and the function [latex]g\left(x\right)=x+1[/latex] are identical for all values of [latex]x\ne 1.[/latex] The graphs of these two functions are shown in [link].

The graphs of [latex]f\left(x\right)[/latex] and [latex]g\left(x\right)[/latex] are identical for all [latex]x\ne 1.[/latex] Their limits at 1 are equal.

The limit has the form [latex]\underset>\frac,[/latex] where [latex]\underset>f\left(x\right)=0[/latex] and [latex]\underset>g\left(x\right)=0.[/latex] (In this case, we say that [latex]f\left(x\right)\textg\left(x\right)[/latex] has the indeterminate form [latex]0\text0\text<.)>[/latex] The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Problem-Solving Strategy: Calculating a Limit When [latex]f\left(x\right)\textg\left(x\right)[/latex] 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 [latex]h\left(x\right)=f\left(x\right)\textg\left(x\right)[/latex] for all [latex]x\ne a[/latex] over some interval containing a. To do this, we may need to try one or more of the following steps:
   1. If [latex]f\left(x\right)[/latex] and [latex]g\left(x\right)[/latex] 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 [latex]f\left(x\right)\textg\left(x\right)[/latex] is a complex fraction, we begin by simplifying it.

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

   Evaluating a Limit by Factoring and Canceling

   Step 1. The function [latex]f\left(x\right)=\frac^-3x>^-5x-3>[/latex] is undefined for [latex]x=3.[/latex] In fact, if we substitute 3 into the function we get [latex]0\text0,[/latex] which is undefined. Factoring and canceling is a good strategy:

   [latex]\underset>\frac^-3x>^-5x-3>=\underset>\frac<\left(x-3\right)\left(2x+1\right)>[/latex]

   Step 2. For all [latex]x\ne 3,\frac^-3x>^-5x-3>=\frac.[/latex] Therefore,

   [latex]\underset>\frac<\left(x-3\right)\left(2x+1\right)>=\underset>\frac.[/latex]

   Step 3. Evaluate using the limit laws:

   [latex]\underset>\frac=\frac.[/latex]

   Follow the steps in the Problem-Solving Strategy and [link].

   Evaluating a Limit by Multiplying by a Conjugate

   Step 1.[latex]\frac-1>[/latex] has the form [latex]0\text0[/latex] at −1. Let’s begin by multiplying by [latex]\sqrt+1,[/latex] the conjugate of [latex]\sqrt-1,[/latex] on the numerator and denominator:

   [latex]\underset>\frac-1>=\underset>\frac-1>·\frac+1>+1>.[/latex]

   Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the [latex]\left(x+1\right)[/latex] in the denominator cancels out in the end:

   [latex]=\underset>\frac<\left(x+1\right)\left(\sqrt+1\right)>.[/latex]

   Step 3. Then we cancel:

   [latex]=\underset>\frac+1>.[/latex]

   Step 4. Last, we apply the limit laws:

   [latex]\underset>\frac+1>=\frac.[/latex]

   Follow the steps in the Problem-Solving Strategy and [link].

   Evaluating a Limit by Simplifying a Complex Fraction

   Step 1.[latex]\frac-\frac>[/latex] has the form [latex]0\text0[/latex] at 1. We simplify the algebraic fraction by multiplying by [latex]2\left(x+1\right)\text2\left(x+1\right):[/latex]

   [latex]\underset>\frac-\frac>=\underset>\frac-\frac>·\frac<2\left(x+1\right)><2\left(x+1\right)>.[/latex]

   Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor [latex]\left(x-1\right)\text[/latex]

   [latex]=\underset>\frac<2-\left(x+1\right)><2\left(x-1\right)\left(x+1\right)>.[/latex]

   Step 3. Then, we simplify the numerator:

   [latex]=\underset>\frac<2\left(x-1\right)\left(x+1\right)>.[/latex]

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

   [latex]=\underset>\frac<-\left(x-1\right)><2\left(x-1\right)\left(x+1\right)>.[/latex]

   Step 5. Then, we cancel the common factors of [latex]\left(x-1\right)\text[/latex]

   [latex]=\underset>\frac<2\left(x+1\right)>.[/latex]

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

   [latex]\underset>\frac<2\left(x+1\right)>=-\frac.[/latex]

   Follow the steps in the Problem-Solving Strategy and [link].

   [link] 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.

   Evaluating a Limit When the Limit Laws Do Not Apply

   Both [latex]1\textx[/latex] and [latex]5\textx\left(x-5\right)[/latex] 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

   [latex]\phantom>\begin\hfill \frac+\frac& =\frac\hfill \\ & =\frac.\hfill \end[/latex]Use the same technique as [link]. Don’t forget to factor [latex]^-2x-3[/latex] before getting a common denominator.
   Evaluating a One-Sided Limit Using the Limit Laws

   Evaluate each of the following limits, if possible.

   [link] illustrates the function [latex]f\left(x\right)=\sqrt[/latex] and aids in our understanding of these limits.

   The graph shows the function [latex]f\left(x\right)=\sqrt.[/latex]

   

   1. The function [latex]f\left(x\right)=\sqrt[/latex] is defined over the interval [latex]\left[3,\text\infty \right).[/latex] Since this function is not defined to the left of 3, we cannot apply the limit laws to compute [latex]\underset^>>\sqrt.[/latex] In fact, since [latex]f\left(x\right)=\sqrt[/latex] is undefined to the left of 3, [latex]\underset^>>\sqrt[/latex] does not exist.
   2. Since [latex]f\left(x\right)=\sqrt[/latex] is defined to the right of 3, the limit laws do apply to [latex]\underset^>>\sqrt.[/latex] By applying these limit laws we obtain [latex]\underset^>>\sqrt=0.[/latex]

   In [link] 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.

   Evaluating a Two-Sided Limit Using the Limit Laws
   1. [latex]\underset^>>f\left(x\right)[/latex]
   2. [latex]\underset^>>f\left(x\right)[/latex]
   3. [latex]\underset>f\left(x\right)[/latex]

   [link] illustrates the function [latex]f\left(x\right)[/latex] and aids in our understanding of these limits.

   This graph shows a function [latex]f\left(x\right).[/latex]

   

   1. Since [latex]f\left(x\right)=4x-3[/latex] for all x in [latex]\left(\text\infty ,2\right),[/latex] replace [latex]f\left(x\right)[/latex] in the limit with [latex]4x-3[/latex] and apply the limit laws:
   [latex]\underset^>>f\left(x\right)=\underset^>>\left(4x-3\right)=5.[/latex] [latex]\underset^>>f\left(x\right)=\underset^>><\left(x-3\right)>^=1.[/latex]

   

   [latex]\underset^>>f\left(x\right)=-1[/latex]

   Use the method in [link] to evaluate the limit.

   We now turn our attention to evaluating a limit of the form [latex]\underset>\frac,[/latex] where [latex]\underset>f\left(x\right)=K,[/latex] where [latex]K\ne 0[/latex] and [latex]\underset>g\left(x\right)=0.[/latex] That is, [latex]f\left(x\right)\textg\left(x\right)[/latex] has the form [latex]K\text0,K\ne 0[/latex] at a.

   Evaluating a Limit of the Form [latex]K\text0,K\ne 0[/latex] Using the Limit Laws

   Step 1. After substituting in [latex]x=2,[/latex] we see that this limit has the form [latex]-1\text0.[/latex] That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of [latex]\frac[/latex] becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

   [latex]\underset^>>\frac^-2x>=\underset^>>\frac.[/latex]

   Step 2. Since [latex]x-2[/latex] is the only part of the denominator that is zero when 2 is substituted, we then separate [latex]1\text\left(x-2\right)[/latex] from the rest of the function:

   [latex]=\underset^>>\frac·\frac.[/latex]

   Step 3.[latex]\underset^>>\frac=-\frac[/latex] and [latex]\underset^>>\frac=\text\infty .[/latex] Therefore, the product of [latex]\left(x-3\right)\textx[/latex] and [latex]1\text\left(x-2\right)[/latex] has a limit of [latex]\text[/latex]

   [latex]\underset^>>\frac^-2x>=\text\infty .[/latex]

   Use the methods from [link].

   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. [link] illustrates this idea.

   The Squeeze Theorem applies when [latex]f\left(x\right)\le g\left(x\right)\le h\left(x\right)[/latex] and [latex]\underset>f\left(x\right)=\underset>h\left(x\right).[/latex]

   

   The Squeeze Theorem

   Let [latex]f\left(x\right),g\left(x\right),[/latex] and [latex]h\left(x\right)[/latex] be defined for all [latex]x\ne a[/latex] over an open interval containing a. If

   [latex]f\left(x\right)\le g\left(x\right)\le h\left(x\right)[/latex]

   for all [latex]x\ne a[/latex] in an open interval containing a and

   [latex]\underset>f\left(x\right)=L=\underset>h\left(x\right)[/latex]

   where L is a real number, then [latex]\underset>g\left(x\right)=L.[/latex]

   Applying the Squeeze Theorem

   Apply the squeeze theorem to evaluate [latex]\underset>x\phantom>\text\phantom>x.[/latex]

   Because [latex]-1\le \text\phantom>x\le 1[/latex] for all x, we have [latex]-x\le x\phantom>\text\phantom>x\le x[/latex] for [latex]x\ge 0[/latex] and [latex]-x\ge xcosx\ge x[/latex] for [latex]x\le 0[/latex] (if x is negative the direction of the inequalities changes when we multiply). Since [latex]\underset>\left(-x\right)=0=\underset>x,[/latex] from the squeeze theorem, we obtain [latex]\underset>x\phantom>\text\phantom>x=0.[/latex] The graphs of [latex]f\left(x\right)=-x,g\left(x\right)=x\phantom>\text\phantom>x,[/latex] and [latex]h\left(x\right)=x[/latex] are shown in [link].

   The graphs of [latex]f\left(x\right),g\left(x\right),[/latex] and [latex]h\left(x\right)[/latex] are shown around the point [latex]x=0.[/latex]

   

   Use the squeeze theorem to evaluate [latex]\underset>^\text\frac.[/latex]

   Use the fact that [latex]-^\le ^\text\phantom>\left(1\textx\right)\le ^[/latex] to help you find two functions such that [latex]^\text\phantom>\left(1\textx\right)[/latex] is squeezed between them.

   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 [latex]\underset>\text\phantom>\theta .[/latex] Consider the unit circle shown in [link]. In the figure, we see that [latex]\text\phantom>\theta [/latex] 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 [latex]0,0<\text\phantom>\theta The sine function is shown as a line on the unit circle.

   

   Because [latex]\underset^>>0=0[/latex] and [latex]\underset^>>\theta =0,[/latex] by using the squeeze theorem we conclude that

   [latex]\underset^>>\text\phantom>\theta =0.[/latex]

   To see that [latex]\underset^>>\text\phantom>\theta =0[/latex] as well, observe that for [latex]-\frac<\pi >\theta <\frac<\pi >[/latex] and hence, [latex]0<\text\phantom>\left(-\theta \right)\theta .[/latex] Consequently, [latex]0<-\text\phantom>\theta \theta .[/latex] It follows that [latex]0>\text\phantom>\theta >\theta .[/latex] An application of the squeeze theorem produces the desired limit. Thus, since [latex]\underset^>>\text\phantom>\theta =0[/latex] and [latex]\underset^>>\text\phantom>\theta =0,[/latex]

   [latex]\underset>\text\phantom>\theta =0.[/latex]

   Next, using the identity [latex]\text\phantom>\theta =\sqrt>^\theta >[/latex] for [latex]-\frac<\pi >,[/latex] we see that

   [latex]\underset>\text\phantom>\theta =\underset>\sqrt>^\theta >=1.[/latex]

   We now take a look at a limit that plays an important role in later chapters—namely, [latex]\underset>\frac\phantom>\theta >.[/latex] To evaluate this limit, we use the unit circle in [link]. Notice that this figure adds one additional triangle to [link]. We see that the length of the side opposite angle θ in this new triangle is [latex]\text\phantom>\theta .[/latex] Thus, we see that for [latex]0,\text\phantom>\theta <\theta <\text\phantom>\theta .[/latex]

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

   

   By dividing by [latex]\text\phantom>\theta [/latex] in all parts of the inequality, we obtain

   [latex]1\phantom>\theta >\phantom>\theta >.[/latex]Equivalently, we have
   [latex]1>\frac\phantom>\theta >>\text\phantom>\theta .[/latex]\frac\phantom>\theta >=1.[/latex]

   In [link] we use this limit to establish [latex]\underset>\frac\phantom>\theta >=0.[/latex] This limit also proves useful in later chapters.

   Evaluating an Important Trigonometric Limit

   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:

   [latex]\underset>\frac\phantom>\theta >=0.[/latex]

   Multiply numerator and denominator by [latex]1+\text\phantom>\theta .[/latex]

   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:

   Express the height h and the base b of the isosceles triangle in [link] in terms of [latex]\theta [/latex] and r.

   

   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, [latex]\underset>f\left(x\right)=f\left(a\right).[/latex]
   • 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

   [latex]\underset>\text\phantom>\theta =0[/latex] [latex]\underset>\text\phantom>\theta =1[/latex] [latex]\underset>\frac\phantom>\theta >=1[/latex]

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

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

   In the following exercises, use direct substitution to show that each limit leads to the indeterminate form [latex]0\text0.[/latex] Then, evaluate the limit.

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

   In the following exercises, assume that [latex]\underset>f\left(x\right)=4,\underset>g\left(x\right)=9,[/latex] and [latex]\underset>h\left(x\right)=6.[/latex] Use these three facts and the limit laws to evaluate each limit.

   [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.

   1. [latex]\underset^>>f\left(x\right)[/latex]
   2. [latex]\underset^>>f\left(x\right)[/latex]

   

   1. [latex]\underset^>>g\left(x\right)[/latex]
   2. [latex]\underset^>>g\left(x\right)[/latex]
   1. [latex]\underset^>>h\left(x\right)[/latex]
   2. [latex]\underset^>>h\left(x\right)[/latex]

   

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

   

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

   [T] True or False? If [latex]2x-1\le g\left(x\right)\le ^-2x+3,[/latex] then [latex]\underset>g\left(x\right)=0.[/latex]

   The limit is zero.

   

   [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: [latex]E\left(r\right)=\frac<4\pi <\epsilon >_^>,[/latex] 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 [latex]\frac<4\pi <\epsilon >_>[/latex] is Coulomb’s constant: [latex]8.988\phantom>×\phantom>^\phantom>\text·>^\text>^.[/latex]

   1. Use a graphing calculator to graph [latex]E\left(r\right)[/latex] given that the charge of the particle is [latex]q=^.[/latex]
   2. Evaluate [latex]\underset^>>E\left(r\right).[/latex] What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

   

   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.

   [T] The density of an object is given by its mass divided by its volume: [latex]\rho =m\textV.[/latex]

   1. Use a calculator to plot the volume as a function of density [latex]\left(V=m\text\rho \right),[/latex] assuming you are examining something of mass 8 kg ([latex]m=8\text<).>[/latex]
   2. Evaluate [latex]\underset^<+>>>V\left(\rho \right)[/latex] and explain the physical meaning.

   Glossary

   constant multiple law for limits the limit law [latex]\underset>cf\left(x\right)=c·\underset>f\left(x\right)=cL[/latex] difference law for limits the limit law [latex]\underset>\left(f\left(x\right)-g\left(x\right)\right)=\underset>f\left(x\right)-\underset>g\left(x\right)=L-M[/latex] limit laws the individual properties of limits; for each of the individual laws, let [latex]f\left(x\right)[/latex] and [latex]g\left(x\right)[/latex] be defined for all [latex]x\ne a[/latex] over some open interval containing a; assume that L and M are real numbers so that [latex]\underset>f\left(x\right)=L[/latex] and [latex]\underset>g\left(x\right)=M;[/latex] let c be a constant power law for limits the limit law [latex]\underset><\left(f\left(x\right)\right)>^=<\left(\underset>f\left(x\right)\right)>^=^[/latex] for every positive integer n product law for limits the limit law [latex]\underset>\left(f\left(x\right)·g\left(x\right)\right)=\underset>f\left(x\right)·\underset>g\left(x\right)=L·M[/latex] quotient law for limits the limit law [latex]\underset>\frac=\frac<\underset>f\left(x\right)><\underset>g\left(x\right)>=\frac[/latex] for [latex]M\ne 0[/latex] root law for limits the limit law [latex]\underset>\sqrt[n]=\sqrt[n]<\underset>f\left(x\right)>=\sqrt[n][/latex] for all L if n is odd and for [latex]L\ge 0[/latex] if n is even squeeze theorem states that if [latex]f\left(x\right)\le g\left(x\right)\le h\left(x\right)[/latex] for all [latex]x\ne a[/latex] over an open interval containing a and [latex]\underset>f\left(x\right)=L=\underset>h\left(x\right)[/latex] where L is a real number, then [latex]\underset>g\left(x\right)=L[/latex] sum law for limits The limit law [latex]\underset>\left(f\left(x\right)+g\left(x\right)\right)=\underset>f\left(x\right)+\underset>g\left(x\right)=L+M[/latex]