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2.5: Limit Laws and Techniques for Computing Limits

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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 previosuly and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Basic Limit Results

For any real number a and any constant c,

  1. limxax=a
  2. limxac=c

Example 2.5.1: Evaluating a Basic Limit

Evaluate each of the following limits using Note.

  1. limx2x
  2. limx25

Solution:

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

Limit Laws

Let f(x) and g(x) be defined for all xa over some open interval containing a. Assume that L and M are real numbers such that limxaf(x)=L and 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+M

  • Difference law for limits:

limxa(f(x)g(x))=limxaf(x)limxag(x)=LM

  • Constant multiple law for limits:

limxacf(x)=climxaf(x)=cL

  • Product law for limits:

limxa(f(x)g(x))=limxaf(x)limxag(x)=LM

  • Quotient law for limits:

limxaf(x)g(x)=limxaf(x)limxag(x)=LM

for M0.

  • Power law for limits:

limxa(f(x))n=(limxaf(x))n=Ln

for every positive integer n.

  • Root law for limits:

limxanf(x)=nlimxaf(x)=nL

for all L if n is odd and for L0 if n is even.

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

Example 2.5.2A: Evaluating a Limit Using Limit Laws

Use the limit laws to evaluate limx3(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.

limx3(4x+2) = limx34x+limx32 Apply the sum law.

=4limx3x+limx32 Apply the constant multiple law.

=4(3)+2=10. Apply the basic limit results and simplify.

Notice this is equivalent to substituting 3 for x in the original function. One just needs to be careful that the limit exists at this point.

Example 2.5.2B: Using Limit Laws Repeatedly

Use the limit laws to evaluate 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, make sure that (2)3+40.

=2limx2x23limx2x+limx21limx2x3+limx24 Apply the sum law and constant multiple law

=2(limx2x)23limx2x+limx21(limx2x)3+limx24 Apply the power law.

=2(4)3(2)+1(2)3+4=14. Apply the basic limit laws and simplify.

Notice this is equivalent to substituting 2 for x in the original function. One just needs to be careful that the limit exists at this point.

Exercise 2.5.2

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

Hint

Begin by applying the product law.

Or just substitute 6 for x in the original function. One just needs to be careful that the limit exists at this point.

Answer

1110

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). 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.

Limits of Polynomial and Rational Functions

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

limxap(x)=p(a)

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

when q(a)0.

To see that this theorem holds, consider the polynomial

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)

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

then

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

Example 2.5.3: Evaluating a Limit of a Rational Function

Evaluate the limx32x23x+15x+4.

Solution

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

limx32x23x+15x+4=1019.

Exercise 2.5.3

Evaluate limx2(3x32x+7).

Hint

Use LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS as reference

Answer

−13

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) to exist when f(a) is undefined. The following observation allows us to evaluate many limits of this type:

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

limxaf(x)=limxag(x).

To understand this idea better, consider the limit limx1x21x1.

The function

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

and the function g(x)=x+1 are identical for all values of x1. The graphs of these two functions are shown in Figure 2.5.1.

CNX_Calc_Figure_02_03_001.jpeg

Figure 2.5.1: The graphs of f(x) and g(x) are identical for all x1. Their limits at 1 are equal.

We see that

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

The limit has the form limxaf(x)g(x), where limxaf(x)=0 and limxag(x)=0. (In this case, we say that f(x)/g(x) has the indeterminate form 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) 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) for all xa over some interval containing a. To do this, we may need to try one or more of the following steps:
    1. If f(x) and 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) 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 illustrates the factor-and-cancel technique; Example shows multiplying by a conjugate. In Example, we look at simplifying a complex fraction.

Example 2.5.4: Evaluating a Limit by Factoring and Canceling

Evaluate limx3x23x2x25x3.

Solution

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

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

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

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

Step 3. Evaluate using the limit laws:

limx3x2x+1=37.

Exercise 2.5.4

Evaluate limx3x2+4x+3x29.

Hint

Follow the steps in the Problem-Solving Strategy

Answer

13

Example 2.5.5: Evaluating a Limit by Multiplying by a Conjugate

Evaluate limx1x+21x+1.

Solution

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

limx1x+21x+1=limx1x+21x+1x+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) in the denominator cancels out in the end:

=limx1x+1(x+1)(x+2+1).

Step 3. Then we cancel:

=limx11x+2+1.

Step 4. Last, we apply the limit laws:

limx11x+2+1=12.

Exercise 2.5.5

Evaluate limx5x12x5.

Hint

Follow the steps in the Problem-Solving Strategy

Answer

14

Example 2.5.6: Evaluating a Limit by Simplifying a Complex Fraction

Evaluate limx11x+112x1.

 
Solution

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

limx11x+112x1=limx11x+112x12(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):

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

Step 3. Then, we simplify the numerator:

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

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

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

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

=limx112(x+1).

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

limx112(x+1)=12(1+1)=14.

Exercise 2.5.6

Evaluate limx31x+2+1x+3.

Hint

Follow the steps in the Problem-Solving Strategy and

Answer

−1

Example 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 2.5.7: Evaluating a Limit When the Limit Laws Do Not Apply

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

Solution:

Both 1/x and 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).

Thus,

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

Exercise 2.5.7

Evaluate limx3(1x34x22x3).

Hint

Use the same technique as Example 2.5.7. Don’t forget to factor x22x3 before getting a common denominator.

Answer

14

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), we require the function h(x) to be defined over an open interval of the form (b,a); for a limit of the form limxa+h(x), we require the function h(x) to be defined over an open interval of the form (a,c). Example illustrates this point.

Example 2.5.8A: Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

  1. limx3x3
  2. limx3+x3

Solution

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

2.3.1.png

Figure 2.5.2: The graph shows the function f(x)=x3.

a. The function f(x)=x3 is defined over the interval [3,+). Since this function is not defined to the left of 3, we cannot apply the limit laws to compute limx3x3. In fact, since f(x)=x3 is undefined to the left of 3, limx3x3 does not exist.

b. Since f(x)=x3 is defined to the right of 3, the limit laws do apply to limx3+x3. By applying these limit laws we obtain limx3+x3=0.

In 2.5.9 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 2.5.8B: Evaluating a Two-Sided Limit Using the Limit Laws

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

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

Solution

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

2.3.2.png

Figure 2.5.3: This graph shows a function f(x).

a. Since f(x)=4x3 for all x in (,2), replace f(x) in the limit with 4x3 and apply the limit laws:

limx2f(x)=limx2(4x3)=5

b. Since f(x)=(x3)2for all x in (2,+), replace f(x) in the limit with (x3)2 and apply the limit laws:

limx2+f(x)=limx2(x3)2=1.

c. Since limx2f(x)=5 and limx2+f(x)=1, we conclude that limx2f(x) does not exist.

Exercise 2.5.8

Graph f(x)={x2ifx<12ifx=1x3ifx>1 and evaluate limx1f(x).

Hint

Use the method in Example to evaluate the limit.

Answer


clipboard_e71ceb22878206659cfeb8211baf900cb.png
Note: this graph is not quite right; there should be a point at (-1,2)

limx1f(x)=1

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

Example 2.5.9: Evaluating a Limit of the Form K/0,K0 Using the Limit Laws

Evaluate limx2x3x22x.

Solution:

Step 1. After substituting in x=2, we see that this limit has the form 1/0. That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of \dfrac{x−3}{x(x−2)} becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

limx2x3x22x=limx2x3x(x2).

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

=limx2x3x1x2.

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

limx2x3x22x=+.

Exercise 2.5.9

Evaluate limx1x+2(x1)2.

Solution

Use the methods from Example 2.5.10.

Answer

+∞

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 2.5.4 illustrates this idea.

CNX_Calc_Figure_02_03_005.jpeg

Figure 2.5.4: The Squeeze Theorem applies when f(x)g(x)h(x) and limxaf(x)=limxah(x).

The Squeeze Theorem

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

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

for all xa in an open interval containing a and

limxaf(x)=L=limxah(x)

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

Example 2.5.10: Applying the Squeeze Theorem

Apply the squeeze theorem to evaluate limx0xcosx.

Solution

Because 1cosx1 for all x, we have xxcosxx for x0 and xxcosxx for x0 (if x is negative the direction of the inequalities changes when we multiply). Since limx0(x)=0=limx0x, from the squeeze theorem, we obtain limx0xcosx=0. The graphs of f(x)=x,g(x)=xcosx, and h(x)=x are shown in Figure 2.5.5.

2.3.4.png

Figure 2.5.5: The graphs of f(x),g(x), and h(x) are shown around the point x=0.

Exercise 2.5.10

Use the squeeze theorem to evaluate limx0x2sin1x.

Hint

Use the fact that x2x2sin(1/x)x2 to help you find two functions such that x2sin(1/x) is squeezed between them.

Answer

0

limθ0sinθθ

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. Consider the unit circle shown in Figure 2.5.6. In the figure, we see that sinθ 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.

CNX_Calc_Figure_02_03_007.jpeg

Figure 2.5.6: The sine function is shown as a line on the unit circle.

Because limθ0+0=0 and limx0+θ=0, by using the squeeze theorem we conclude that

limθ0+sinθ=0.

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

limθ0sinθ=0

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

limθ0cosθ=limθ01sin2θ=1.

We now take a look at a limit that plays an important role in later chapters—namely, limθ0sinθθ. To evaluate this limit, we use the unit circle in Figure 2.5.6. Notice that this figure adds one additional triangle to Figure 2.5.7. We see that the length of the side opposite angle θ in this new triangle is tanθ. Thus, we see that for 0<θ<π2, sinθ<θ<tanθ.

CNX_Calc_Figure_02_03_008.jpeg

Figure 2.5.7: The sine and tangent functions are shown as lines on the unit circle.

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

1<θsinθ<1cosθ.

Equivalently, we have

1>sinθθ>cosθ.

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

limθ0sinθθ=1.

limθ01cosθθ

In Example 2.5.11 we use this limit to establish limθ01cosθθ=0. This limit also proves useful in later chapters.

Example 2.5.11: Evaluating an Important Trigonometric Limit

Evaluate 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=limθ01cosθθ1+cosθ1+cosθ

=limθ01cos2θθ(1+cosθ)

=limθ0sin2θθ(1+cosθ)

=limθ0sinθθsinθ1+cosθ

=102=0.

Therefore,

limθ01cosθθ=0.

Exercise 2.5.11

Evaluate limθ01cosθsinθ.

hint

Multiply numerator and denominator by 1+cosθ.

Answer

0

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 2.5.6 in terms of θ and r.

CNX_Calc_Figure_02_03_009.jpeg

Figure 2.5.6

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 for 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).

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

Key Concepts

  • The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each time.
  • For polynomials and rational functions, limxaf(x)=f(a).
  • You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying a complex fraction.
  • The squeeze theorem allows you to find the limit of a function if the function is always greater than one function and less than another function with limits that are known.

Key Equations

  • Basic Limit Results

limxax=alimxac=c

  • Important Limits

limθ0sinθ=0

limθ0cosθ=1

limθ0sinθθ=1

limθ01cosθθ=0

Glossary

constant multiple law for limits
the limit law limxacf(x)=climxaf(x)=cL
difference law for limits
the limit law limxa(f(x)g(x))=limxaf(x)limxag(x)=LM
limit laws
the individual properties of limits; for each of the individual laws, let f(x) and g(x) be defined for all xa over some open interval containing a; assume that L and M are real numbers so that limxaf(x)=L and limxag(x)=M; let c be a constant
power law for limits
the limit law limxa(f(x))n=(limxaf(x))n=Ln for every positive integer n
product law for limits
the limit law limxa(f(x)g(x))=limxaf(x)limxag(x)=LM
quotient law for limits
the limit law limxaf(x)g(x)=limxaf(x)limxag(x)=LM for M≠0
root law for limits
the limit law limxanf(x)=nlimxaf(x)=nL for all L if n is odd and for L0 if n is even
squeeze theorem
states that if f(x)g(x)h(x) for all xa over an open interval containing a and limxaf(x)=L=limxah(x) where L is a real number, then limxag(x)=L
sum law for limits
The limit law limxa(f(x)+g(x))=limxaf(x)+limxag(x)=L+M

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


This page titled 2.5: Limit Laws and Techniques for Computing Limits is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by OpenStax.

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