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Calculus Volume 1

2.3 The Limit Laws

Calculus Volume 12.3 The Limit Laws
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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

  • 2.3.1. Recognize the basic limit laws.
  • 2.3.2. Use the limit laws to evaluate the limit of a function.
  • 2.3.3. Evaluate the limit of a function by factoring.
  • 2.3.4. Use the limit laws to evaluate the limit of a polynomial or rational function.
  • 2.3.5. Evaluate the limit of a function by factoring or by using conjugates.
  • 2.3.6. Evaluate the limit of a function by using the squeeze theorem.

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

Evaluating Limits with the Limit Laws

The first two limit laws were stated in Two Important Limits and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Theorem 2.4

Basic Limit Results

For any real number a and any constant c,

  1. limxax=alimxax=a
    2.14
  2. limxac=climxac=c
    2.15

Example 2.13

Evaluating a Basic Limit

Evaluate each of the following limits using Basic Limit Results.

  1. limx2xlimx2x
  2. limx25limx25

Solution

  1. The limit of x as x approaches a is a: limx2x=2.limx2x=2.
  2. The limit of a constant is that constant: limx25=5.limx25=5.

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

Theorem 2.5

Limit Laws

Let f(x)f(x) and g(x)g(x) be defined for all xaxa over some open interval containing a. Assume that L and M are real numbers such that limxaf(x)=Llimxaf(x)=L and limxag(x)=M.limxag(x)=M. Let c be a constant. Then, each of the following statements holds:

Sum law for limits: limxa(f(x)+g(x))=limxaf(x)+limxag(x)=L+Mlimxa(f(x)+g(x))=limxaf(x)+limxag(x)=L+M

Difference law for limits: limxa(f(x)g(x))=limxaf(x)limxag(x)=LMlimxa(f(x)g(x))=limxaf(x)limxag(x)=LM

Constant multiple law for limits: limxacf(x)=c·limxaf(x)=cLlimxacf(x)=c·limxaf(x)=cL

Product law for limits: limxa(f(x)·g(x))=limxaf(x)·limxag(x)=L·Mlimxa(f(x)·g(x))=limxaf(x)·limxag(x)=L·M

Quotient law for limits: limxaf(x)g(x)=limxaf(x)limxag(x)=LMlimxaf(x)g(x)=limxaf(x)limxag(x)=LM for M0M0

Power law for limits: limxa(f(x))n=(limxaf(x))n=Lnlimxa(f(x))n=(limxaf(x))n=Ln for every positive integer n.

Root law for limits: limxaf(x)n=limxaf(x)n=Lnlimxaf(x)n=limxaf(x)n=Ln for all L if n is odd and for L0L0 if n is even.

We now practice applying these limit laws to evaluate a limit.

Example 2.14

Evaluating a Limit Using Limit Laws

Use the limit laws to evaluate limx−3(4x+2).limx−3(4x+2).

Solution

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

limx−3(4x+2)=limx−34x+limx−32Apply the sum law.=4·limx−3x+limx−32Apply the constant multiple law.=4·(−3)+2=−10.Apply the basic limit results and simplify.limx−3(4x+2)=limx−34x+limx−32Apply the sum law.=4·limx−3x+limx−32Apply the constant multiple law.=4·(−3)+2=−10.Apply the basic limit results and simplify.

Example 2.15

Using Limit Laws Repeatedly

Use the limit laws to evaluate limx22x23x+1x3+4.limx22x23x+1x3+4.

Solution

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

limx22x23x+1x3+4=limx2(2x23x+1)limx2(x3+4)Apply the quotient law, making sure that.(2)3+40=2·limx2x23·limx2x+limx21limx2x3+limx24Apply the sum law and constant multiple law.=2·(limx2x)23·limx2x+limx21(limx2x)3+limx24Apply the power law.=2(4)3(2)+1(2)3+4=14.Apply the basic limit laws and simplify.limx22x23x+1x3+4=limx2(2x23x+1)limx2(x3+4)Apply the quotient law, making sure that.(2)3+40=2·limx2x23·limx2x+limx21limx2x3+limx24Apply the sum law and constant multiple law.=2·(limx2x)23·limx2x+limx21(limx2x)3+limx24Apply the power law.=2(4)3(2)+1(2)3+4=14.Apply the basic limit laws and simplify.

Checkpoint 2.11

Use the limit laws to evaluate limx6(2x1)x+4.limx6(2x1)x+4. In each step, indicate the limit law applied.

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that limxaf(x)=f(a).limxaf(x)=f(a). This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.

Theorem 2.6

Limits of Polynomial and Rational Functions

Let p(x)p(x) and q(x)q(x) be polynomial functions. Let a be a real number. Then,

limxap(x)=p(a)limxap(x)=p(a)
limxap(x)q(x)=p(a)q(a)whenq(a)0.limxap(x)q(x)=p(a)q(a)whenq(a)0.

To see that this theorem holds, consider the polynomial p(x)=cnxn+cn1xn1++c1x+c0.p(x)=cnxn+cn1xn1++c1x+c0. By applying the sum, constant multiple, and power laws, we end up with

limxap(x)=limxa(cnxn+cn1xn1++c1x+c0)=cn(limxax)n+cn1(limxax)n1++c1(limxax)+limxac0=cnan+cn1an1++c1a+c0=p(a).limxap(x)=limxa(cnxn+cn1xn1++c1x+c0)=cn(limxax)n+cn1(limxax)n1++c1(limxax)+limxac0=cnan+cn1an1++c1a+c0=p(a).

It now follows from the quotient law that if p(x)p(x) and q(x)q(x) are polynomials for which q(a)0,q(a)0, then

limxap(x)q(x)=p(a)q(a).limxap(x)q(x)=p(a)q(a).

Example 2.16 applies this result.

Example 2.16

Evaluating a Limit of a Rational Function

Evaluate the limx32x23x+15x+4.limx32x23x+15x+4.

Solution

Since 3 is in the domain of the rational function f(x)=2x23x+15x+4,f(x)=2x23x+15x+4, we can calculate the limit by substituting 3 for x into the function. Thus,

limx32x23x+15x+4=1019.limx32x23x+15x+4=1019.
Checkpoint 2.12

Evaluate limx−2(3x32x+7).limx−2(3x32x+7).

Additional Limit Evaluation Techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for limxaf(x)limxaf(x) to exist when f(a)f(a) is undefined. The following observation allows us to evaluate many limits of this type:

If for all xa,f(x)=g(x)xa,f(x)=g(x) over some open interval containing a, then limxaf(x)=limxag(x).limxaf(x)=limxag(x).

To understand this idea better, consider the limit limx1x21x1.limx1x21x1.

The function

f(x)=x21x1=(x1)(x+1)x1f(x)=x21x1=(x1)(x+1)x1

and the function g(x)=x+1g(x)=x+1 are identical for all values of x1.x1. The graphs of these two functions are shown in Figure 2.24.

Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 – 1) / (x – 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.
Figure 2.24 The graphs of f(x)f(x) and g(x)g(x) are identical for all x1.x1. Their limits at 1 are equal.

We see that

limx1x21x1=limx1(x1)(x+1)x1=limx1(x+1)=2.limx1x21x1=limx1(x1)(x+1)x1=limx1(x+1)=2.

The limit has the form limxaf(x)g(x),limxaf(x)g(x), where limxaf(x)=0limxaf(x)=0 and limxag(x)=0.limxag(x)=0. (In this case, we say that f(x)/g(x)f(x)/g(x) has the indeterminate form 0/0.)0/0.) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Problem-Solving Strategy: Calculating a Limit When f(x)/g(x)f(x)/g(x) has the Indeterminate Form 0/0
  1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
  2. We then need to find a function that is equal to h(x)=f(x)/g(x)h(x)=f(x)/g(x) for all xaxa over some interval containing a. To do this, we may need to try one or more of the following steps:
    1. If f(x)f(x) and g(x)g(x) are polynomials, we should factor each function and cancel out any common factors.
    2. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.
    3. If f(x)/g(x)f(x)/g(x) is a complex fraction, we begin by simplifying it.
  3. Last, we apply the limit laws.

The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19, we look at simplifying a complex fraction.

Example 2.17

Evaluating a Limit by Factoring and Canceling

Evaluate limx3x23x2x25x3.limx3x23x2x25x3.

Solution

Step 1. The function f(x)=x23x2x25x3f(x)=x23x2x25x3 is undefined for x=3.x=3. In fact, if we substitute 3 into the function we get 0/0,0/0, which is undefined. Factoring and canceling is a good strategy:

limx3x23x2x25x3=limx3x(x3)(x3)(2x+1)limx3x23x2x25x3=limx3x(x3)(x3)(2x+1)

Step 2. For all x3,x23x2x25x3=x2x+1.x3,x23x2x25x3=x2x+1. Therefore,

limx3x(x3)(x3)(2x+1)=limx3x2x+1.limx3x(x3)(x3)(2x+1)=limx3x2x+1.

Step 3. Evaluate using the limit laws:

limx3x2x+1=37.limx3x2x+1=37.
Checkpoint 2.13

Evaluate limx−3x2+4x+3x29.limx−3x2+4x+3x29.

Example 2.18

Evaluating a Limit by Multiplying by a Conjugate

Evaluate limx−1x+21x+1.limx−1x+21x+1.

Solution

Step 1. x+21x+1x+21x+1 has the form 0/00/0 at −1. Let’s begin by multiplying by x+2+1,x+2+1, the conjugate of x+21,x+21, on the numerator and denominator:

limx−1x+21x+1=limx−1x+21x+1·x+2+1x+2+1.limx−1x+21x+1=limx−1x+21x+1·x+2+1x+2+1.

Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the (x+1)(x+1) in the denominator cancels out in the end:

=limx−1x+1(x+1)(x+2+1).=limx−1x+1(x+1)(x+2+1).

Step 3. Then we cancel:

=limx−11x+2+1.=limx−11x+2+1.

Step 4. Last, we apply the limit laws:

limx−11x+2+1=12.limx−11x+2+1=12.
Checkpoint 2.14

Evaluate limx5x12x5.limx5x12x5.

Example 2.19

Evaluating a Limit by Simplifying a Complex Fraction

Evaluate limx11x+112x1.limx11x+112x1.

Solution

Step 1. 1x+112x11x+112x1 has the form 0/00/0 at 1. We simplify the algebraic fraction by multiplying by 2(x+1)/2(x+1):2(x+1)/2(x+1):

limx11x+112x1=limx11x+112x1·2(x+1)2(x+1).limx11x+112x1=limx11x+112x1·2(x+1)2(x+1).

Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor (x1):(x1):

=limx12(x+1)2(x1)(x+1).=limx12(x+1)2(x1)(x+1).

Step 3. Then, we simplify the numerator:

=limx1x+12(x1)(x+1).=limx1x+12(x1)(x+1).

Step 4. Now we factor out −1 from the numerator:

=limx1(x1)2(x1)(x+1).=limx1(x1)2(x1)(x+1).

Step 5. Then, we cancel the common factors of (x1):(x1):

=limx1−12(x+1).=limx1−12(x+1).

Step 6. Last, we evaluate using the limit laws:

limx1−12(x+1)=14.limx1−12(x+1)=14.
Checkpoint 2.15

Evaluate limx−31x+2+1x+3.limx−31x+2+1x+3.

Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.

Example 2.20

Evaluating a Limit When the Limit Laws Do Not Apply

Evaluate limx0(1x+5x(x5)).limx0(1x+5x(x5)).

Solution

Both 1/x1/x and 5/x(x5)5/x(x5) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that

1x+5x(x5)=x5+5x(x5)=xx(x5).1x+5x(x5)=x5+5x(x5)=xx(x5).

Thus,

limx0(1x+5x(x5))=limx0xx(x5)=limx01x5=15.limx0(1x+5x(x5))=limx0xx(x5)=limx01x5=15.
Checkpoint 2.16

Evaluate limx3(1x34x22x3).limx3(1x34x22x3).

Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form limxah(x),limxah(x), we require the function h(x)h(x) to be defined over an open interval of the form (b,a);(b,a); for a limit of the form limxa+h(x),limxa+h(x), we require the function h(x)h(x) to be defined over an open interval of the form (a,c).(a,c). Example 2.21 illustrates this point.

Example 2.21

Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

  1. limx3x3limx3x3
  2. limx3+x3limx3+x3

Solution

Figure 2.25 illustrates the function f(x)=x3f(x)=x3 and aids in our understanding of these limits.

A graph of the function f(x) = sqrt(x-3). Visually, the function looks like the top half of a parabola opening to the right with vertex at (3,0).
Figure 2.25 The graph shows the function f(x)=x3.f(x)=x3.
  1. The function f(x)=x3f(x)=x3 is defined over the interval [3,+).[3,+). Since this function is not defined to the left of 3, we cannot apply the limit laws to compute limx3x3.limx3x3. In fact, since f(x)=x3f(x)=x3 is undefined to the left of 3, limx3x3limx3x3 does not exist.
  2. Since f(x)=x3f(x)=x3 is defined to the right of 3, the limit laws do apply to limx3+x3.limx3+x3. By applying these limit laws we obtain limx3+x3=0.limx3+x3=0.

In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Example 2.22

Evaluating a Two-Sided Limit Using the Limit Laws

For f(x)={4x3ifx<2(x3)2ifx2,f(x)={4x3ifx<2(x3)2ifx2, evaluate each of the following limits:

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

Solution

Figure 2.26 illustrates the function f(x)f(x) and aids in our understanding of these limits.

The graph of a piecewise function with two segments. For x<2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x>=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).
Figure 2.26 This graph shows a function f(x).f(x).
  1. Since f(x)=4x3f(x)=4x3 for all x in (,2),(,2), replace f(x)f(x) in the limit with 4x34x3 and apply the limit laws:
    limx2f(x)=limx2(4x3)=5.limx2f(x)=limx2(4x3)=5.
  2. Since f(x)=(x3)2f(x)=(x3)2 for all x in (2,+),(2,+), replace f(x)f(x) in the limit with (x3)2(x3)2 and apply the limit laws:
    limx2+f(x)=limx2+(x3)2=1.limx2+f(x)=limx2+(x3)2=1.
  3. Since limx2f(x)=5limx2f(x)=5 and limx2+f(x)=1,limx2+f(x)=1, we conclude that limx2f(x)limx2f(x) does not exist.
Checkpoint 2.17

Graph f(x)={x2ifx<12ifx=−1x3ifx>1f(x)={x2ifx<12ifx=−1x3ifx>1 and evaluate limx−1f(x).limx−1f(x).

We now turn our attention to evaluating a limit of the form limxaf(x)g(x),limxaf(x)g(x), where limxaf(x)=K,limxaf(x)=K, where K0K0 and limxag(x)=0.limxag(x)=0. That is, f(x)/g(x)f(x)/g(x) has the form K/0,K0K/0,K0 at a.

Example 2.23

Evaluating a Limit of the Form K/0,K0K/0,K0 Using the Limit Laws

Evaluate limx2x3x22x.limx2x3x22x.

Solution

Step 1. After substituting in x=2,x=2, we see that this limit has the form −1/0.−1/0. That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of x3x(x2)x3x(x2) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

limx2x3x22x=limx2x3x(x2).limx2x3x22x=limx2x3x(x2).

Step 2. Since x2x2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x2)1/(x2) from the rest of the function:

=limx2x3x·1x2.=limx2x3x·1x2.

Step 3. limx2x3x=12limx2x3x=12 and limx21x2=.limx21x2=. Therefore, the product of (x3)/x(x3)/x and 1/(x2)1/(x2) has a limit of +∞:+∞:

limx2x3x22x=+.limx2x3x22x=+.
Checkpoint 2.18

Evaluate limx1x+2(x1)2.limx1x+2(x1)2.

The Squeeze Theorem

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. Figure 2.27 illustrates this idea.

A graph of three functions over a small interval. All three functions curve. Over this interval, the function g(x) is trapped between the functions h(x), which gives greater y values for the same x values, and f(x), which gives smaller y values for the same x values. The functions all approach the same limit when x=a.
Figure 2.27 The Squeeze Theorem applies when f(x)g(x)h(x)f(x)g(x)h(x) and limxaf(x)=limxah(x).limxaf(x)=limxah(x).
Theorem 2.7

The Squeeze Theorem

Let f(x),g(x),f(x),g(x), and h(x)h(x) be defined for all xaxa over an open interval containing a. If

f(x)g(x)h(x)f(x)g(x)h(x)

for all xaxa in an open interval containing a and

limxaf(x)=L=limxah(x)limxaf(x)=L=limxah(x)

where L is a real number, then limxag(x)=L.limxag(x)=L.

Example 2.24

Applying the Squeeze Theorem

Apply the squeeze theorem to evaluate limx0xcosx.limx0xcosx.

Solution

Because −1cosx1−1cosx1 for all x, we have |x|xcosx|x||x|xcosx|x|. Since limx0(|x|)=0=limx0|x|,limx0(|x|)=0=limx0|x|, from the squeeze theorem, we obtain limx0xcosx=0.limx0xcosx=0. The graphs of f(x)=|x|,g(x)=xcosx,f(x)=|x|,g(x)=xcosx, and h(x)=|x|h(x)=|x| are shown in Figure 2.28.

The graph of three functions: h(x) = x, f(x) = -x, and g(x) = xcos(x). The first, h(x) = x, is a linear function with slope of 1 going through the origin. The second, f(x), is also a linear function with slope of −1; going through the origin. The third, g(x) = xcos(x), curves between the two and goes through the origin. It opens upward for x>0 and downward for x>0.
Figure 2.28 The graphs of f(x),g(x),f(x),g(x), and h(x)h(x) are shown around the point x=0.x=0.
Checkpoint 2.19

Use the squeeze theorem to evaluate limx0x2sin1x.limx0x2sin1x.

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is limθ0sinθ.limθ0sinθ. Consider the unit circle shown in Figure 2.29. In the figure, we see that sinθsinθ is the y-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for 0<θ<π2,0<sinθ<θ.0<θ<π2,0<sinθ<θ.

A diagram of the unit circle in the x,y plane – it is a circle with radius 1 and center at the origin. A specific point (cos(theta), sin(theta)) is labeled in quadrant 1 on the edge of the circle. This point is one vertex of a right triangle inside the circle, with other vertices at the origin and (cos(theta), 0).  As such, the lengths of the sides are cos(theta) for the base and sin(theta) for the height, where theta is the angle created by the hypotenuse and base. The radian measure of angle theta is the length of the arc it subtends on the unit circle. The diagram shows that for 0 < theta < pi/2,  0 < sin(theta) < theta.
Figure 2.29 The sine function is shown as a line on the unit circle.

Because limθ0+0=0limθ0+0=0 and limθ0+θ=0,limθ0+θ=0, by using the squeeze theorem we conclude that

limθ0+sinθ=0.limθ0+sinθ=0.

To see that limθ0sinθ=0limθ0sinθ=0 as well, observe that for π2<θ<0,0<θ<π2π2<θ<0,0<θ<π2 and hence, 0<sin(θ)<θ.0<sin(θ)<θ. Consequently, 0<sinθ<θ.0<sinθ<θ. It follows that 0>sinθ>θ.0>sinθ>θ. An application of the squeeze theorem produces the desired limit. Thus, since limθ0+sinθ=0limθ0+sinθ=0 and limθ0sinθ=0,limθ0sinθ=0,

limθ0sinθ=0.limθ0sinθ=0.
2.16

Next, using the identity cosθ=1sin2θcosθ=1sin2θ for π2<θ<π2,π2<θ<π2, we see that

limθ0cosθ=limθ01sin2θ=1.limθ0cosθ=limθ01sin2θ=1.
2.17

We now take a look at a limit that plays an important role in later chapters—namely, limθ0sinθθ.limθ0sinθθ. To evaluate this limit, we use the unit circle in Figure 2.30. Notice that this figure adds one additional triangle to Figure 2.30. We see that the length of the side opposite angle θ in this new triangle is tanθ.tanθ. Thus, we see that for 0<θ<π2,sinθ<θ<tanθ.0<θ<π2,sinθ<θ<tanθ.

The same diagram as the previous one. However, the triangle is expanded. The base is now from the origin to (1,0). The height goes from (1,0) to (1, tan(theta)). The hypotenuse goes from the origin to (1, tan(theta)). As such, the height is now tan(theta). It shows that for 0 < theta < pi/2, sin(theta) < theta < tan(theta).
Figure 2.30 The sine and tangent functions are shown as lines on the unit circle.

By dividing by sinθsinθ in all parts of the inequality, we obtain

1<θsinθ<1cosθ.1<θsinθ<1cosθ.

Equivalently, we have

1>sinθθ>cosθ.1>sinθθ>cosθ.

Since limθ0+1=1=limθ0+cosθ,limθ0+1=1=limθ0+cosθ, we conclude that limθ0+sinθθ=1.limθ0+sinθθ=1. By applying a manipulation similar to that used in demonstrating that limθ0sinθ=0,limθ0sinθ=0, we can show that limθ0sinθθ=1.limθ0sinθθ=1. Thus,

limθ0sinθθ=1.limθ0sinθθ=1.
2.18

In Example 2.25 we use this limit to establish limθ01cosθθ=0.limθ01cosθθ=0. This limit also proves useful in later chapters.

Example 2.25

Evaluating an Important Trigonometric Limit

Evaluate limθ01cosθθ.limθ01cosθθ.

Solution

In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

limθ01cosθθ=limθ01cosθθ·1+cosθ1+cosθ=limθ01cos2θθ(1+cosθ)=limθ0sin2θθ(1+cosθ)=limθ0sinθθ·sinθ1+cosθ=1·02=0.limθ01cosθθ=limθ01cosθθ·1+cosθ1+cosθ=limθ01cos2θθ(1+cosθ)=limθ0sin2θθ(1+cosθ)=limθ0sinθθ·sinθ1+cosθ=1·02=0.

Therefore,

limθ01cosθθ=0.limθ01cosθθ=0.
2.19
Checkpoint 2.20

Evaluate limθ01cosθsinθ.limθ01cosθsinθ.

Student Project

Deriving the Formula for the Area of a Circle

Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased. He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have predicted about the limit.

We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular polygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you can obtain the area of the circle. To see this, carry out the following steps:

  1. Express the height h and the base b of the isosceles triangle in Figure 2.31 in terms of θθ and r.
    A diagram of a circle with an inscribed polygon – namely, an octagon. An isosceles triangle is drawn with one of the sides of the octagon as the base and center of the circle/octagon as the top vertex. The height h goes from the center of the base b to the center, and each of the legs is also radii r of the circle. The angle created by the height h and one of the legs r is labeled as theta.
    Figure 2.31
  2. Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r.
    (Substitute (1/2)sinθ(1/2)sinθ for sin(θ/2)cos(θ/2)sin(θ/2)cos(θ/2) in your expression.)
  3. If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve this for n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)
  4. Find an expression for the area of the n-sided polygon in terms of r and θ.
  5. To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. (Hint: limθ0(sinθ)θ=1).limθ0(sinθ)θ=1).

The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration.

Section 2.3 Exercises

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

83.

limx0(4x22x+3)limx0(4x22x+3)

84.

limx1x3+3x2+547xlimx1x3+3x2+547x

85.

limx−2x26x+3limx−2x26x+3

86.

limx−1(9x+1)2limx−1(9x+1)2

In the following exercises, use direct substitution to evaluate each limit.

87.

limx7x2limx7x2

88.

limx−2(4x21)limx−2(4x21)

89.

limx011+sinxlimx011+sinx

90.

limx2e2xx2limx2e2xx2

91.

limx127xx+6limx127xx+6

92.

limx3lne3xlimx3lne3x

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0/0.0/0. Then, evaluate the limit.

93.

limx4x216x4limx4x216x4

94.

limx2x2x22xlimx2x2x22x

95.

limx63x182x12limx63x182x12

96.

limh0(1+h)21hlimh0(1+h)21h

97.

limt9t9t3limt9t9t3

98.

limh01a+h1ah,limh01a+h1ah, where a is a real-valued constant

99.

limθπsinθtanθlimθπsinθtanθ

100.

limx1x31x21limx1x31x21

101.

limx1/22x2+3x22x1limx1/22x2+3x22x1

102.

limx−3x+41x+3limx−3x+41x+3

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 2.23 to simplify the function to help determine the limit.

103.

limx−22x2+7x4x2+x2limx−22x2+7x4x2+x2

104.

limx−2+2x2+7x4x2+x2limx−2+2x2+7x4x2+x2

105.

limx12x2+7x4x2+x2limx12x2+7x4x2+x2

106.

limx1+2x2+7x4x2+x2limx1+2x2+7x4x2+x2

In the following exercises, assume that limx6f(x)=4,limx6g(x)=9,limx6f(x)=4,limx6g(x)=9, and limx6h(x)=6.limx6h(x)=6. Use these three facts and the limit laws to evaluate each limit.

107.

limx62f(x)g(x)limx62f(x)g(x)

108.

limx6g(x)1f(x)limx6g(x)1f(x)

109.

limx6(f(x)+13g(x))limx6(f(x)+13g(x))

110.

limx6(h(x))32limx6(h(x))32

111.

limx6g(x)f(x)limx6g(x)f(x)

112.

limx6x·h(x)limx6x·h(x)

113.

limx6[(x+1)·f(x)]limx6[(x+1)·f(x)]

114.

limx6(f(x)·g(x)h(x))limx6(f(x)·g(x)h(x))

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

115.

f(x)={x2,x3x+4,x>3f(x)={x2,x3x+4,x>3

  1. limx3f(x)limx3f(x)
  2. limx3+f(x)limx3+f(x)
116.

g(x)={x31,x01,x>0g(x)={x31,x01,x>0

  1. limx0g(x)limx0g(x)
  2. limx0+g(x)limx0+g(x)
117.

h(x)={x22x+1,x<23x,x2h(x)={x22x+1,x<23x,x2

  1. limx2h(x)limx2h(x)
  2. limx2+h(x)limx2+h(x)

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

Two graphs of piecewise functions. The upper is f(x), which has two linear segments. The first is a line with negative slope existing for x < -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x > -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4).
118.

limx−3+(f(x)+g(x))limx−3+(f(x)+g(x))

119.

limx−3(f(x)3g(x))limx−3(f(x)3g(x))

120.

limx0f(x)g(x)3limx0f(x)g(x)3

121.

limx−52+g(x)f(x)limx−52+g(x)f(x)

122.

limx1(f(x))2limx1(f(x))2

123.

limx1f(x)g(x)3limx1f(x)g(x)3

124.

limx−7(x·g(x))limx−7(x·g(x))

125.

limx−9[x·f(x)+2·g(x)]limx−9[x·f(x)+2·g(x)]

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions f(x),g(x),f(x),g(x), and h(x)h(x) when possible.

126.

[T] True or False? If 2x1g(x)x22x+3,2x1g(x)x22x+3, then limx2g(x)=0.limx2g(x)=0.

127.

[T] limθ0θ2cos(1θ)limθ0θ2cos(1θ)

128.

limx0f(x),limx0f(x), where f(x)={0,xrationalx2,xirrrationalf(x)={0,xrationalx2,xirrrational

129.

[T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: E(r)=q4πε0r2,E(r)=q4πε0r2, where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and 14πε014πε0 is Coulomb’s constant: 8.988×109N·m2/C2.8.988×109N·m2/C2.

  1. Use a graphing calculator to graph E(r)E(r) given that the charge of the particle is q=10−10.q=10−10.
  2. Evaluate limr0+E(r).limr0+E(r). What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?
130.

[T] The density of an object is given by its mass divided by its volume: ρ=m/V.ρ=m/V.

  1. Use a calculator to plot the volume as a function of density (V=m/ρ),(V=m/ρ), assuming you are examining something of mass 8 kg (m=8).m=8).
  2. Evaluate limρ0+V(ρ)limρ0+V(ρ) and explain the physical meaning.
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