Gregory Hartman, Ph.D. Contributing Authors Troy Siemers, Ph.D. Brian Heinold, Ph.D. Dimplekumar Chalishajar, Ph.D. Editor Jennifer Bowen, Ph.D.

Transcription

1 AP E XCalculus I Version 30 Custom Edition for Math 00, Universit of Lethbridge Gregor Hartman, PhD Department of Applied Mathematics Virginia Militar Institute Contributing Authors Tro Siemers, PhD Department of Applied Mathematics Virginia Militar Institute Brian Heinold, PhD Department of Mathematics and Computer Science Mount Saint Mar s Universit Dimplekumar Chalishajar, PhD Department of Applied Mathematics Virginia Militar Institute Editor Jennifer Bowen, PhD Department of Mathematics and Computer Science The College of Wooster Sean Fitzpatrick, PhD Department of Mathematics and Computer Science Universit of Lethbridge (for U of L custom edition)

2 Copright 05 Gregor Hartman Licensed to the public under Creative Commons Attribution-Noncommercial 40 International Public License This version edited from the original b Sean Fitzpatrick for use in Math 00, Universit of Lethbridge

3 Contents Table of Contents Preface iii v Limits An Introduction To Limits Finding Limits Analticall 8 3 One Sided Limits 8 4 Continuit 4 5 Limits Involving Infinit 33 Derivatives 43 Instantaneous Rates of Change: The Derivative 43 Interpretations of the Derivative 55 3 Basic Differentiation Rules 6 4 The Product and Quotient Rules 67 3 The Graphical Behavior of Functions 77 3 Etreme Values 77 3 Increasing and Decreasing Functions Concavit and the Second Derivative 9 34 Curve Sketching Antiderivatives and Indefinite Integration 04 Inde

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5 Preface Preface to the Math 00 Custom Edition Since most of the remarks I made in the preface to the Precalculus tetbook appl here, I ll be brief The tet that follows is an adaptation of the first tetbook in the APEX Calculus series b Hartman et al I have edited their original L A TEXsource code to fit our course In particular, I have omitted an sections or chapters that we will not cover in Math 00 Those interested in the original, unabridged tetbook can obtain it b accessing the authors website, which is given in the original preface, which follows below A Note on Using this Tet Sean Fitzpatrick Universit of Lethbridge June 5, 05 Thank ou for reading this short preface Allow us to share a few ke points about the tet so that ou ma better understand what ou will find beond this page This tet is Part I of a three tet series on Calculus The first part covers material taught in man Calc courses: limits, derivatives, and the basics of integration, found in Chapters through 6 The second tet covers material often taught in Calc : integration and its applications, along with an introduction to sequences, series and Talor Polnomials, found in Chapters 5 through 8 The third tet covers topics common in Calc 3 or multivariable calc: parametric equations, polar coordinates, vector valued functions, and functions of more than one variable, found in Chapters 9 through 3 All three are available separatel for free at wwwapecalculuscom These three tets are intended to work together and make one cohesive tet, APEX Calculus, which can also be downloaded from the website Printing the entire tet as one volume makes for a large, heav, cumbersome book One can certainl onl print the pages the currentl need, but some prefer to have a nice, bound cop of the tet Therefore this tet has been split into these three manageable parts, each of which can be purchased for under $5 at Amazoncom A result of this splitting is that sometimes a concept is said to be eplored in a later section, though that section does not actuall appear in this particular tet Also, the inde makes reference to topics and page numbers that do not appear in this tet This is done intentionall to show the reader what topics are available for stud Downloading the pdf of APEX Calculus will ensure that ou have all the content For Students: How to Read this Tet Mathematics tetbooks have a reputation for being hard to read High level mathematical writing often seeks to sa much with few words, and this stle often seeps into tets of lower level topics This book was written with the goal of being easier to read than man other calculus tetbooks, without becoming too verbose

6 Contents Each chapter and section starts with an introduction of the coming material, hopefull setting the stage for wh ou should care, and ends with a look ahead to see how the just learned material helps address future problems Please read the tet; it is written to eplain the concepts of Calculus There are numerous eamples to demonstrate the meaning of definitions, the truth of theorems, and the application of mathematical techniques When ou encounter a sentence ou don t understand, read it again If it still doesn t make sense, read on anwa, as sometimes confusing sentences are eplained b later sentences You don t have to read ever equation The eamples generall show all the steps needed to solve a problem Sometimes reading through each step is helpful; sometimes it is confusing When the steps are illustrating a new technique, one probabl should follow each step closel to learn the new technique When the steps are showing the mathematics needed to find a number to be used later, one can usuall skip ahead and see how that number is being used, instead of getting bogged down in reading how the number was found Most proofs have been omitted In mathematics, proving something is alwas true is etremel important, and entails much more than testing to see if it works twice However, students often are confused b the details of a proof, or become concerned that the should have been able to construct this proof on their own To alleviate this potential problem, we do not include the proofs to most theorems in the tet The interested reader is highl encouraged to find proofs online or from their instructor In most cases, one is ver capable of understanding what a theorem means and how to appl it without knowing full wh it is true

7 AP E X Affordable Print and Electronic texts AP E X is a consortium of authors who collaborate to produce high qualit, low cost tetbooks The current tetbook writing paradigm is facing a potential revolution as desktop publishing and electronic formats increase in popularit However, writing a good tetbook is no eas task, as the time requirements alone are substantial It takes countless hours of work to produce tet, write eamples and eercises, edit and publish Through collaboration, however, the cost to an individual can be lessened, allowing us to create tets that we freel distribute electronicall and sell in printed form for an incredibl low cost Having said that, nothing is entirel free; someone alwas bears some cost This tet cost the authors of this book their time, and that was not enough APEX Calculus would not eist had not the Virginia Militar Institute, through a generous Jackson Hope grant, given the lead author significant time awa from teaching so he could focus on this tet Each tet is available as a free pdf, protected b a Creative Commons Attribution - Noncommercial 40 copright That means ou can give the pdf to anone ou like, print it in an form ou like, and even edit the original content and redistribute it If ou do the latter, ou must clearl reference this work and ou cannot sell our edited work for mone We encourage others to adapt this work to fit their own needs One might add sections that are missing or remove sections that our students won t need The source files can be found at githubcom/apexcalculus You can learn more at wwwvmiedu/apex

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9 : Limits Calculus means a method of calculation or reasoning When one computes the sales ta on a purchase, one emplos a simple calculus When one finds the area of a polgonal shape b breaking it up into a set of triangles, one is using another calculus Proving a theorem in geometr emplos et another calculus Despite the wonderful advances in mathematics that had taken place into the first half of the 7 th centur, mathematicians and scientists were keenl aware of what the could not do (This is true even toda) In particular, two important concepts eluded master b the great thinkers of that time: area and rates of change Area seems innocuous enough; areas of circles, rectangles, parallelograms, etc, are standard topics of stud for students toda just as the were then However, the areas of arbitrar shapes could not be computed, even if the boundar of the shape could be described eactl Rates of change were also important When an object moves at a constant rate of change, then distance = rate time But what if the rate is not constant can distance still be computed? Or, if distance is known, can we discover the rate of change? It turns out that these two concepts were related Two mathematicians, Sir Isaac Newton and Gottfried Leibniz, are credited with independentl formulating a sstem of computing that solved the above problems and showed how the were connected Their sstem of reasoning was a calculus However, as the power and importance of their discover took hold, it became known to man as the calculus Toda, we generall shorten this to discuss calculus The foundation of the calculus is the limit It is a tool to describe a particular behavior of a function This chapter begins our stud of the limit b approimating its value graphicall and numericall After a formal definition of the limit, properties are established that make finding limits tractable Once the limit is understood, then the problems of area and rates of change can be approached An Introduction To Limits We begin our stud of limits b considering eamples that demonstrate ke concepts that will be eplained as we progress Figure : sin()/ near = Consider the function = sin When is near the value, what value (if an) is near? While our question is not precisel formed (what constitutes near the value?), the answer does not seem difficult to find One might think first to look at a graph of this function to approimate the appropriate values Consider Figure, where = sin is graphed For values of near, it seems that takes on values near 085 In fact, when =, then = sin 084, so it makes sense that when is near, will be near 084 Consider this again at a different value for When is near 0, what value (if an) is near? B considering Figure, one can see that it seems that takes on values near But what happens when = 0? We have sin The epression 0/0 has no value; it is indeterminate Such an epression gives Figure : sin()/ near = 0

10 Chapter Limits no information about what is going on with the function nearb We cannot find out how behaves near = 0 for this function simpl b letting = 0 Finding a limit entails understanding how a function behaves near a particular value of Before continuing, it will be useful to establish some notation Let = f(); that is, let be a function of for some function f The epression the limit of as approaches describes a number, often referred to as L, that nears as nears We write all this as sin()/ Figure 3: Values of sin()/ with near sin()/ not defined Figure 4: Values of sin()/ with near lim = lim f() = L This is not a complete definition (that will come in the net section); this is a pseudo-definition that will allow us to eplore the idea of a limit Above, where f() = sin()/, we approimated sin sin lim 084 and lim 0 (We approimated these limits, hence used the smbol, since we are working with the pseudo-definition of a limit, not the actual definition) Once we have the true definition of a limit, we will find limits analticall; that is, eactl using a variet of mathematical tools For now, we will approimate limits both graphicall and numericall Graphing a function can provide a good approimation, though often not ver precise Numerical methods can provide a more accurate approimation We have alread approimated limits graphicall, so we now turn our attention to numerical approimations Consider again lim sin()/ To approimate this limit numericall, we can create a table of and f() values where is near This is done in Figure 3 Notice that for values of near, we have sin()/ near 084 The = row is in bold to highlight the fact that when considering limits, we are not concerned with the value of the function at that particular value; we are onl concerned with the values of the function when is near Now approimate lim 0 sin()/ numericall We alread approimated the value of this limit as graphicall in Figure The table in Figure 4 shows the value of sin()/ for values of near 0 Ten places after the decimal point are shown to highlight how close to the value of sin()/ gets as takes on values ver near 0 We include the = 0 row in bold again to stress that we are not concerned with the value of our function at = 0, onl on the behavior of the function near 0 This numerical method gives confidence to sa that is a good approimation of lim 0 sin()/; that is, lim sin()/ 0 Later we will be able to prove that the limit is eactl We now consider several eamples that allow us eplore different aspects of the limit concept Eample Approimating the value of a limit Use graphical and numerical methods to approimate lim Solution To graphicall approimate the limit, graph = ( 6)/( )

11 An Introduction To Limits on a small interval that contains 3 To numericall approimate the limit, create a table of values where the values are near 3 This is done in Figures 5 and 6, respectivel The graph shows that when is near 3, the value of is ver near 03 B considering values of near 3, we see that = 094 is a better approimation The graph and the table impl that lim This eample ma bring up a few questions about approimating limits (and the nature of limits themselves) If a graph does not produce as good an approimation as a table, wh bother with it? How man values of in a table are enough? In the previous eample, could we have just used = 300 and found a fine approimation? Graphs are useful since the give a visual understanding concerning the behavior of a function Sometimes a function ma act erraticall near certain values which is hard to discern numericall but ver plain graphicall Since graphing utilities are ver accessible, it makes sense to make proper use of them Since tables and graphs are used onl to approimate the value of a limit, there is not a firm answer to how man data points are enough Include enough so that a trend is clear, and use values (when possible) both less than and greater than the value in question In Eample, we used both values less than and greater than 3 Had we used just = 300, we might have been tempted to conclude that the limit had a value of 03 While this is not far off, we could do better Using values on both sides of 3 helps us identif trends Figure 5: Graphicall approimating a limit in Eample not defined Figure 6: Numericall approimating a limit in Eample 05 Eample Approimating the value of a limit Graphicall and numericall approimate the limit of f() as approaches 0, where { + < 0 f() = + > 0 Solution Again we graph f() and create a table of its values near = 0 to approimate the limit Note that this is a piecewise defined function, so it behaves differentl on either side of 0 Figure 7 shows a graph of f(), and on either side of 0 it seems the values approach Note that f(0) is not actuall defined, as indicated in the graph with the open circle The table shown in Figure 8 shows values of f() for values of near 0 It is clear that as takes on values ver near 0, f() takes on values ver near It turns out that if we let = 0 for either piece of f(), is returned; this is significant and we ll return to this idea later The graph and table allow us to sa that lim 0 f() ; in fact, we are probabl ver sure it equals Figure 7: Graphicall approimating a limit in Eample f() Figure 8: Numericall approimating a limit in Eample 3

12 Chapter Limits Figure 9: Observing no limit as in Eample 3 f() Figure 0: Values of f() near = in Eample Figure : Observing no limit as in Eample 4 f() Figure : Values of f() near = in Eample 4 Identifing When Limits Do Not Eist A function ma not have a limit for all values of That is, we cannot sa lim c f() = L for some numbers L for all values of c, for there ma not be a number that f() is approaching There are three was in which a limit ma fail to eist The function f() ma approach different values on either side of c The function ma grow without upper or lower bound as approaches c 3 The function ma oscillate as approaches c We ll eplore each of these in turn Eample 3 Different Values Approached From Left and Right Eplore wh lim f() does not eist, where { f() = + 3 > Solution A graph of f() around = and a table are given Figures 9 and 0, respectivel It is clear that as approaches, f() does not seem to approach a single number Instead, it seems as though f() approaches two different numbers When considering values of less than (approaching from the left), it seems that f() is approaching ; when considering values of greater than (approaching from the right), it seems that f() is approaching Recognizing this behavior is important; we ll stud this in greater depth later Right now, it suffices to sa that the limit does not eist since f() is not approaching one value as approaches Eample 4 The Function Grows Without Bound Eplore wh lim /( ) does not eist Solution A graph and table of f() = /( ) are given in Figures and, respectivel Both show that as approaches, f() grows larger and larger We can deduce this on our own, without the aid of the graph and table If is near, then ( ) is ver small, and: = ver large number ver small number Since f() is not approaching a single number, we conclude that does not eist lim ( ) Eample 5 The Function Oscillates Eplore wh lim sin(/) does not eist 0 Solution Two graphs of f() = sin(/) are given in Figures 3 Figure 3(a) shows f() on the interval [, ]; notice how f() seems to oscillate near = 0 One might think that despite the oscillation, as approaches 4

13 An Introduction To Limits 0, f() approaches 0 However, Figure 3(b) zooms in on sin(/), on the interval [ 0, 0] Here the oscillation is even more pronounced Finall, in the table in Figure 3(c), we see sin()/ evaluated for values of near 0 As approaches 0, f() does not appear to approach an value It can be shown that in realit, as approaches 0, sin(/) takes on all values between and infinitel man times! Because of this oscillation, lim 0 sin(/) does not eist (a) (b) (c) 0 sin(/) Figure 3: Observing that f() = sin(/) has no limit as 0 in Eample 5 Limits of Difference Quotients We have approimated limits of functions as approached a particular number We will consider another important kind of limit after eplaining a few ke ideas Let f() represent the position function, in feet, of some particle that is moving in a straight line, where is measured in seconds Let s sa that when =, the particle is at position 0 ft, and when = 5, the particle is at 0 ft Another wa of epressing this is to sa f() = 0 and f(5) = 0 Since the particle traveled 0 feet in 4 seconds, we can sa the particle s average velocit was 5 ft/s We write this calculation using a quotient of differences, or, a difference quotient: 0 0 f f(5) f() 5 = 0 4 = 5ft/s This difference quotient can be thought of as the familiar rise over run used to compute the slopes of lines In fact, that is essentiall what we are doing: given two points on the graph of f, we are finding the slope of the secant line through those two points See Figure 4 Now consider finding the average speed on another time interval We again start at =, but consider the position of the particle h seconds later That is, consider the positions of the particle when = and when = + h The difference quotient is now f( + h) f() ( + h) = f( + h) f() h 4 6 Figure 4: Interpreting a difference quotient as the slope of a secant line Let f() = 5 + 5; note that f() = 0 and f(5) = 0, as in our discussion We can compute this difference quotient for all values of h (even 5

14 Chapter Limits 0 0 f negative values!) ecept h = 0, for then we get 0/0, the indeterminate form introduced earlier For all values h 0, the difference quotient computes the average velocit of the particle over an interval of time of length h starting at = For small values of h, ie, values of h close to 0, we get average velocities over ver short time periods and compute secant lines over small intervals See Figure 5 This leads us to wonder what the limit of the difference quotient is as h approaches 0 That is, f (a) f (b) 4 6 (c) Figure 5: Secant lines of f() at = and = + h, for shrinking values of h (ie, h 0) f( + h) f() lim =? h 0 h As we do not et have a true definition of a limit nor an eact method for computing it, we settle for approimating the value While we could graph the difference quotient (where the -ais would represent h values and the -ais would represent values of the difference quotient) we settle for making a table See Figure 6 The table gives us reason to assume the value of the limit is about 85 Proper understanding of limits is ke to understanding calculus With limits, we can accomplish seemingl impossible mathematical things, like adding up an infinite number of numbers (and not get infinit) and finding the slope of a line between two points, where the two points are actuall the same point These are not just mathematical curiosities; the allow us to link position, velocit and acceleration together, connect cross-sectional areas to volume, find the work done b a variable force, and much more Unfortunatel, the precise definition of the limit, and most of the applications mentioned in the paragraph above, are beond what we can cover in this course Instead, we will settle for the following imprecise definition: Definition Informal Definition of the Limit Let I be an open interval containing c, and let f be a function defined on I, ecept possibl at c We sa that the limit of f(), as approaches c, is L, and write lim f() = L, c if we can make the value of f() arbitraril close to L b choosing c sufficientl close to c f(+h) f() h h The formal definition of the limit, which we will not discuss, makes precise the meaning of the phrases arbitraril close and sufficientl close The problem with the definition we have given is that, while it gives an intuitive understanding of the meaning of the limit, it s of no use for proving theorems about limits In the net section we will state (but not prove) several theorems about limits which will allow use to compute their values analticall, without recourse to tables of values Figure 6: The difference quotient evaluated at values of h near 0 6

15 Eercises Terms and Concepts In our own words, what does it mean to find the limit of f() as approaches 3? An epression of the form 0 is called 0 3 T/F: The limit of f() as approaches 5 is f(5) 4 Describe three situations where lim c f() does not eist 5 In our own words, what is a difference quotient? Problems In Eercises 6 6, approimate the given limits both numericall and graphicall 6 lim lim lim lim lim lim lim f(), where { + f() = 3 5 > 3 lim f(), where 3 { + 3 f() = + > 3 4 lim f(), where 0 { cos 0 f() = > 0 5 lim π/ f(), where { sin f() = cos π/ > π/ In Eercises 6 4, a function f and a value a are given Approimate the limit of the difference quotient, f(a + h) f(a) lim, using h = ±0, ±00 h 0 h 6 f() = 7 +, a = 3 7 f() = , a = 8 f() = + 3 7, a = 9 f() = +, a = 0 f() = 4 + 5, a = 3 f() = ln, a = 5 f() = sin, a = π 3 f() = cos, a = π 7

16 Chapter Limits Finding Limits Analticall In Section we eplored the concept of the limit without a strict denition, meaning we could onl make approimations Proving that these approimations are correct requires a rigorous denition of limits, which is beond the scope of this course Suppose that lim f() = and lim g() = 3 What is lim (f()+ g())? Intuition tells us that the limit should be 5, as we epect limits to behave in a nice wa The following theorem (whose proof requires the rigorous denition referred to above) states that alread established limits do behave nicel Theorem Basic Limit Properties Let b, c, L and K be real numbers, let n be a positive integer, and let f and g be functions with the following limits: The following limits hold lim f() = L and lim g() = K c c Constants: lim b = b c Identit lim = c c 3 Sums/Dierences: lim(f() ± g()) = L ± K c 4 Scalar Multiples: lim b f() = bl c 5 Products: lim f() g() = LK c 6 Quotients: lim f()/g() = L/K, (K 0) c 7 Powers: lim f() n = L n c 8 Roots: lim n f() = n L c 9 Compositions: Adjust our previousl given limit situation to: lim f() = L and lim g() = K c L Then lim c g(f()) = K We make a note about Propert #8: when n is even, L must be greater than 0 If n is odd, then the statement is true for all L We appl the theorem to an eample Eample 6 Let Using basic limit properties lim f() =, lim g() = 3 and p() = Find the following limits: lim ( f() + g() ) lim ( 5f() + g() ) 3 lim p() 8

17 Finding Limits Analticall Solution Using the Sum/Dierence rule, we know that lim ( f() + g() ) = + 3 = 5 Using ( the Scalar Multiple and Sum/Dierence rules, we nd that 5f() + g() ) = = 9 lim 3 Here we combine the Power, Scalar Multiple, Sum/Dierence and Constant Rules We show quite a few steps, but in general these can be omitted: lim p() = lim ( ) = lim 3 lim 5 + lim 7 = = 9 Part 3 of the previous eample demonstrates how the limit of a quadratic polnomial can be determined using the properties of Theorem Not onl that, recognize that lim p() = 9 = p(); ie, the limit at was found just b plugging into the function This holds true for all polnomials, and also for rational functions (which are quotients of polnomials), as stated in the following theorem Limits of Polnomial and Rational Func- Theorem tions Let p() and q() be polnomials and c a real number Then: lim c p() = p(c) p() lim c q() = p(c), where q(c) 0 q(c) Eample 7 Finding a limit of a rational function Using Theorem, nd lim Solution Using Theorem, we can quickl state that lim = 3( ) 5( ) + ( ) 4 ( ) + 3 = 9 3 = 3 Using approimations (or worse - the rigorous denition) to deal with a limit such as lim = 4 9

18 Chapter Limits can be annoing, since it seems fairl obvious The previous theorems state that man functions behave in such an obvious fashion, as demonstrated b the rational function in Eample 7 Polnomial and rational functions are not the onl functions to behave in such a predictable wa The following theorem gives a list of functions whose behavior is particularl nice in terms of limits In the net section, we will give a formal name to these functions that behave nicel Theorem 3 Special Limits Let c be a real number in the domain of the given function and let n be a positive integer The following limits hold: lim c sin = sin c lim c cos = cos c 3 lim tan = tan c c 7 lim a = a c (a > 0) c 8 lim c ln = ln c 9 lim c n = n c 4 lim c csc = csc c 5 lim c sec = sec c 6 lim c cot = cot c Eample 8 Evaluating limits analticall Evaluate the following limits lim π cos lim 3 (sec tan ) 3 lim cos sin π/ 4 lim e ln 5 lim 0 sin Solution This is a straightforward application of Theorem 3 lim π cos = cos π = We can approach this in at least two was First, b directl appling Theorem 3, we have: lim 3 (sec tan ) = sec 3 tan 3 Using the Pthagorean Theorem, this last epression is ; therefore lim 3 (sec tan ) = We can also use the Pthagorean Theorem from the start 0 lim 3 (sec tan ) = lim =, 3 using the Constant limit rule Either wa, we nd the limit is 3 Appling the Product limit rule of Theorem and Theorem 3 gives lim cos sin = cos(π/) sin(π/) = 0 = 0 π/

19 Finding Limits Analticall 4 Again, we can approach this in two was First, we can use the eponential/logarithmic identit that e ln = and evaluate lim e ln = lim = We can also use the Composition limit rule of Theorem Using Theorem 3, we have lim ln = ln = 0 Appling the Composition rule, lim eln = lim e = e 0 = 0 Both approaches are valid, giving the same result 5 We encountered this limit in Section Appling our theorems, we attempt to nd the limit as sin lim 0 sin This, of course, violates a condition of Theorem, as the limit of the denominator is not allowed to be 0 Therefore, we are still unable to evaluate this limit with tools we currentl have at hand The section could have been titled Using Known Limits to Find Unknown Limits B knowing certain limits of functions, we can nd limits involving sums, products, powers, etc, of these functions We further the development of such comparative tools with the Squeeze Theorem, a clever and intuitive wa to nd the value of some limits Before stating this theorem formall, suppose we have functions f, g and h where g alwas takes on values between f and h; that is, for all in an interval, f() g() h() If f and h have the same limit at c, and g is alwas squeezed between them, then g must have the same limit as well That is what the Squeeze Theorem states Theorem 4 Squeeze Theorem Let f, g and h be functions on an open interval I containing c such that for all in I, f() g() h() If then lim f() = L = lim h(), c c lim g() = L c It can take some work to gure out appropriate functions b which to squeeze the given function of which ou are tring to evaluate a limit However, that is generall the onl place work is necessar; the theorem makes the evaluating the limit part ver simple We use the Squeeze Theorem in the following eample to nall prove sin that lim 0 =

20 Chapter Limits Eample 9 Using the Squeeze Theorem Use the Squeeze Theorem to show that sin lim 0 = (, tan θ) (cos θ, sin θ) θ (, 0) Solution We begin b considering the unit circle Each point on the unit circle has coordinates (cos θ, sin θ) for some angle θ as shown in Figure 7 Using similar triangles, we can etend the line from the origin through the point to the point (, tan θ), as shown (Here we are assuming that 0 θ π/ Later we will show that we can also consider θ 0) Figure 7 shows three regions have been constructed in the rst quadrant, two triangles and a sector of a circle, which are also drawn below The area of the large triangle is tan θ; the area of the sector is θ/; the area of the triangle contained inside the sector is sin θ It is then clear from the diagram that Figure 7: The unit circle and related triangles tan θ sin θ θ θ θ tan θ Multipl all terms b sin θ, giving θ cos θ θ sin θ Taking reciprocals reverses the inequalities, giving sin θ cos θ sin θ θ (These inequalities hold for all values of θ near 0, even negative values, since cos( θ) = cos θ and sin( θ) = sin θ) Now take limits sin θ lim cos θ lim lim θ 0 θ 0 θ θ 0 sin θ cos 0 lim θ 0 θ sin θ lim θ 0 θ sin θ Clearl this means that lim = θ 0 θ Two notes about the previous eample are worth mentioning First, one might be discouraged b this application, thinking I would never have come up with that on m own This is too hard! Don't be discouraged;

21 Finding Limits Analticall within this tet we will guide ou in our use of the Squeeze Theorem As one gains mathematical maturit, clever proofs like this are easier and easier to create Second, this limit tells us more than just that as approaches 0, sin()/ approaches Both and sin are approaching 0, but the ratio of and sin approaches, meaning that the are approaching 0 in essentiall the same wa Another wa of viewing this is: for small, the functions = and = sin are essentiall indistinguishable We include this special limit, along with three others, in the following theorem Theorem 5 Special Limits lim 0 sin = cos lim = lim ( + ) = e 0 e 4 lim = 0 A short word on how to interpret the latter three limits We know that as goes to 0, cos goes to So, in the second limit, both the numerator and denominator are approaching 0 However, since the limit is 0, we can interpret this as saing that cos is approaching faster than is approaching 0 In the third limit, inside the parentheses we have an epression that is approaching (though never equaling ), and we know that raised to an power is still At the same time, the power is growing toward innit What happens to a number near raised to a ver large power? In this particular case, the result approaches Euler's number, e, approimatel 78 In the fourth limit, we see that as 0, e approaches just as fast as 0, resulting in a limit of Our nal theorem for this section will be motivated b the following eample Eample 0 Using algebra to evaluate a limit Evaluate the following limit: 3 lim Solution We begin b attempting to appl Theorem 3 and substituting for in the quotient This gives: Figure 8: Graphing f in Eample 0 to understand a limit lim = = 0 0 and indeterminate form We cannot appl the theorem B graphing the function, as in Figure 8, we see that the function seems to be linear, impling that the limit should be eas to evaluate, 3

22 Chapter Limits Recognize that the numerator of our quotient can be factored: ( )( + ) = The function is not dened when =, but for all other, = ( )( + ) = ( )( + ) = + Clearl lim + = Recall that when considering limits, we are not concerned with the value of the function at, onl the value the function approaches as approaches Since ( )/( ) and + are the same at all points ecept =, the both approach the same value as approaches Therefore we can conclude that lim = The ke to the above eample is that the functions = ( )/( ) and = + are identical ecept at = Since limits describe a value the function is approaching, not the value the function actuall attains, the limits of the two functions are alwas equal Theorem 6 Point Limits of Functions Equal At All But One Let g() = f() for all in an open interval, ecept possibl at c, and let lim c g() = L for some real number L Then lim f() = L c The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form g()/f() and directl evaluating the limit g() lim returns 0/0, then ( c) is a factor of both g() and f() One c f() can then use algebra to factor this term out, cancel, then appl Theorem 6 We demonstrate this once more Eample Evaluating a limit using Theorem Evaluate lim Solution We begin b appling Theorem 3 and substituting 3 for This returns the familiar indeterminate form of 0/0 Since the numerator and denominator are each polnomials, we know that ( 3) is factor of each Using whatever method is most comfortable to ou, factor out ( 3) from each (using polnomial division, snthetic division, a computer algebra sstem, etc) We nd that = ( 3)( + ) ( 3)( + 9 5) 4

23 Finding Limits Analticall We can cancel the ( 3) terms as long as 3 Using Theorem 6 we conclude: lim = lim 3 ( 3)( + ) ( 3)( + 9 5) = lim 3 ( + ) ( + 9 5) = 0 40 = 4 We end this section b revisiting a limit rst seen in Section, a limit of a dierence quotient Let f() = 5 + 5; we approimated the f( + h) f() limit lim 85 We formall evaluate this limit in the h 0 h following eample Eample Evaluating the limit of a dierence quotient Let f() = 5 f( + h) f() + 5; nd lim h 0 h Solution Since f is a polnomial, our rst attempt should be to emplo Theorem 3 and substitute 0 for h However, we see that this gives us 0/0 Knowing that we have a rational function hints that some algebra will help Consider the following steps: ( f( + h) f() 5( + h) + 5( + h) 5() + 5() ) lim = lim h 0 h h 0 h 5( + h + h ) h 0 = lim h 0 h 5h + 85h = lim h 0 h h( 5h + 85) = lim h 0 h = lim ( 5h + 85) (using Theorem 6, as h 0) h 0 = 85 (using Theorem 3) This matches our previous approimation This section contains several valuable tools for evaluating limits One of the main results of this section is Theorem 3; it states that man functions that we use regularl behave in a ver nice, predictable wa In the net section we give a name to this nice behavior; we label such functions as continuous Dening that term will require us to look again at what a limit is and what causes limits to not eist 5

24 Eercises Terms and Concepts Eplain in our own words, without using ɛ-δ formalit, wh lim c b = b Eplain in our own words, without using ɛ-δ formalit, wh lim c = c 3 What does the tet mean when it sas that certain functions behavior is nice in terms of limits? What, in particular, is nice? 4 Sketch a graph that visuall demonstrates the Squeeze Theorem 5 You are given the following information: (a) lim f() = 0 (b) lim g() = 0 (c) lim f()/g() = What can be said about the relative sizes of f() and g() as approaches? Problems Using: lim f() = 6 9 lim lim f() = 9 6 lim g() = 3 6 g() = 3 9 evaluate the limits given in Eercises 6 3, where possible If it is not possible to know, state so 6 lim 9 (f() + g()) 7 lim 9 (3f()/g()) ( ) f() g() 8 lim 9 g() ( ) f() 9 lim 6 3 g() 0 lim 9 g ( f() ) lim 6 f ( g() ) lim 6 g ( f(f()) ) 3 lim 6 f()g() f () + g () Using: lim f() = lim lim f() = 0 lim g() = π 0 g() = 0 evaluate the limits given in Eercises 4 7, where possible If it is not possible to know, state so 4 lim f() g() 5 lim 0 cos ( g() ) 6 lim f()g() 7 lim g ( 5f() ) In Eercises 8 3, evaluate the given limit 8 lim ( ) lim π 5 0 lim cos sin π/4 lim 0 ln lim lim csc π/6 4 lim 0 ln( + ) 5 lim π lim π lim lim lim lim lim lim 6 7 6

25 Use the Squeeze Theorem in Eercises 33 36, where appropriate, to evaluate the given limit ( ) 33 lim sin 0 ( ) 34 lim sin cos 0 35 lim f(), where 3 f() 3 36 lim 3 + f(), where 6 9 f() on [0, 3] Eercises challenge our understanding of limits but can be evaluated using the knowledge gained in this section 37 lim 0 sin 3 38 lim 0 sin lim 0 ln( + ) sin 40 lim, where is measured in degrees, not radians 0 7

26 Chapter Limits 3 One Sided Limits We introduced the concept of a limit gentl, approimating their values graphicall and numericall The previous section gave us tools (which we call theorems) that allow us to compute limits with greater ease Chief among the results were the facts that polnomials and rational, trigonometric, eponential and logarithmic functions (and their sums, products, etc) all behave nicel In this section we rigorousl dene what we mean b nicel In Section we eplored the three was in which limits of functions failed to eist: The function approached dierent values from the left and right, The function grows without bound, and 3 The function oscillates In this section we eplore in depth the concepts behind # b introducing the one-sided limit We begin with denitions that are ver similar to the denition of the limit given at the end of Section, but the notation is slightl dierent and c is replaced with either < c or > c Denition One Sided Limits Left-Hand Limit Let I be an open interval containing c, and let f be a function dened on I, ecept possibl at c We sa that limit of f(), as approaches c from the left, is L, or, the lefthand limit of f at c is L, and write lim f() = L, c if we can make the value of f() arbitraril close to L b choosing < c sucientl close to c Right-Hand Limit Let I be an open interval containing c, and let f be a function dened on I, ecept possibl at c We sa that the limit of f(), as approaches c from the right, is L, or, the righthand limit of f at c is L, and write lim f() = L, c + if we can make the value of f() sucientl close to L b choosing > c sucientl close to c 8 Practicall speaking, when evaluating a left-hand limit, we consider onl values of to the left of c, ie, where < c The admittedl imperfect notation c is used to impl that we look at values of to the left of c The notation has nothing to do with positive or negative values of either or c A similar statement holds for evaluating righthand limits; there we consider onl values of to the right of c, ie, > c We can use the theorems from previous sections to help us evaluate these limits; we just restrict our view to one side of c

27 3 One Sided Limits We practice evaluating left and right-hand limits through a series of eamples Eample 3 { Evaluating one sided limits 0 Let f() =, as shown in Figure 9 Find each of 3 < < the following: lim f() 5 lim 0 + f() lim + f() 3 lim f() 4 f() 6 f(0) 7 lim f() 8 f() Solution For these problems, the visual aid of the graph is likel more eective in evaluating the limits than using f itself Therefore we will refer often to the graph As goes to from the left, we see that f() is approaching the value of Therefore lim f() = Figure 9: A graph of f in Eample 3 As goes to from the right, we see that f() is approaching the value of Recall that it does not matter that there is an open circle there; we are evaluating a limit, not the value of the function Therefore lim f() = + 3 The limit of f as approaches does not eist, as discussed in the rst section The function does not approach one particular value, but two dierent values from the left and the right 4 Using the denition and b looking at the graph we see that f() = 5 As goes to 0 from the right, we see that f() is also approaching 0 Therefore lim f() = 0 Note we cannot consider a left-hand 0 + limit at 0 as f is not dened for values of < 0 6 Using the denition and the graph, f(0) = 0 7 As goes to from the left, we see that f() is approaching the value of Therefore lim f() = 8 The graph and the denition of the function show that f() is not dened Note how the left and right-hand limits were dierent at = This, of course, causes the limit to not eist The following theorem states what is fairl intuitive: the limit eists precisel when the left and right-hand limits are equal 9

28 Chapter Limits Theorem 7 Limits and One Sided Limits Let f be a function dened on an open interval I containing c Then lim c f() = L if, and onl if, lim c c f() = L and lim f() = L + The phrase if, and onl if means the two statements are equivalent: the are either both true or both false If the limit equals L, then the left and right hand limits both equal L If the limit is not equal to L, then at least one of the left and right-hand limits is not equal to L (it ma not even eist) One thing to consider in Eamples 3 6 is that the value of the function ma/ma not be equal to the value(s) of its left/right-hand limits, even when these limits agree Eample 4 { Evaluating limits of a piecewisedened function 0 < < Let f() = ( ), as shown in Figure 0 Evaluate < < the following lim f() lim + f() 3 lim f() 4 f() 5 lim 0 + f() 6 f(0) 7 lim f() 8 f() Solution f and its graph Again we will evaluate each using both the denition of Figure 0: A graph of f from Eample 4 As approaches from the left, we see that f() approaches Therefore lim f() = As approaches from the right, we see that again f() approaches Therefore lim f() = + 3 The limit of f as approaches eists and is, as f approaches from both the right and left Therefore lim f() = 4 f() is not dened Note that is not in the domain of f as dened b the problem, which is indicated on the graph b an open circle when = 5 As goes to 0 from the right, f() approaches So lim 0 + f() = 6 f(0) is not dened as 0 is not in the domain of f 7 As goes to from the left, f() approaches 0 So lim f() = 0 0

29 3 One Sided Limits 8 f() is not dened as is not in the domain of f Eample 5 { Evaluating limits of a piecewisedened function ( ) 0, Let f() =, as shown in Figure Evaluate the = following 05 lim f() lim + f() 3 lim f() 4 f() Solution It is clear b looking at the graph that both the left and right-hand limits of f, as approaches, is 0 Thus it is also clear that the limit is 0; ie, lim f() = 0 It is also clearl stated that f() = Figure : Graphing f in Eample 5 Eample 6 { Evaluating limits of a piecewisedened function 0 Let f() =, as shown in Figure Evaluate the < following lim f() lim + f() 3 lim f() 4 f() Solution It is clear from the denition of the function and its graph that all of the following are equal: lim f() = lim f() = lim f() = f() = + 05 Figure : Graphing f in Eample 6 In Eamples 3 6 we were asked to nd both lim f() and f() Consider the following table: lim f() Eample 3 does not eist Eample 4 not dened Eample 5 0 Eample 6 Onl in Eample 6 do both the function and the limit eist and agree This seems nice; in fact, it seems normal This is in fact an important situation which we eplore in the net section, entitled Continuit In short, a continuous function is one in which when a function approaches a value as c (ie, when lim f() = L), it actuall attains that value c at c Such functions behave nicel as the are ver predictable

30 Eercises 3 Terms and Concepts What are the three was in which a limit ma fail to eist? 7 5 T/F: If lim f() = 5, then lim f() = T/F: If lim f() = 5, then lim f() = T/F: If lim f() = 5, then lim f() = 5 (a) (b) lim f() lim f() + (c) lim f() (d) f() (e) (f) lim f() lim f() 0 + Problems 5 In Eercises 5, evaluate each epression using the given graph of f() (a) (b) lim f() lim f() + (c) lim f() (d) f() 5 05 (a) (b) lim f() lim f() + (c) lim f() 05 5 (d) f() (e) (f) lim f() 0 lim f() (a) (b) 5 05 lim f() lim f() (c) lim f() (d) f() (a) (b) lim f() lim + f() (c) lim f() (d) f() (e) (f) lim f() lim + f() (a) (b) lim 0 f() lim f() (c) lim 0 f() (d) f(0)

31 (a) (b) (c) 4 3 lim f() lim f() + lim f() (d) f( ) Let 3 a 3 be an integer (a) (b) lim f() a lim f() a + 4 (e) (f) 3 4 lim f() lim f() + (g) lim f() (h) f() 3 4 (c) lim a f() (d) f(a) In Eercises 3, evaluate the given limits of the piecewise defined functions f 3 f() = (a) (b) 4 f() = (a) (b) { + 5 > lim f() lim f() + (c) lim f() (d) f() { + 5 < 0 sin 0 lim f() 0 lim f() 0 + (c) lim 0 f() (d) f(0) < 5 f() = > (a) (b) (c) lim f() lim f() + lim f() (d) f( ) (e) (f) lim f() lim f() + (g) lim f() (h) f() { cos < π 6 f() = sin π (a) (b) lim f() π lim f() π + (c) lim π f() (d) f(π) { cos < a 7 f() = sin a, where a is a real number (a) (b) lim f() a lim f() a + + < 8 f() = = > (a) (b) lim f() lim f() + (c) lim a f() (d) f(a) (c) lim f() (d) f() < 9 f() = + = > (a) (b) lim f() lim f() + (c) lim f() (d) f() { a( b) + c < b 0 f() = a( b) + c b, where a, b and c are real numbers (a) (b) f() = (a) (b) Review lim f() b lim b + f() { 0 0 = 0 lim f() 0 lim f() 0 + (c) lim b f() (d) f(b) (c) lim 0 f() (d) f(0) Evaluate the limit: lim Evaluate the limit: lim Evaluate the limit: lim Evaluate the limit: lim 3 3

32 Chapter Limits 4 Continuit As we have studied limits, we have gained the intuition that limits measure where a function is heading That is, if lim f() = 3, then as is close to, f() is close to 3 We have seen, though, that this is not necessaril a good indicator of what f() actuall this This can be problematic; functions can tend to one value but attain another This section focuses on functions that do not ehibit such behavior Denition 3 Continuous Function Let f be a function dened on an open interval I containing c f is continuous at c if lim c f() = f(c) f is continuous on I if f is continuous at c for all values of c in I If f is continuous on (, ), we sa f is continuous everwhere 5 A useful wa to establish whether or not a function f is continuous at c is to verif the following three things: lim c f() eists, f(c) is dened, and 3 lim c f() = f(c) Eample 7 Finding intervals of continuit Let f be dened as shown in Figure 3 Give the interval(s) on which f is continuous 05 We proceed b eamining the three criteria for continuit Solution 3 Figure 3: A graph of f in Eample 7 The limits lim c f() eists for all c between 0 and 3 f(c) is dened for all c between 0 and 3, ecept for c = We know immediatel that f cannot be continuous at = 3 The limit lim c f() = f(c) for all c between 0 and 3, ecept, of course, for c = Figure 4: A graph of the step function in Eample 8 We conclude that f is continuous at ever point of (0, 3) ecept at = Therefore f is continuous on (0, ) (, 3) Eample 8 Finding intervals of continuit The oor function, f() =, returns the largest integer smaller than the input (For eample, f(π) = π = 3) The graph of f in Figure 4 demonstrates wh this is often called a step function Give the intervals on which f is continuous Solution We eamine the three criteria for continuit 4

33 4 Continuit The limits lim c f() do not eist at the jumps from one step to the net, which occur at all integer values of c Therefore the limits eist for all c ecept when c is an integer The function is dened for all values of c 3 The limit lim c f() = f(c) for all values of c where the limit eist, since each step consists of just a line We conclude that f is continuous everwhere ecept at integer values of c So the intervals on which f is continuous are, (, ), (, 0), (0, ), (, ), Our denition of continuit on an interval species the interval is an open interval We can etend the denition of continuit to closed intervals b considering the appropriate one-sided limits at the endpoints Denition 4 Continuit on Closed Intervals Let f be dened on the closed interval [a, b] for some real numbers a, b f is continuous on [a, b] if: f is continuous on (a, b), lim f() = f(a) and a + 3 lim b f() = f(b) We can make the appropriate adjustments to talk about continuit on halfopen intervals such as [a, b) or (a, b] if necessar Using this new denition, we can adjust our answer in Eample 8 b stating that the oor function is continuous on the following halfopen intervals, [, ), [, 0), [0, ), [, ), This can tempt us to conclude that f is continuous everwhere; after all, if f is continuous on [0, ) and [, ), isn't f also continuous on [0, )? Of course, the answer is no, and the graph of the oor function immediatel conrms this Continuous functions are important as the behave in a predictable fashion: functions attain the value the approach Because continuit is so important, most of the functions ou have likel seen in the past are continuous on their domains This is demonstrated in the following eample where we eamine the intervals of continuit of a variet of common functions Eample 9 Determining intervals on which a function is continuous For each of the following functions, give the domain of the function and the interval(s) on which it is continuous 5