Section 2.3The Limit Laws

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

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

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

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 \(\underset{x\to a}{\text{lim}}f(x)=f(a).\) 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.

To see that this theorem holds, consider the polynomial \(p(x)={c}_{n}{x}^{n}+{c}_{n-1}{x}^{n-1}+⋯+{c}_{1}x+{c}_{0}.\) By applying the sum, constant multiple, and power laws, we end up with

\[\begin{array}{ll}\underset{x\to a}{\text{lim}}p(x) & =\underset{x\to a}{\text{lim}}({c}_{n}{x}^{n}+{c}_{n-1}{x}^{n-1}+⋯+{c}_{1}x+{c}_{0}) \\ & ={c}_{n}{(\underset{x\to a}{\text{lim}}x)}^{n}+{c}_{n-1}{(\underset{x\to a}{\text{lim}}x)}^{n-1}+⋯+{c}_{1}(\underset{x\to a}{\text{lim}}x)+\underset{x\to a}{\text{lim}}{c}_{0} \\ & ={c}_{n}{a}^{n}+{c}_{n-1}{a}^{n-1}+⋯+{c}_{1}a+{c}_{0} \\ & =p(a).\end{array}\]

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

\[\underset{x\to a}{\text{lim}}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}.\]

Example 4 applies this result.

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 \(\underset{x\to a}{\text{lim}}f(x)\) to exist when \(f(a)\) is undefined. The following observation allows us to evaluate many limits of this type:

If for all \(x\ne a,f(x)=g(x)\) over some open interval containing a, then \(\underset{x\to a}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}g(x).\)

To understand this idea better, consider the limit \(\underset{x\to 1}{\text{lim}}\frac{{x}^{2}-1}{x-1}.\)

The function

\[\begin{array}{ll}f(x) & =\frac{{x}^{2}-1}{x-1} \\ & =\frac{(x-1)(x+1)}{x-1}\end{array}\]

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

We see that

\[\begin{array}{ll}\underset{x\to 1}{\text{lim}}\frac{{x}^{2}-1}{x-1} & =\underset{x\to 1}{\text{lim}}\frac{(x-1)(x+1)}{x-1} \\ & =\underset{x\to 1}{\text{lim}}(x+1) \\ & =2.\end{array}\]

The limit has the form \(\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)},\) where \(\underset{x\to a}{\text{lim}}f(x)=0\) and \(\underset{x\to a}{\text{lim}}g(x)=0.\) (In this case, we say that \(f(x)\text{/}g(x)\) has the indeterminate form \(0\text{/}0\text{.)}\) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

The next examples demonstrate the use of this Problem-Solving Strategy. Example 5 illustrates the factor-and-cancel technique; Example 6 shows multiplying by a conjugate. In Example 7, we look at simplifying a complex fraction.

Example 8 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.

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 \(\underset{x\to {a}^{-}}{\text{lim}}h(x),\) we require the function \(h(x)\) to be defined over an open interval of the form \((b,a);\) for a limit of the form \(\underset{x\to {a}^{+}}{\text{lim}}h(x),\) we require the function \(h(x)\) to be defined over an open interval of the form \((a,c).\) Example 9 illustrates this point.

Example 9

Evaluate each of the following limits, if possible.

• \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}\)
• \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}\)

Think about the domain of the square root function near x=3 before applying any limit law.

Show solution

Figure 2 illustrates the function \(f(x)=\sqrt{x-3}\) and aids in our understanding of these limits.

• The function \(f(x)=\sqrt{x-3}\) is defined over the interval \([3,\text{+}\infty ).\) Since this function is not defined to the left of 3, we cannot apply the limit laws to compute \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}.\) In fact, since \(f(x)=\sqrt{x-3}\) is undefined to the left of 3, \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}\) does not exist.
• Since \(f(x)=\sqrt{x-3}\) is defined to the right of 3, the limit laws do apply to \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}.\) By applying these limit laws we obtain \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}=0.\)

In Example 10 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 10

For \(f(x)=\left\{\begin{array}{ll}4x-3 & \text{if}\,x<2 \\ {(x-3)}^{2} & \text{if}\,x\ge 2\end{array}\right.,\) evaluate each of the following limits:

• \(\underset{x\to {2}^{-}}{\text{lim}}f(x)\)
• \(\underset{x\to {2}^{+}}{\text{lim}}f(x)\)
• \(\underset{x\to 2}{\text{lim}}f(x)\)

Evaluate the two one-sided limits separately using each formula's own domain, then compare them.

Show solution

Figure 3 illustrates the function \(f(x)\) and aids in our understanding of these limits.

• Since \(f(x)=4x-3\) for all x in \((\text{-}\infty ,2),\) replace \(f(x)\) in the limit with \(4x-3\) and apply the limit laws:
  

  \[\underset{x\to {2}^{-}}{\text{lim}}f(x)=\underset{x\to {2}^{-}}{\text{lim}}(4x-3)=5.\]

• Since \(f(x)={(x-3)}^{2}\) for all x in \((2,\text{+}\infty ),\) replace \(f(x)\) in the limit with \({(x-3)}^{2}\) and apply the limit laws:
  

  \[\underset{x\to {2}^{+}}{\text{lim}}f(x)=\underset{x\to {2}^{+}}{\text{lim}}{(x-3)}^{2}=1.\]

• Since \(\underset{x\to {2}^{-}}{\text{lim}}f(x)=5\) and \(\underset{x\to {2}^{+}}{\text{lim}}f(x)=1,\) we conclude that \(\underset{x\to 2}{\text{lim}}f(x)\) does not exist.

Try It #7

Graph \(f(x)=\left\{\begin{array}{l}-x-2\,\text{if}\,x<\text{-}1 \\ 2\,\text{if}\,x=-1 \\ {x}^{3}\,\text{if}\,x>\text{-}1\end{array}\right.\) and evaluate \(\underset{x\to {-1}^{-}}{\text{lim}}f(x).\)

Show answer

-1, which crossed the x axis and y axis at the origin.">
\(\underset{x\to {-1}^{-}}{\text{lim}}f(x)=-1\)

Did you get it?I got it ✓Not yet — review

We now turn our attention to evaluating a limit of the form \(\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)},\) where \(\underset{x\to a}{\text{lim}}f(x)=K,\) where \(K\ne 0\) and \(\underset{x\to a}{\text{lim}}g(x)=0.\) That is, \(f(x)\text{/}g(x)\) has the form \(K\text{/}0,K\ne 0\) at a.

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 4 illustrates this idea.

Example 12

Apply the squeeze theorem to evaluate \(\underset{x\to 0}{\text{lim}}x\,\text{cos}\,x.\)

Squeeze x cos x between two functions you already know the limit of at 0.

Show solution

Because \(-1\le \text{cos}\,x\le 1\) for all x, we have \(-|x|\le x\,\text{cos}\,x\le |x|\) . Since \(\underset{x\to 0}{\text{lim}}(-|x|)=0=\underset{x\to 0}{\text{lim}}|x|,\) from the squeeze theorem, we obtain \(\underset{x\to 0}{\text{lim}}x\,\text{cos}\,x=0.\) The graphs of \(f(x)=-|x|,g(x)=x\,\text{cos}\,x,\) and \(h(x)=|x|\) are shown in Figure 5.

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 \(\underset{θ\to 0}{\text{lim}}\text{sin}\,θ.\) Consider the unit circle shown in Figure 6. In the figure, we see that \(\text{sin}\,θ\) 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 \(0<θ<\frac{\pi }{2},0<\text{sin}\,θ<θ.\)

Because \(\underset{θ\to {0}^{+}}{\text{lim}}0=0\) and \(\underset{θ\to {0}^{+}}{\text{lim}}θ=0,\) by using the squeeze theorem we conclude that

\[\underset{θ\to {0}^{+}}{\text{lim}}\text{sin}\,θ=0.\]

To see that \(\underset{θ\to {0}^{-}}{\text{lim}}\text{sin}\,θ=0\) as well, observe that for \(-\frac{\pi }{2}<θ<0,0<\text{-}θ<\frac{\pi }{2}\) and hence, \(0<\text{sin}\,(-θ)<\text{-}θ.\) Consequently, \(0<-\text{sin}\,θ<\text{-}θ.\) It follows that \(0>\text{sin}\,θ>θ.\) An application of the squeeze theorem produces the desired limit. Thus, since \(\underset{θ\to {0}^{+}}{\text{lim}}\text{sin}\,θ=0\) and \(\underset{θ\to {0}^{-}}{\text{lim}}\text{sin}\,θ=0,\)

\[\underset{θ\to 0}{\text{lim}}\,\text{sin}\,θ=0.\]

Next, using the identity \(\text{cos}\,θ=\sqrt{1-{\text{sin}}^{2}θ}\) for \(-\frac{\pi }{2}<θ<\frac{\pi }{2},\) we see that

\[\underset{θ\to 0}{\text{lim}}\,\text{cos}\,θ=\underset{θ\to 0}{\text{lim}}\sqrt{1-{\text{sin}}^{2}θ}=1.\]

We now take a look at a limit that plays an important role in later chapters—namely, \(\underset{θ\to 0}{\text{lim}}\frac{\text{sin}\,θ}{θ}.\) To evaluate this limit, we use the unit circle in Figure 7. Notice that this figure adds one additional triangle to Figure 7. We see that the length of the side opposite angle θ in this new triangle is \(\text{tan}\,θ.\) Thus, we see that for \(0<θ<\frac{\pi }{2},\text{sin}\,θ<θ<\text{tan}\,θ.\)

By dividing by \(\text{sin}\,θ\) in all parts of the inequality, we obtain

\[1<\frac{θ}{\text{sin}\,θ}<\frac{1}{\text{cos}\,θ}.\]

Equivalently, we have

\[1>\frac{\text{sin}\,θ}{θ}>\text{cos}\,θ.\]

Since \(\underset{θ\to {0}^{+}}{\text{lim}}1=1=\underset{θ\to {0}^{+}}{\text{lim}}\text{cos}\,θ,\) we conclude that \(\underset{θ\to {0}^{+}}{\text{lim}}\frac{\text{sin}\,θ}{θ}=1.\) By applying a manipulation similar to that used in demonstrating that \(\underset{θ\to {0}^{-}}{\text{lim}}\text{sin}\,θ=0,\) we can show that \(\underset{θ\to {0}^{-}}{\text{lim}}\frac{\text{sin}\,θ}{θ}=1.\) Thus,

\[\underset{θ\to 0}{\text{lim}}\frac{\text{sin}\,θ}{θ}=1.\]

In Example 13 we use this limit to establish \(\underset{θ\to 0}{\text{lim}}\frac{1-\text{cos}\,θ}{θ}=0.\) This limit also proves useful in later chapters.

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, \(\underset{x\to a}{\text{lim}}f(x)=f(a).\)
• 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

Section Exercises

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

1

\(\underset{x\to 0}{\text{lim}}(4{x}^{2}-2x+3)\)

Show answer

Use constant multiple law and difference law: \(\underset{x\to 0}{\text{lim}}(4{x}^{2}-2x+3)=4\underset{x\to 0}{\text{lim}}{x}^{2}-2\underset{x\to 0}{\text{lim}}x+\underset{x\to 0}{\text{lim}}3=3\)

2

\(\underset{x\to 1}{\text{lim}}\frac{{x}^{3}+3{x}^{2}+5}{4-7x}\)

3

\(\underset{x\to -2}{\text{lim}}\sqrt{{x}^{2}-6x+3}\)

Show answer

Use root law: \(\underset{x\to -2}{\text{lim}}\sqrt{{x}^{2}-6x+3}=\sqrt{\underset{x\to -2}{\text{lim}}({x}^{2}-6x+3)}=\sqrt{19}\)

4

\(\underset{x\to -1}{\text{lim}}{(9x+1)}^{2}\)

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

5

\(\underset{x\to 7}{\text{lim}}{x}^{2}\)

Show answer

49

6

\(\underset{x\to -2}{\text{lim}}(4{x}^{2}-1)\)

7

\(\underset{x\to 0}{\text{lim}}\frac{1}{1+\text{sin}\,x}\)

Show answer

1

8

\(\underset{x\to 2}{\text{lim}}{e}^{2x-{x}^{2}}\)

9

\(\underset{x\to 1}{\text{lim}}\frac{2-7x}{x+6}\)

Show answer

\(-\frac{5}{7}\)

10

\(\underset{x\to 3}{\text{lim}}\text{ln}{e}^{3x}\)

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form \(0\text{/}0.\) Then, evaluate the limit.

11

\(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}\)

Show answer

\(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}=\frac{16-16}{4-4}=\frac{0}{0};\) then, \(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}=\underset{x\to 4}{\text{lim}}\frac{(x+4)(x-4)}{x-4}=8\)

12

\(\underset{x\to 2}{\text{lim}}\frac{x-2}{{x}^{2}-2x}\)

13

\(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}\)

Show answer

\(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0};\) then, \(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}=\underset{x\to 6}{\text{lim}}\frac{3(x-6)}{2(x-6)}=\frac{3}{2}\)

14

\(\underset{h\to 0}{\text{lim}}\frac{{(1+h)}^{2}-1}{h}\)

15

\(\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}\)

Show answer

\(\underset{x\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}=\frac{9-9}{3-3}=\frac{0}{0};\) then, \(\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to 9}{\text{lim}}(\sqrt{t}+3)=6\)

16

\(\underset{h\to 0}{\text{lim}}\frac{\frac{1}{a+h}-\frac{1}{a}}{h},\) where a is a non-zero real-valued constant

17

\(\underset{θ\to \pi }{\text{lim}}\frac{\text{sin}\,θ}{\text{tan}\,θ}\)

Show answer

\(\underset{θ\to \pi }{\text{lim}}\frac{\text{sin}\,θ}{\text{tan}\,θ}=\frac{\text{sin}\,\pi }{\text{tan}\,\pi }=\frac{0}{0};\) then, \(\underset{θ\to \pi }{\text{lim}}\frac{\text{sin}\,θ}{\text{tan}\,θ}=\underset{θ\to \pi }{\text{lim}}\frac{\text{sin}\,θ}{\frac{\text{sin}\,θ}{\text{cos}\,θ}}=\underset{θ\to \pi }{\text{lim}}\text{cos}\,θ=-1\)

18

\(\underset{x\to 1}{\text{lim}}\frac{{x}^{3}-1}{{x}^{2}-1}\)

19

\(\underset{x\to 1\text{/}2}{\text{lim}}\frac{2{x}^{2}+3x-2}{2x-1}\)

Show answer

\(\underset{x\to 1\text{/}2}{\text{lim}}\frac{2{x}^{2}+3x-2}{2x-1}=\frac{\frac{1}{2}+\frac{3}{2}-2}{1-1}=\frac{0}{0};\) then, \(\underset{x\to 1\text{/}2}{\text{lim}}\frac{2{x}^{2}+3x-2}{2x-1}=\underset{x\to 1\text{/}2}{\text{lim}}\frac{(2x-1)(x+2)}{2x-1}=\frac{5}{2}\)

20

\(\underset{x\to -3}{\text{lim}}\frac{\sqrt{x+4}-1}{x+3}\)

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example 11 to simplify the function to help determine the limit.

21

\(\underset{x\to {-2}^{-}}{\text{lim}}\frac{2{x}^{2}+7x-4}{{x}^{2}+x-2}\)

Show answer

−∞

22

\(\underset{x\to {-2}^{+}}{\text{lim}}\frac{2{x}^{2}+7x-4}{{x}^{2}+x-2}\)

23

\(\underset{x\to {1}^{-}}{\text{lim}}\frac{2{x}^{2}+7x-4}{{x}^{2}+x-2}\)

Show answer

−∞

24

\(\underset{x\to {1}^{+}}{\text{lim}}\frac{2{x}^{2}+7x-4}{{x}^{2}+x-2}\)

In the following exercises, assume that \(\underset{x\to 6}{\text{lim}}f(x)=4,\underset{x\to 6}{\text{lim}}g(x)=9,\) and \(\underset{x\to 6}{\text{lim}}h(x)=6.\) Use these three facts and the limit laws to evaluate each limit.

25

\(\underset{x\to 6}{\text{lim}}2f(x)g(x)\)

Show answer

\(\underset{x\to 6}{\text{lim}}2f(x)g(x)=2\underset{x\to 6}{\text{lim}}f(x)\underset{x\to 6}{\text{lim}}g(x)=72\)

26

\(\underset{x\to 6}{\text{lim}}\frac{g(x)-1}{f(x)}\)

27

\(\underset{x\to 6}{\text{lim}}(f(x)+\frac{1}{3}g(x))\)

Show answer

\(\underset{x\to 6}{\text{lim}}(f(x)+\frac{1}{3}g(x))=\underset{x\to 6}{\text{lim}}f(x)+\frac{1}{3}\underset{x\to 6}{\text{lim}}g(x)=7\)

28

\(\underset{x\to 6}{\text{lim}}\frac{{(h(x))}^{3}}{2}\)

29

\(\underset{x\to 6}{\text{lim}}\sqrt{g(x)-f(x)}\)

Show answer

\(\underset{x\to 6}{\text{lim}}\sqrt{g(x)-f(x)}=\sqrt{\underset{x\to 6}{\text{lim}}g(x)-\underset{x\to 6}{\text{lim}}f(x)}=\sqrt{5}\)

30

\(\underset{x\to 6}{\text{lim}}x·h(x)\)

31

\(\underset{x\to 6}{\text{lim}}[(x+1)·f(x)]\)

Show answer

\(\underset{x\to 6}{\text{lim}}[(x+1)f(x)]=(\underset{x\to 6}{\text{lim}}(x+1))(\underset{x\to 6}{\text{lim}}f(x))=28\)

32

\(\underset{x\to 6}{\text{lim}}(f(x)·g(x)-h(x))\)

[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.

33

\(f(x)=\left\{\begin{array}{ll}{x}^{2}, & x\le 3 \\ x+4, & x>3\end{array}\right.\)

• \(\underset{x\to {3}^{-}}{\text{lim}}f(x)\)
• \(\underset{x\to {3}^{+}}{\text{lim}}f(x)\)

Show answer

3. There is an open circle at (3, 7), and the slope is 1.">
a. 9; b. 7

34

\(g(x)=\left\{\begin{array}{ll}{x}^{3}-1, & x\le 0 \\ 1, & x>0\end{array}\right.\)

• \(\underset{x\to {0}^{-}}{\text{lim}}g(x)\)
• \(\underset{x\to {0}^{+}}{\text{lim}}g(x)\)

35

\(h(x)=\left\{\begin{array}{ll}{x}^{2}-2x+1, & x<2 \\ 3-x, & x\ge 2\end{array}\right.\)

• \(\underset{x\to {2}^{-}}{\text{lim}}h(x)\)
• \(\underset{x\to {2}^{+}}{\text{lim}}h(x)\)

Show answer

= 2. It has a slope of -1 and an x intercept at (3,0).">
a. 1; b. 1

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

36

\(\underset{x\to {-3}^{+}}{\text{lim}}(f(x)+g(x))\)

37

\(\underset{x\to {-3}^{-}}{\text{lim}}(f(x)-3g(x))\)

Show answer

\(\underset{x\to {-3}^{-}}{\text{lim}}(f(x)-3g(x))=\underset{x\to {-3}^{-}}{\text{lim}}f(x)-3\underset{x\to {-3}^{-}}{\text{lim}}g(x)=0+6=6\)

38

\(\underset{x\to 0}{\text{lim}}\frac{f(x)g(x)}{3}\)

39

\(\underset{x\to -5}{\text{lim}}\frac{2+g(x)}{f(x)}\)

Show answer

\(\underset{x\to -5}{\text{lim}}\frac{2+g(x)}{f(x)}=\frac{2+(\underset{x\to -5}{\text{lim}}g(x))}{\underset{x\to -5}{\text{lim}}f(x)}=\frac{2+0}{2}=1\)

40

\(\underset{x\to 1}{\text{lim}}{(f(x))}^{2}\)

41

\(\underset{x\to 1}{\text{lim}}\sqrt[3]{f(x)-g(x)}\)

Show answer

\(\underset{x\to 1}{\text{lim}}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\underset{x\to 1}{\text{lim}}f(x)-\underset{x\to 1}{\text{lim}}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}\)

42

\(\underset{x\to -7}{\text{lim}}(x·g(x))\)

43

\(\underset{x\to -9}{\text{lim}}[x·f(x)+2·g(x)]\)

Show answer

\(\underset{x\to -9}{\text{lim}}(xf(x)+2g(x))=(\underset{x\to -9}{\text{lim}}x)(\underset{x\to -9}{\text{lim}}f(x))+2\underset{x\to -9}{\text{lim}}(g(x))=(-9)(6)+2(4)=-46\)

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

44

[T] True or False? If \(2x-1\le g(x)\le {x}^{2}-2x+3,\) then \(\underset{x\to 2}{\text{lim}}g(x)=0.\)

45

[T] \(\underset{θ\to 0}{\text{lim}}{θ}^{2}\text{cos}\,(\frac{1}{θ})\)

Show answer

The limit is zero.

46

\(\underset{x\to 0}{\text{lim}}f(x),\) where \(f(x)=\left\{\begin{array}{ll}0, & x\,\text{rational} \\ {x}^{2}, & x\,\text{irrational}\end{array}\right.\)

47

[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: \(E(r)=\frac{q}{4\pi {ε}_{0}{r}^{2}},\) 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 \(\frac{1}{4\pi {ε}_{0}}\) is Coulomb’s constant: \(8.988\,\times \,{10}^{9}\,\text{N}·{\text{m}}^{2}\text{/}{\text{C}}^{2}.\)

• Use a graphing calculator to graph \(E(r)\) given that the charge of the particle is \(q={10}^{-10}.\)
• Evaluate \(\underset{r\to {0}^{+}}{\text{lim}}E(r).\) What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

Show answer

a.

b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

48

[T] The density of an object is given by its mass divided by its volume: \(ρ=m\text{/}V.\)

• Use a calculator to plot the volume as a function of density \((V=m\text{/}ρ),\) assuming you are examining something of mass 8 kg ( \(m=8\text{).}\)
• Evaluate \(\underset{ρ\to {0}^{+}}{\text{lim}}V(ρ)\) and explain the physical meaning.

Glossary

constant multiple law for limits

the limit law \(\underset{x\to a}{\text{lim}}cf(x)=c·\underset{x\to a}{\text{lim}}f(x)=cL\)

difference law for limits

the limit law \(\underset{x\to a}{\text{lim}}(f(x)-g(x))=\underset{x\to a}{\text{lim}}f(x)-\underset{x\to a}{\text{lim}}g(x)=L-M\)

limit laws

the individual properties of limits; for each of the individual laws, let \(f(x)\) and \(g(x)\) be defined for all \(x\ne a\) over some open interval containing a; assume that L and M are real numbers so that \(\underset{x\to a}{\text{lim}}f(x)=L\) and \(\underset{x\to a}{\text{lim}}g(x)=M;\) let c be a constant

power law for limits

the limit law \(\underset{x\to a}{\text{lim}}{(f(x))}^{n}={(\underset{x\to a}{\text{lim}}f(x))}^{n}={L}^{n}\) for every positive integer n

product law for limits

the limit law \(\underset{x\to a}{\text{lim}}(f(x)·g(x))=\underset{x\to a}{\text{lim}}f(x)·\underset{x\to a}{\text{lim}}g(x)=L·M\)

quotient law for limits

the limit law \(\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}=\frac{L}{M}\) for \(M\ne 0\)

root law for limits

the limit law \(\underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\text{lim}}f(x)}=\sqrt[n]{L}\) for all L if n is odd and for \(L\ge 0\) if n is even

squeeze theorem

states that if \(f(x)\le g(x)\le h(x)\) for all \(x\ne a\) over an open interval containing a and \(\underset{x\to a}{\text{lim}}f(x)=L=\underset{x\to a}{\text{lim}}h(x)\) where L is a real number, then \(\underset{x\to a}{\text{lim}}g(x)=L\)

sum law for limits

The limit law \(\underset{x\to a}{\text{lim}}(f(x)+g(x))=\underset{x\to a}{\text{lim}}f(x)+\underset{x\to a}{\text{lim}}g(x)=L+M\)