

The graph of f(x)=1/xf(x)=1/x in Figure 2.19 confirms these conclusions.

The graph of the function f(x) = 1/x. The function curves asymptotically towards x=0 and y=0 in quadrants one and three.
Figure 2.19 The graph of f(x)=1/xf(x)=1/x confirms that the limit as x approaches 0 does not exist.
Checkpoint 2.8

Evaluate each of the following limits, if possible. Use a table of functional values and graph f(x)=1/x2f(x)=1/x2 to confirm your conclusion.

  1. limx01x2limx01x2
  2. limx0+1x2limx0+1x2
  3. limx01x2limx01x2

It is useful to point out that functions of the form f(x)=1/(xa)n,f(x)=1/(xa)n, where n is a positive integer, have infinite limits as x approaches a from either the left or right (Figure 2.20). These limits are summarized in Infinite Limits from Positive Integers.

Two graphs side by side of f(x) = 1 / (x-a)^n. The first graph shows the case where n is an odd positive integer, and the second shows the case where n is an even positive integer. In the first, the graph has two segments. Each curve asymptotically towards the x axis, also known as y=0, and x=a. The segment to the left of x=a is below the x axis, and the segment to the right of x=a is above the x axis. In the second graph, both segments are above the x axis.
Figure 2.20 The function f(x)=1/(xa)nf(x)=1/(xa)n has infinite limits at a.
Theorem 2.3

Infinite Limits from Positive Integers

If n is a positive even integer, then

limxa1(xa)n=+.limxa1(xa)n=+.

If n is a positive odd integer, then

limxa+1(xa)n=+limxa+1(xa)n=+

and

limxa1(xa)n=.limxa1(xa)n=.

We should also point out that in the graphs of f(x)=1/(xa)n,f(x)=1/(xa)n, points on the graph having x-coordinates very near to a are very close to the vertical line x=a.x=a. That is, as x approaches a, the points on the graph of f(x)f(x) are closer to the line x=a.x=a. The line x=ax=a is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:

Definition

Let f(x)f(x) be a function. If any of the following conditions hold, then the line x=ax=a is a vertical asymptote of f(x).f(x).

limxaf(x)=+or−∞limxa+f(x)=+or−∞orlimxaf(x)=+or−∞limxaf(x)=+or−∞limxa+f(x)=+or−∞orlimxaf(x)=+or−∞

Example 2.10

Finding a Vertical Asymptote

Evaluate each of the following limits using Infinite Limits from Positive Integers. Identify any vertical asymptotes of the function f(x)=1/(x+3)4.f(x)=1/(x+3)4.

  1. limx−31(x+3)4limx−31(x+3)4
  2. limx−3+1(x+3)4limx−3+1(x+3)4
  3. limx−31(x+3)4limx−31(x+3)4

Solution

We can use Infinite Limits from Positive Integers directly.

  1. limx−31(x+3)4=+limx−31(x+3)4=+
  2. limx−3+1(x+3)4=+limx−3+1(x+3)4=+
  3. limx−31(x+3)4=+limx−31(x+3)4=+

The function f(x)=1/(x+3)4f(x)=1/(x+3)4 has a vertical asymptote of x=−3.x=−3.

Checkpoint 2.9

Evaluate each of the following limits. Identify any vertical asymptotes of the function f(x)=1(x2)3.f(x)=1(x2)3.

  1. limx21(x2)3limx21(x2)3
  2. limx2+1(x2)3limx2+1(x2)3
  3. limx21(x2)3limx21(x2)3

In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points.

Example 2.11

Behavior of a Function at Different Points

Use the graph of f(x)f(x) in Figure 2.21 to determine each of the following values:

  1. limx−4f(x);limx−4+f(x);limx−4f(x);f(−4)limx−4f(x);limx−4+f(x);limx−4f(x);f(−4)
  2. limx−2f(x);limx−2+f(x);limx−2f(x);f(−2)limx−2f(x);limx−2+f(x);limx−2f(x);f(−2)
  3. limx1f(x);limx1+f(x);limx1f(x);f(1)limx1f(x);limx1+f(x);limx1f(x);f(1)
  4. limx3f(x);limx3+f(x);limx3f(x);f(3)limx3f(x);limx3+f(x);limx3f(x);f(3)
The graph of a function f(x) described by the above limits and values. There is a smooth curve for values below x=-2; at (-2, 3), there is an open circle. There is a smooth curve between (-2, 1] with a closed circle at (1,6). There is an open circle at (1,3), and a smooth curve stretching from there down asymptotically to negative infinity along x=3. The function also curves asymptotically along x=3 on the other side, also stretching to negative infinity. The function then changes concavity in the first quadrant around y=4.5 and continues up.
Figure 2.21 The graph shows f(x).f(x).

Solution

Using Infinite Limits from Positive Integers and the graph for reference, we arrive at the following values:

  1. limx−4f(x)=0;limx−4+f(x)=0;limx−4f(x)=0;f(−4)=0limx−4f(x)=0;limx−4+f(x)=0;limx−4f(x)=0;f(−4)=0
  2. limx−2f(x)=3.;limx−2+f(x)=3;limx−2f(x)=3;f(−2)limx−2f(x)=3.;limx−2+f(x)=3;limx−2f(x)=3;f(−2) is undefined
  3. limx1f(x)=6;limx1+f(x)=3;limx1f(x)limx1f(x)=6;limx1+f(x)=3;limx1f(x) DNE; f(1)=6f(1)=6
  4. limx3f(x)=;limx3+f(x)=;limx3f(x)=;f(3)limx3f(x)=;limx3+f(x)=;limx3f(x)=;f(3) is undefined
Checkpoint 2.10

Evaluate limx1f(x)limx1f(x) for f(x)f(x) shown here:

A graph of a piecewise function. The first segment curves from the third quadrant to the first, crossing through the second quadrant. Where the endpoint would be in the first quadrant is an open circle. The second segment starts at a closed circle a few units below the open circle. It curves down from quadrant one to quadrant four.

Example 2.12

Chapter Opener: Einstein’s Equation

A picture of a futuristic spaceship speeding through deep space.
Figure 2.22 (credit: NASA)

In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object can travel. Given Einstein’s equation for the mass of a moving object, what is the value of this bound?

Solution

Our starting point is Einstein’s equation for the mass of a moving object,

m=m01v2c2,m=m01v2c2,

where m0m0 is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at high speeds, we can graph the ratio of masses m/m0m/m0 as a function of the ratio of speeds, v/cv/c (Figure 2.23).

A graph showing the ratio of masses as a function of the ratio of speed in Einstein’s equation for the mass of a moving object. The x axis is the ratio of the speeds, v/c. The y axis is the ratio of the masses, m/m0. The equation of the function is m = m0 / sqrt(1 –  v2 / c2 ). The graph is only in quadrant 1. It starts at (0,1) and curves up gently until about 0.8, where it increases seemingly exponentially; there is a vertical asymptote at v/c (or x) = 1.
Figure 2.23 This graph shows the ratio of masses as a function of the ratio of speeds in Einstein’s equation for the mass of a moving object.

We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speed of light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at v/c=1.v/c=1. We can try a few values of this ratio to test this idea.

vcvc 1v2c21v2c2 mm0mm0
0.99 0.1411 7.089
0.999 0.0447 22.37
0.9999 0.0141 70.71
Table 2.8 Ratio of Masses and Speeds for a Moving Object

Thus, according to Table 2.8, if an object with mass 100 kg is traveling at 0.9999c, its mass becomes 7071 kg. Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light.

Section 2.2 Exercises

For the following exercises, consider the function f(x)=x21|x1|.f(x)=x21|x1|.

30.

[T] Complete the following table for the function. Round your solutions to four decimal places.

x f(x)f(x) x f(x)f(x)
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.
31.

What do your results in the preceding exercise indicate about the two-sided limit limx1f(x)?limx1f(x)? Explain your response.

For the following exercises, consider the function f(x)=(1+x)1/x.f(x)=(1+x)1/x.

32.

[T] Make a table showing the values of f for x=−0.01,−0.001,−0.0001,−0.00001x=−0.01,−0.001,−0.0001,−0.00001 and for x=0.01,0.001,0.0001,0.00001.x=0.01,0.001,0.0001,0.00001. Round your solutions to five decimal places.

x f(x)f(x) x f(x)f(x)
−0.01 a. 0.01 e.
−0.001 b. 0.001 f.
−0.0001 c. 0.0001 g.
−0.00001 d. 0.00001 h.
33.

What does the table of values in the preceding exercise indicate about the function f(x)=(1+x)1/x?f(x)=(1+x)1/x?

34.

To which mathematical constant does the limit in the preceding exercise appear to be getting closer?

In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.

35.

[T] limx0sin2xx;±0.1,±0.01,±0.001,±.0001limx0sin2xx;±0.1,±0.01,±0.001,±.0001

x sin2xxsin2xx x sin2xxsin2xx
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.
36.

[T] limx0sin3xxlimx0sin3xx ±0.1, ±0.01, ±0.001, ±0.0001

X sin3xxsin3xx x sin3xxsin3xx
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.
37.

Use the preceding two exercises to conjecture (guess) the value of the following limit: limx0sinaxxlimx0sinaxx for a, a positive real value.

[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

38.

limx2x24x2+x6limx2x24x2+x6

x x24x2+x6x24x2+x6 x x24x2+x6x24x2+x6
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.
39.

limx1(12x)limx1(12x)

x 12x12x x 12x12x
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.
40.

limx051e1/xlimx051e1/x

x 51e1/x51e1/x x 51e1/x51e1/x
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.
41.

limz0z1z2(z+3)limz0z1z2(z+3)

z z1z2(z+3)z1z2(z+3) z z1z2(z+3)z1z2(z+3)
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.
42.

limt0+costtlimt0+costt

t costtcostt
0.1 a.
0.01 b.
0.001 c.
0.0001 d.
43.

limx212xx24limx212xx24

x 12xx2412xx24 x 12xx2412xx24
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.

[T] In the following exercises, set up a table of values and round to eight significant digits. Based on the table of values, make a guess about what the limit is. Then, use a calculator to graph the function and determine the limit. Was the conjecture correct? If not, why does the method of tables fail?

44.

limθ0sin(πθ)limθ0sin(πθ)

θ sin(πθ)sin(πθ) θ sin(πθ)sin(πθ)
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.
45.

limα0+1αcos(πα)limα0+1αcos(πα)

a 1αcos(πα)1αcos(πα)
0.1 a.
0.01 b.
0.001 c.
0.0001 d.

In the following exercises, consider the graph of the function y=f(x)y=f(x) shown here. Which of the statements about y=f(x)y=f(x) are true and which are false? Explain why a statement is false.

A graph of a piecewise function with three segments and a point. The first segment is a curve opening upward with vertex at (-8, -6). This vertex is an open circle, and there is a closed circle instead at (-8, -3).  The segment ends at (-2,3), where there is a closed circle. The second segment stretches up asymptotically to infinity along x=-2, changes direction to increasing at about (0,1.25), increases until about (2.25, 3), and decreases until (6,2), where there is an open circle. The last segment starts at (6,5), increases slightly, and then decreases into quadrant four, crossing the x axis at (10,0). All of the changes in direction are smooth curves.
46.

limx10f(x)=0limx10f(x)=0

47.

limx−2+f(x)=3limx−2+f(x)=3

48.

limx−8f(x)=f(−8)limx−8f(x)=f(−8)

49.

limx6f(x)=5limx6f(x)=5

In the following exercises, use the following graph of the function y=f(x)y=f(x) to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first segment exists for x <=1, and the second segment exists for x > 1. The first segment is linear with a slope of 1 and goes through the origin. Its endpoint is a closed circle at (1,1). The second segment is also linear with a slope of -1. It begins with the open circle at (1,2).
50.

limx1f(x)limx1f(x)

51.

limx1+f(x)limx1+f(x)

52.

limx1f(x)limx1f(x)

53.

limx2f(x)limx2f(x)

54.

f(1)f(1)

In the following exercises, use the graph of the function y=f(x)y=f(x) shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first is a linear function for x < 0. There is an open circle at (0,1), and its slope is -1. The second segment is the right half of a parabola opening upward. Its vertex is a closed circle at (0, -4), and it goes through the point (2,0).
55.

limx0f(x)limx0f(x)

56.

limx0+f(x)limx0+f(x)

57.

limx0f(x)limx0f(x)

58.

limx2f(x)limx2f(x)

In the following exercises, use the graph of the function y=f(x)y=f(x) shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with three segments, all linear. The first exists for x < -2, has a slope of 1, and ends at the open circle at (-2, 0). The second exists over the interval [-2, 2], has a slope of -1, goes through the origin, and has closed circles at its endpoints (-2, 2) and (2,-2). The third exists for x>2, has a slope of 1, and begins at the open circle (2,2).
59.

limx−2f(x)limx−2f(x)

60.

limx−2+f(x)limx−2+f(x)

61.

limx−2f(x)limx−2f(x)

62.

limx2f(x)limx2f(x)

63.

limx2+f(x)limx2+f(x)

64.

limx2f(x)limx2f(x)

In the following exercises, use the graph of the function y=g(x)y=g(x) shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first exists for x>=0 and is the left half of an upward opening parabola with vertex at the closed circle (0,3). The second exists for x>0 and is the right half of a downward opening parabola with vertex at the open circle (0,0).
65.

limx0g(x)limx0g(x)

66.

limx0+g(x)limx0+g(x)

67.

limx0g(x)limx0g(x)

In the following exercises, use the graph of the function y=h(x)y=h(x) shown here to find the values, if possible. Estimate when necessary.

A graph of a function with two curves approaching 0 from quadrant 1 and quadrant 3. The curve in quadrant one appears to be the top half of a parabola opening to the right of the y axis along the x axis with vertex at the origin. The curve in quadrant three appears to be the left half of a parabola opening downward with vertex at the origin.
68.

limx0h(x)limx0h(x)

69.

limx0+h(x)limx0+h(x)

70.

limx0h(x)limx0h(x)

In the following exercises, use the graph of the function y=f(x)y=f(x) shown here to find the values, if possible. Estimate when necessary.

A graph with a curve and a point. The point is a closed circle at (0,-2). The curve is part of an upward opening parabola with vertex at (1,-1). It exists for x > 0, and there is a closed circle at the origin.
71.

limx0f(x)limx0f(x)

72.

limx0+f(x)limx0+f(x)

73.

limx0f(x)limx0f(x)

74.

limx1f(x)limx1f(x)

75.

limx2f(x)limx2f(x)

In the following exercises, sketch the graph of a function with the given properties.

76.

limx2f(x)=1,limx4f(x)=3,limx4+f(x)=6,x=4limx2f(x)=1,limx4f(x)=3,limx4+f(x)=6,x=4 is not defined.

77.

limxf(x)=0,limx−1f(x)=,limxf(x)=0,limx−1f(x)=, limx−1+f(x)=,limx0f(x)=f(0),f(0)=1,limxf(x)=limx−1+f(x)=,limx0f(x)=f(0),f(0)=1,limxf(x)=

78.

limxf(x)=2,limx3f(x)=,limxf(x)=2,limx3f(x)=, limx3+f(x)=,limxf(x)=2,f(0)=−13limx3+f(x)=,limxf(x)=2,f(0)=−13

79.

limxf(x)=2,limx−2f(x)=,limxf(x)=2,limx−2f(x)=, limxf(x)=2,f(0)=0limxf(x)=2,f(0)=0

80.

limxf(x)=0,limx−1f(x)=,limx−1+f(x)=,limxf(x)=0,limx−1f(x)=,limx−1+f(x)=, f(0)=−1,limx1f(x)=,limx1+f(x)=,limxf(x)=0f(0)=−1,limx1f(x)=,limx1+f(x)=,limxf(x)=0

81.

Shock waves arise in many physical applications, ranging from supernovas to detonation waves. A graph of the density of a shock wave with respect to distance, x, is shown here. We are mainly interested in the location of the front of the shock, labeled xSFxSF in the diagram.

A graph in quadrant one of the density of a shockwave with three labeled points: p1 and p2 on the y axis, with p1 > p2, and xsf on the x axis. It consists of y= p1 from 0 to xsf, x = xsf from y= p1 to y=p2, and y=p2 for values greater than or equal to xsf.
  1. Evaluate limxxSF+ρ(x).limxxSF+ρ(x).
  2. Evaluate limxxSFρ(x).limxxSFρ(x).
  3. Evaluate limxxSFρ(x).limxxSFρ(x). Explain the physical meanings behind your answers.
82.

A track coach uses a camera with a fast shutter to estimate the position of a runner with respect to time. A table of the values of position of the athlete versus time is given here, where x is the position in meters of the runner and t is time in seconds. What is limt2x(t)?limt2x(t)? What does it mean physically?

t (sec) x (m)
1.75 4.5
1.95 6.1
1.99 6.42
2.01 6.58
2.05 6.9
2.25 8.5
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