MATH CALCULUS FALL Scientia Imperii Decus et Tutamen 1

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1 MATH CALCULUS FALL 2018 Scientia Imperii Decus et Tutamen 1 Robert R. Kallman University of North Texas Department of Mathematics 1155 Union Circle Denton, Texas office telephone: office fax: office robert.kallman@unt.edu July 15, Taken from the coat of arms of Imperial College London.

2 FALL 2018 COURSE: MATH , CALCULUS I. PREREQUISITES: MATH 1650 (Pre-Calculus) or {MATH 1600 (Trigonometry) and MATH 1610 (Functions, Graphs and Applications)}. CLASS MEETS: We have GAB 461, MWF, 10:00 a.m. - 10:50 a.m. and GAB 461, TR, 9:30 a.m. - 10:50 a.m. reserved for our use. In despite of the fact that this is a 4 hour class, initially we will meet MTWRF, 10:00 a.m. - 10:50 a.m. in the assigned classroom. An exception to this is that tests will be given on Tuesday or Thursday, 9:30 a.m. - 10:50 a.m. This will give us some flexibility if we need to cancel some future classes and/or if we become ill. Later on these meeting days and times might well be modified. FINAL EXAM DATE AND TIME: The final is scheduled for Saturday, December 8, 2018 in GAB 461, 8:00 a.m. - 10:00 a.m. TEXT: James Stewart, Calculus, Eighth Edition, Cengage Learning, Boston, MA 02210, copyright 2016, 2012, ISBN INSTRUCTOR: Robert R. Kallman, 315 GAB [office], [office telephone], [fax], kallman@unt.edu [ ] OFFICE HOURS: Monday - Friday, 8:00 a.m. - 9:30 a.m. & Monday, Wednesday, 11:00 a.m. - 12:50 p.m. and before and after any class ATTENDANCE POLICY: Mandatory. Specifically for TAMS students: if you are absent for any reason, you are required to file an absence report with Dr. Fleming of the TAMS Academic Office. ELECTRONIC DEVICES: No electronic devices of any sort are to be on let alone used during the class. Repeated flouting of this will result in a grade penalty. HOMEWORK: Homework will be assigned and some designated subset of it will be graded. The designated homework assigned during a given week are to be handed in at the beginning of class on the Wednesday of the following week. Late homeworks will not be accepted under any circumstances. Each homework problem will receive a grade of 0, 1, or 2 points. Failure to turn in a homework set on time will result in a grade of -1 for that homework set. Take the homework seriously, for your performance on the homework will constitute 40 ACADEMIC INTEGRITY: There is no reason why TAMS students should not demonstrate complete academic integrity at all times, particularly on tests. Any transgression of this will result in a grade of zero on the test and a grade of F for the course. Consistent with this policy the instructor will retain xerox copies of a random sampling of all tests. GRADING POLICY: Grades will be based on the total number of points accrued from the assigned graded homeworks, from three in class 80 minute examinations (5 problems plus 1 bonus question), given circa late September, late October and late November, and from an in class 120 minute final (8 problems plus 1 bonus question). I will retain xerox copies of a sample of these tests. The number of points per test problem and final problem will normally be 10. There is no excuse for missing a test and no makeup tests will be given under any circumstances. A student missing a test will receive a grade of -1 on that test. If a 1

3 student is unavoidably absent from a test and makes arrangements with the instructor well before the test date, then the grade assigned to the missing test will be prorated by the student s performance on the final examination minus 10 points. It is difficult a priori to determine the precise break points for the final grades. However, the golden rule in determining the final assigned grade is that if the number of points earned by person A is to the number of points earned by person B, then person A has a grade which is to the grade of person B. The only possible exception is that you must take the final examination and receive a passing grade on the final in order to get a grade greater than F. TOPICS: The topics to be covered can be found in most of Chapter 1 - Chapter 5 plus a few extra topics. It is a rather ambitious goal to cover these topics in some depth. This will require considerable work on the part of the students and the instructor. Some supplementary notes will be handed out. At 5 class days per week we will have 71 class days (plus a reading day and a final day) to cover 42 sections. This should give us plenty of slack for three tests and lots of time for review if we stick to 5 classes per week. APPROXIMATE ITINERARY: The following is a first attempt, very rough approximation to what our schedule will be. This will perhaps be dynamically reconfigured as the semester progresses since it is of course impossible to make such a schedule with hard-and-fast rules, particularly since we are starting the academic year relatively late and ending so early. Test #1, Thursday, October 4, 2018, covering Chapters 1-2. Test #2, Thursday, November 1, 2018, covering Chapters 3-4. Test #3, Tuesday, November 27, 2018, covering Chapter 5. Final, Saturday, December 8, 2018, cumulative, Chapter 1-5. ASK QUESTIONS in class so that we may all benefit. If you need help, it is your responsibility to seek me out. See me during my office hours. Empirical evidence suggests that there is a strong correlation between the amount of work done by the student and his/her final grade. STUDENTS WITH DISABILITIES: It is the responsibility of students with certified disabilities to provide the instructor with appropriate documentation from the Dean of Students Office. STUDENT BEHAVIOR IN THE CLASSROOM: The Powers That Be have strongly suggested that students be given the following statement: Student behavior that interferes with an instructor s ability to conduct a class or other students opportunity to learn is unacceptable and disruptive and will not be tolerated in any instructional forum at UNT. Students engaging in unacceptable behavior will be directed to leave the classroom and the instructor may refer the student to the Center for Student Rights and Responsibilities to consider whether the student s conduct violated the Code of Student Conduct. The university s expectations for student conduct apply to all instructional forums, including university and electronic classroom, labs, discussion groups, field trips, etc. The Code of Student Conduct can be found at 2

4 In other words, cause trouble in the classroom and you will probably be cast into the Darkness and sent to the KGB. 3

5 How to Study for This Class Attend every class. Pay attention in class, take careful notes, and ask questions if needed. The evening of every class go over your classroom notes, list topics on which you have questions or need clarification, read the relevant section of the textbook, do the assigned homework to be graded, look over the additional homeworks to verify that you understand how to do them and make note of those additional homeworks that you do not understand to ask about them during the next class. It is important that you put a great deal of effort into the homework, both those to be turned in and those that are less formally assigned. One becomes adroit at any human activity - e.g., hitting a fast ball, throwing a slider, making foul shots or jump shots, or driving off a tee - only with a great deal of practice. The same applies to calculus. Do not waste your time memorizing endless lists of derivatives and antiderivatives. This in fact is counterproductive. Instead, know a few basic computational techniques (e.g., product rule for differentiation, chain rule for differentiation, sin = cos, etc.), and try to understand the big picture and concepts involved in problem solving. All of the problems encountered in this class should be first approached by asking oneself what is a reasonable way to proceed. Then given the proper path or direction, you can then solve the problems by small, logical steps that inevitably lead one to the final solution. 4

6 Science is the belief in the ignorance of experts. Richard Phillips Feynman (5/11/1918-2/15/1988) 5

7 Symbols. = for all or for every, = there exists, = in or an element of, / = not in or not an element of, = such that, = = implied by, = = implies, = if and only. Naive Set Theory. We should be familiar with the primitive notion of set and x A, A = {x S(x)} or A = {a 1, a 2,... a n } (note {x} = {x, x}), the empty set, A B or B A is equivalent to x A = x B, A = B is equivalent to A B and B A which is equivalent to x A x B. A B = {x x A or x B}, A B = {x x A and x B}, A B = {x x A and x / B}, A B = (A B) (B A), the symmetric difference of A and B, if A U, where U is the universal set under discourse, then A c = U A and A B = A B c. We have (A c ) c = A and therefore A = B if and only if A c = B c. We also have (A B) c = A c B c and (A B) c = A c B c. We have the commutative laws A B = B A and A B = B A, the associative laws (A B) C = A (B C) and (A B) C = A (B C), and the distributive laws A (B C) = (A B) (A C) and A (B C) = (A B) (A C). An elementary but quite hard fact to prove is that the symmetric difference operation is associative, viz., (A B) C = A (B C). I ll be quite impressed if you can prove this. Proposition 1. Prove that A = A, A B = B A, A c B c = A B, (A B) c = A c B = A B c, A (B C) = (A B) (A C), and (A B) C = A (B C). This proves that P(U) is a commutative ring with identity with = and + =. A B = (A B c ) (B A c ). From this formula A = A, A A =, and A B = B A follow easily. A c B c = (A c B cc ) (B c A cc ) = (A B c ) (B A c ) = A B. (A B) c = ((A B c ) (B A c )) c = (A B c ) c (B A c ) c = (A c B) (B c A) = (A c B c ) (A c A) (B B c ) (B A) = (A B cc ) (B c A c ) = A B c. Therefore (A B) c = (B A) c = B A c = A c B. A (B C) = A ((B C c ) (C B c )) = (A B C c ) (A C B c )) = ((A B) (A c C c )) ((A C) (A c B c )) = ((A B) (A C) c ) ((A C) (A B) c )) = (A B) (A C). (A B) C = ((A B) C c ) (C (A B) c ) = ((A B) C c ) (C (A B c )) = (((A B c ) (B A c )) C c ) (((A B) (B c A c )) C) = (A B c C c ) (A c B C c ) (A B C) (A c B c C). This last expression is obviously invariant under permutations of A, B and C. Hence, (A B) C = (C B) A = (B C) A = A (B C). Note that A + = A = A so is an additive identity. Note that A + A = A A =, so A is its own additive inverse. Finally if U is the universal set lurking in the background with respect to which we are taking complements, then A U = A U = A and thus U is a multiplicative identity. It is now simple to combine these observations together with the set theoretic identities already deduced to conclude that P(U) is a commutative ring with identity with * = and + =. Ordered pairs (a, b) = {{a}, {a, b}}. Try and prove the following lemma on your own. 6

8 Lemma 2. (a, b) = (c, d) if and only if a = c and b = d. If a = c and b = d, then certainly (a, b) = {{a}, {a, b}} = {{c}, {c, d}} = (c, d). Conversely, suppose that {{a}, {a, b}} = (a, b) = (c, d) = {{c}, {c, d}}. First, suppose that a = b. Then {{a}} = {{a}, {a}} = {{a}, {a, a}} = {{a}, {a, b}} = {{c}, {c, d}}, {c} {{a}} = {c} = {a} = c {a} = c = a, {c, d} {{a}} = {c, d} = {a} = d {a} = d = a, so a = b = c = d and the exercise is proved in this case. A symmetric argument shows the exercise follows if c = d. Next, suppose that a b and c d and {{a}, {a, b}} = {{c}, {c, d}} = {a} {{c}, {c, d}} = {a} = {c} or {a} = {c, d}. If {a} = {c, d}, then c {a} = c = a and d {a} = d = a, a contradiction since c d. Thus {a} = {c} = c {a} = c = a. {c, d} {{a}, {a, b}} = {c, d} = {a} or {c, d} = {a, b}. If {c, d} = {a} = d {a} = d = a = c, a contradiction. Hence, {c, d} = {a, b} = d {a, b} = d = a or d = b. If d = a = d = a = c, a contradiction. Therefore d = b. Note that {a, b} (a, b) - this is strange. Note also that (a, b) {a, b}. a is called the first coordinate of (a, b) and b is called the second coordinate of (a, b). The Cartesian product of the sets X and Y, denoted X Y, is defined to be {(x, y) x X and y Y }. Motivate with pictures. Informally, given sets X and Y, a function from X into Y is a rule that assigns to each element x X a unique y Y. In this case X is called the domain of the function. Functions are usually denoted by letters such as f, g, h, F, G, etc. f : x y = f(x). The set f(x) = {y Y y = f(x) for some x X} Y is called the range of f, denoted range(f). More formally, a function (or map or mapping) f from X to Y, denoted f : X Y, is a subset f X Y such that for each x X there is some y Y such that (x, y) f and with the additional property that (x, y 1 ) f and (x, y 2 ) f = y 1 = y 2. In other words, for every x X there is a unique y Y such that (x, y) f. In this circumstance y is said to be the value of f at x X and the notation f(x) = y is equivalent to (x, y) f. There is nothing special about the symbol x, any symbol would do, but x is easier to write than a picture of a tyrannosaurus rex. Functions are therefore identified with their graphs. Recall that f : X Y is into (injective, injection, one-to-one) if f(x 1 ) = f(x 2 ) = x 1 = x 2 and f : X Y is onto (surjective, surjection) if range(f) = Y, i.e., for every y Y there is some x X such that f(x) = y (x, y) f, f is a bijection if f is one-to-one and onto. If so, then f 1 = {(y, x) Y X (x, y) f} is a function, f 1 : Y X. Check the identities f f 1 : Y Y is the identity and f 1 f : X X is the identity. If f : X Y is a function and = A X, the restriction of f to A, denoted by f A, f A : A Y, is defined to be f (A Y ). Thus if a A, then (f A)(a) = f(a). Notice that if f : X Y and g : X Y are two functions and f g, then f = g. Composition of functions. If g : X Y and f : Y Z, then the composition of f with g, denoted f g, is defined by (f g)(x) = f(g(x)) Z, so f g : X Z. Representing functions, polynomial functions, even functions, odd functions, ratinal functions, algebraic functions. For this section, there will be no formal homework assigned. Instead read Section 1.1, make sure you understand compositions, look over the chapter s homeworks and let us know if there is anything interesting there. Nonformal homework: look over problems # 62 and # 63 and volunteer to present on the board. 7

9 Polynomials, rational functions with a suitable domain, even functions, odd functions, f + g, fg, f/g with a suitable domain, functions defined by tables (as might arise in physics, engineering, economics, experiments). No exponentials, no logarithms. For this section, there will be no formal homework assigned. Instead read Section 1.2, look over the chapter s homeworks and let us know if there is anything interesting there. 8

10 Trig Functions. Recall the definition of angles, the definition of the six basic trig functions, the Pythagorean Theorem, the Law of Cosines in both the acute and obtuse angle cases, basic identities, the addition formulas, double angle formulas, half angle formulas, addition formulas for tangent, even and odd. In what follows Z is the set of integers. cos(0) = 1 and sin(0) = 0 cos(π/2) = 0 and sin(π/2) = 1 cos(π) = 1 and sin(π) = 0 cos(3π/2) = 0 and sin(3π/2) = 1 cos(2π) = 1 and sin(2π) = 0 1 = cos 2 (A) + sin 2 (A) 1 + tan 2 (A) = sec 2 (A) 1 + cot 2 (A) = csc 2 (A) cos(a + 2πn) = cos(a) for all n Z sin(a + 2πn) = sin(a) for all n Z cos( A) = cos(a) sin( A) = sin(a) tan( A) = tan(a) cos(π/2 A) = sin(a) sin(π/2 A) = cos(a) cos(a ± B) = cos(a) cos(b) sin(a) sin(b) sin(a ± B) = sin(a) cos(b) ± cos(a) sin(b) cos(a + π) = cos(a) and cos(π A) = cos(a) sin(a + π) = sin(a) and sin(π A) = sin(a) 9

11 tan(a + nπ) = tan(a) for all n Z sin(2a) = 2 sin(a) cos(a) cos(2a) = cos 2 (A) sin 2 (A) = 2 cos 2 (A) 1 = 1 2 sin 2 (A) cos 2 (A) = 1+cos(2A) 2 sin 2 (A) = 1 cos(2a) 2 tan 2 (A) = 1 cos(2a) 1+cos(2A) cos 2 (A/2) = 1+cos(A) 2 sin 2 (A/2) = 1 cos(a) 2 tan 2 (A/2) = 1 cos(a) 1+cos(A) tan(a + B) = tan(a)+tan(b) 1 tan(a) tan(b) tan(a B) = tan(a) tan(b) 1+tan(A) tan(b) tan(2a) = 2 tan(a) 1 tan 2 (A) sin(a + B) + sin(a B) = 2 sin(a) cos(b) sin(a + B) sin(a B) = 2 cos(a) sin(b) cos(a + B) + cos(a B) = 2 cos(a) cos(b) cos(a B) cos(a + B) = 2 sin(a) sin(b) sin(a) + sin(b) = 2 sin( 1 2 (A + B)) cos( 1 2 (A B)) sin(a) sin(b) = 2 sin( 1 2 (A B)) cos( 1 2 (A + B)) cos(a) + cos(b) = 2 cos( 1 2 (A + B)) cos( 1 2 (A B)) cos(a) cos(b) = 2 sin( 1 2 (A + B)) sin( 1 2 (A B)) sin(2a) = 2 tan(a) 1+tan 2 (A) cos(2a) = 1 tan2 (A) 1+tan 2 (A) Use the addition laws for the tan function to prove: π/4 = arctan(1/2) + arctan(1/3) 10

12 π = arctan(1) + arctan(2) + arctan(3) π/4 = 4 arctan(1/5) arctan(1/239) Exercise 1. Graded: #1: Compute cos(π/12); #2: Compute tan(3π/8) These homeworks are due at the beginning of class on Wednesday, September 5, Remember that homeworks make up circa 40% of your grade. Look over the homeworks from Section 1.3 and let us know if there is anything interesting there. 11

13 Limits arise naturally. Simple examples, f(x) = x 2, f(t) = t 3. Exercise 2. Graded: Problem #5, page 50. These homeworks are due at the beginning of class on Wednesday, September 5, Remember that homeworks make up circa 40% of your grade. Look over the homeworks from Section 1.4 and let us know if there is anything interesting there. 12

14 Average velocity of a particle on the line, the instantaneous velocity. Difference quotients and the slope of a tangent line. Let f be a function defined on some open interval containing c but not necessarily at c itself. The number L is said to be the limit of f as x approaches c, denoted, lim x c f(x) = L, means that the function values f(x) approach L arbitrarily closely as x approaches c, or equivalently, f(x) is close to L whenever x is close to c but x c. Examples: f(x) = 4x + 5, c = 2 = L = 13. c = 3 = L = 7 f(x) = 2x 2 3x + 5, c = 1 = L = 10 f(x) = x3 2x+4 x 2 +1, c = 3 = L = 25/10 = 5/2 If a continuous graph, just slide along the graph f(x) = x2 9 x 3, c = 3 = L = 6 by factoring f(x) = x3 8 x 2, c = 2 = L = 12 by factoring f(x) = 3x 4 if x 0 and f(0) = 10, c = 0 = L = 4 One-sided limits f(x) = x /x for x 0, c = 0 = L = 1 and L+ = 1 f(x) = tan(x), c = π/2 = L = + and L+ = f(x) = sin(1/x) for x 0, c = 0 = L+, L and L do not exist f(x) = sin(x)/x for x 0, c = 0 = L = 1 f(x) = x2 1 2x+2 2, c = 1 = L = 4 f(x) = x sin(x) x 3, c = 0 and numerically estimate L = 1/6 Iif A R, define the characteristic function of A be defined by χ A (x) = 1 if x A and χ A (x) = 0 otherwise. Then lim x c χ Q (x) does not exist for any c R Exercise 3. Graded: x #1 compute lim 3 1 x 1 x 1 ; #2 compute lim x x 1 x 1 Look over the problems at the end of sections 1.5 and let us know if any seem of interest or are particularly 13

15 difficult. These homeworks are due at the beginning of class on Wednesday, September 5, Remember that homeworks make up circa 40% of your grade. 14

16 Definition 3. For each x R we define the absolute value of x to be the number x given by x = x if x 0 and x = x if x 0. This makes sense because = 0 = 0 = 0. Also note that x 0 for all x R, 1 = 1 and 0 = 0. Lemma 4. If x R, x = x. If x 0, then x = x = ( x) = x. If x < 0, then x = x = x. Lemma 5. If a, b R, then a b if and only if b a b if and only if a b and a b. a a a. In particular, = : 0 a b. If a 0, then b 0 a = a b. If a > 0, then b 0 < a = a b = b a b. = : If a 0, then a = a b. If a < 0, then b a = a b = a b. Finally, a b and a b = 0 = a + ( a) 2b = b 0. Either a or a is a, so a b and a b = a b. On the other hand, b a b = a b and a b. Theorem 6. For all x, y R we have (i) xy = x y ; (ii) x + y x + y ; (iii) x y x y. x 0, y 0 = xy 0 and xy = xy = x y. If x < 0, y 0 = xy 0 and xy = (xy) = ( x)y = x y. If x 0, y < 0, then xy 0 and xy = (xy) = x( y) = x y. If x < 0 and y < 0, then xy > 0 and xy = xy = ( x)( y) = x y. This proves (i). x x x and y y y = ( x + y ) x + y x + y = x + y x + y. This proves (ii). x = (x y) + y x y + y = x y x y. Interchanging x and y we obtain that ( x y ) = y x y x = (y x) = x y. These two inequalities imply that x y x + y. This proves (iii). Definition 7. Suppose that the function f is defined for all x near a except possibly at a. We say that the limit of f(x) as x approaches a equals L and we write lim x a f(x) = L if, for every ϵ > 0, there exists a δ = δ(ϵ) > 0 such that 0 < x a < δ = f(x) L < ϵ. If f is defined for all x near a with x > a, we say that the limit of f(x) as x approaches a from the right equals L+ if, for every ϵ > 0, there exists a δ = δ(ϵ) > 0 such that 0 < x a < δ and x > a = f(x) (L+) < ϵ. 15

17 If f is defined for all x near a with x < a, we say that the limit of f(x) as x approaches a from the left equals L if, for every ϵ > 0, there exists a δ = δ(ϵ) > 0 such that 0 < x a < δ and x < a = f(x) (L ) < ϵ. We also write f(x) L as x a, f(x) L as x a and f(x) L+ as x a+. Example 8. lim x a (2x 1) = 2a 1 Let f(x) = 2x 1, ϵ > 0 and δ = ϵ/2. Then 0 < x a < δ = f(x) (2a 1) = (2x 1) (2a 1) = 2 x a < 2δ = ϵ. Example 9. lim x 1 (2 3x) = 5 Let f(x) = 2 3x, ϵ > 0 and δ = ϵ/3. Then 0 < x+1 = x ( 1) < δ = f(x) 5 = (2 3x) 5 = 3 x + 1 < 3δ = ϵ. Example 10. lim x a x = a Let f(x) = x, ϵ > 0 and δ = ϵ. Then 0 < x a < δ = f(x) a = x a < δ = ϵ. Example 11. lim x a x = a Let f(x) = x, ϵ > 0 and δ = ϵ. Then 0 < x a < δ = f(x) a = x a x a < δ = ϵ. Example 12. If λ R then lim x a λ = λ Let f(x) = λ, ϵ > 0 and δ = 1. Then 0 < x a < δ = f(x) λ = λ λ = 0 < ϵ. Example 13. lim x 2 x 2 = 4 Let f(x) = x 2, ϵ > 0 and 0 < δ < min(ϵ/5, 1/2). Then 0 < x 2 < δ = f(x) 4 = x 2 4 = (x 2)(x 2) + 4(x 2) x 2 x x 2 < δ 2 + 4δ < δ + 4δ = 5δ < ϵ. For another proof, let f(x) = x 2, ϵ > 0 and 0 < δ < min(ϵ/5, 1). Then 0 < x 2 < δ = x + 2 = (x 2)+4 x = 5 and f(x) 4 = x 2 4 = (x+2)(x 2) = x+2 x 2 5 x 2 < 5δ < ϵ. Example 14. lim x 2 x = 2 16

18 Let f(x) = x, ϵ > 0 and δ = 2ϵ. Then x > 0 and 0 < x 2 < δ = f(x) 2 = x 2 = x 2 x+ 2 < x 2 / 2 < δ/ 2 = ϵ. Note that lim x a f(x) = L lim h 0 f(a + h) = L lim x a (f(x) L) = 0 lim x a f(x) L = 0. For example, lim x a f(x) = L lim (x a) 0 f(a + (x a)) = L lim h 0 f(a + h) = L. Example 15. Let f(x) = x3 8 4(x 2) and compute all three limits. If both one-sided limits exist and agree, then the limit exists and agrees with the common value. Example 16. lim x 0 sin(1/x) does not exist Suppose that lim x 0 sin(1/x) = L and let ϵ = 1/2. Then there exists δ > 0 such that 0 < x 0 = x < δ = sin(1/x) L < ϵ. For very large n we have 0 < 1/(2πn ± π/2) < δ and sin(2πn ± π/2) = ±1. But 2 = 1 ( 1) = (1 L) (( 1) L) L 1 + L ( 1) < 2ϵ = 1, a contradiction. Example 17. lim x 0+ x = 0. Let f(x) = x, ϵ > 0 and δ = ϵ 2. Then x > 0 and 0 < x 0 < δ = 0 < x < δ = ϵ 2 = f(x) 0 = x 0 x < ϵ. Example 18. If a > 0, then lim x a x = a Let f(x) = x, ϵ > 0 and δ = aϵ. Then x > 0 and 0 < x a < δ = f(x) a = x a = x a x+ a x a / a < δ/ a = ϵ. Exercise 4. Graded: #1 give an ϵ, δ proof that lim x 3 (6x 7) = 11; #2 give an ϵ, δ proof that lim x 2 1 3x = 5; #3 give an ϵ, δ proof that lim x 3 x 2 = 9; #4 give an ϵ, δ proof that lim x 1 x 3 = 1. Look over the problems at the end of sections 1.6 and let us know if any seem of interest or are particularly difficult. These homeworks are due at the beginning of class on Wednesday, September 12, Remember that homeworks make up circa 40% of your grade. 17

19 In this section, without explicit comment, the domains for f and g include open intervals about a but not necessarily a. Theorem 19 (Uniqueness of Limits). If lim x a f(x) = L and lim x a f(x) = M, then L = M. Suppose that L M > 0 and let ϵ = L M /3. Choose δ 1 > 0 such that 0 < x a < δ 1 = f(x) L < ϵ, choose δ 2 > 0 such that 0 < x a < δ 2 = f(x) M < ϵ and δ = min(δ 1, δ 2 ). Therefore 0 < x a < δ = 0 < L M = (L f(x))+(f(x) M) f(x) L + f(x) M < 2ϵ = 2 L M /3 < L M, a contradiction. Hence, L = M. Lemma 20. Let R be a ring (e.g., this holds if R = R) and let x, y, a and b R. Then xy ab = (x a)(y b) + a(y b) + (x a)b. (x a)(y b) + a(y b) + (x a)b = xy xb ay + ab + ay ab + xb ab = xy ab. Theorem 21. If lim x a f(x) = L and lim x a g(x) = M, then (i) lim x a (f + g)(x) = L + M; (ii) if λ R, then lim x a (λf)(x) = λl; (iii) lim x a (fg)(x) = LM. (i). Let ϵ > 0. Choose δ 1 > 0 such that 0 < x a < δ 1 = f(x) L < ϵ/2, choose δ 2 > 0 such that 0 < x a < δ 2 = g(x) M < ϵ/2 and let δ = min(δ 1, δ 2 ). Then 0 < x a < δ = (f + g)(x) (L + M) = (f(x) L) + (g(x) M) f(x) L + g(x) M < ϵ/2 + ϵ/2 = ϵ. (iii) Let ϵ > 0 and let ϵ = min(ϵ/(2 + L + M ), 1/2). Choose δ 1 > 0 such that 0 < x a < δ 1 = f(x) L < ϵ, choose δ 2 > 0 such that 0 < x a < δ 2 = g(x) M < ϵ and let δ = min(δ 1, δ 2 ). Then 0 < x a < δ = (fg)(x) (LM) = (f(x) L)(g(x) M) + L(g(x) M) + (f(x) L)M f(x) L g(x) M + L g(x) M + M f(x) L < ϵ 2 + ϵ L + ϵ M < ϵ (1 + L + M ) < ϵ. (ii). If λ = 0, then lim x a (λf)(x) = lim x a 0 = 0 = 0L. If λ 0 and ϵ > 0, choose δ > 0 such that x a < δ = f(x) L < ϵ/ λ. Then 0 < x a < δ = (λf)(x) λl = λ f(x) L < λ ϵ/ λ = ϵ. This is also a special case of (iii). Parts (i) and (iii) of this theorem may be easily extended using mathematical induction to arbitrary finite sums and products and linear combinations using a simple induction. In particular this applies to monomials and polynomials. Corollary 22. Let p(x) be a polynomial. Then lim x a p(x) = p(a). Theorem 23. If lim x a g(x) = M with M 0, then lim x a 1/g(x) = 1/M. 18

20 First suppose that M > 0. Given ϵ > 0, first choose δ 1 > 0 so that 0 < x a < δ 1 = g(x) M < M/2 = M/2 < g(x) M < M/2 = 0 < M/2 < g(x) < 3M/2, choose δ 2 > 0 so that 0 < x a < δ 2 = g(x) M < M 2 ϵ/2 and let δ = min(δ 1, δ 2 ). Then 0 < x a < δ = 1/g(x) 1/M = g(x) M /g(x)m < g(x) M /(M 2 /2) < (M 2 /2)ϵ/(M 2 /2) = ϵ. If M < 0, then M > 0 and lim x a g(x) = lim x a g(x) = M = lim x a 1/g(x) = 1/M = lim x a 1/g(x) = lim x a 1/g(x) = ( 1/M) = 1/M. Corollary 24. If lim x a f(x) = L and lim x a g(x) = M 0, then lim x a f(x)/g(x) = L/M. Corollary 25. If lim x a f(x) = L 0 and lim x 0 g(x) = 0 = lim x a f(x) g(x) does not exist. f(x) If lim x a g(x) = K R, then L = lim f(x) x a f(x) = lim x a g(x) g(x) = K 0 = 0, a contradiction. Theorem 26. Suppose that lim x a g(x) = b and lim w b f(w) = L. Then lim x a (f g)(x) = L. Let ϵ > 0. Choose δ > 0 such that 0 < w b < δ = f(w) L < ϵ and then choose δ > 0 so that 0 < x a < δ = g(x) b < δ. Therefore 0 < x a < δ = g(x) b < δ = (f g)(x) L = f(g(x)) L < ϵ. Example 27. Evaluate the limits that exist: x (a) lim 2 x 6 x 3 x 3 ; (x (b) lim 2 3x 4) x 4 x 4 ; x+1 (c) lim x 1 (2x 2 +7x+5) 2 Example 28. Evaluate the following limits: (a) lim x 2 1/x 1/2 x 2 ; (b) lim x 9 x 2 9 x 3 Proposition 29. Let n 2. Then A n B n = (A B) 1 l n An l B l 1. (A B) 1 l n A n l B l 1 = = 1 l n 0 l n 1 = A n + = A n B n A n l+1 B l 1 A n l B l 1 l n 1 1 l n 1 l n A n l B l A n l B l (A n l B l A n l B l ) B n 19

21 Theorem 30. Let n 2 and let a > 0. Then lim x 0+ n x = 0 and lim x a n x = n a. First, let ϵ > 0 and let δ = ϵ n. Then 0 < x = x 0 < δ = 0 < n x < n δ = ϵ = lim n x 0+ x = 0. Next, let a > 0 and ϵ > 0 and let δ = a (n 1)/n ϵ. Then x > 0 and 0 < x a < δ = n x n a = x a / x (n l)/n a (l 1)/n 1 l n x a /a (n 1)/n < a (n 1)/n ϵ/a (n 1)/n = ϵ Corollary 31. Let m, n 1 and a > 0. Then lim x 0+ x m/n = 0 and lim x a x ±m/n = a ±m/n. If r Q, then lim x a x r = a r. Theorem 32 (Squeeze Theorem or Pinching Theorem or Sandwich Theorem). Assume that the three functions f, g and h satisfy f(x) g(x) h(x) for all x near a except possibly at a. Then lim x a f(x) = lim x a h(x) = L = lim x a g(x) = L. Let ϵ > 0. Choose δ 1 > 0 so that 0 < x a < δ 1 = f(x) L < ϵ = L ϵ < f(x). Choose δ 2 > 0 so that 0 < x a < δ 2 = h(x) L < ϵ = h(x) < L + ϵ. Let δ = min(δ 1, δ 2 ). Then 0 < x a < δ = L ϵ < f(x) g(x) h(x) < L + ϵ = g(x) L < ϵ = lim x a g(x) = L. Example 33. lim x 0 sin(x) = 0 and lim x 0 cos(x) = 1 = lim x 0 tan(x) = 0 and lim x 0 sec(x) = 1. Use a simple geometric argument involving the areas of a triangle and the unit disk plus the squeeze theorem to make plausible that lim θ 0+ sin(θ) = 0. Next use the fact that sin(θ) is odd to conclude that lim θ 0 sin(θ) = 0 and therefore lim θ 0 sin(θ) = 0. Now use the identity that 1 cos(θ) = 2 sin 2 (θ/2) to conclude that lim θ 0 cos(θ) = 1. Theorem 34. Let α R. Then lim θ α sin(θ) = sin(α) and lim θ α cos(θ) = cos(α). lim x a sin(x) = sin(a) lim h 0 sin(a + h) = sin(a). Now sin(a + h) = sin(a) cos(h) + cos(a) sin(h) sin(a) 1 + cos(a) 0 = sin(a). One proves that lim x a cos(x) = cos(a) in a similar manner. Lemma 35. lim x a f(x) = 0 lim x a f(x) = 0. 20

22 To prove =, given ϵ > 0, choose δ > 0 so that 0 < x a < ϵ = f(x) 0 = f(x) = f(x) 0 < ϵ = lim x a f(x) = 0. To prove =, given ϵ > 0, choose δ > 0 so that 0 < x a < ϵ = f(x) 0 = f(x) = f(x) 0 < ϵ = lim x a f(x) = 0. Example 36. lim x 0 x 2 sin(1/x) = 0 and lim x 0 x sin(1/x 17 ) = 0. Example 37. lim θ 0 sin(θ) θ = 1 and lim θ 0 1 cos(θ) θ = 0. Let θ > 0 be small and use a geometric argument using the areas of two triangles and a pie-shaped region of the unit disk to obtain the inequality sin(θ) 2 θ tan(θ) 2π π 2 = cos(θ) sin(θ) sin(θ) θ 1 = lim θ 0+ θ = 1 sin( θ) sin(θ) sin(θ) by the Squeeze Theorem. Since lim θ 0+ θ = lim θ 0+ θ = 1, we have lim θ 0 θ = 1. Furthermore, 1 cos(θ) θ = 2 sin 2 (θ/2)/θ = sin(θ/2) sin(θ/2) θ/2 0 1 = 0 as θ 0. Exercise 5. Graded: #1 compute the limit lim x 4 ( 1 x 1 4 )( 1 x 4 ) if it exists; x #2 compute the limit lim 5 1 x 1 x 4 1 ; f(a+h) f(a) #3 fix a number a and evaluate lim h 0 h for f(x) = x 4 ; #4 problem #45, p. 71; #5 problem #46, p. 71. Look over the problems at the end of sections 1.5 and 1.6 and let us know if any seem of interest or are particularly difficult. These homeworks are due at the beginning of class on Wednesday, September 19, Remember that homeworks make up circa 40% of your grade. 21

23 Infinite Limits. Idea, f(x) grows arbitrarily large as x a or x a+ or x a, where a R. Definition 38. Let f be defined near a R but not necessarily at a. We write lim x a f(x) = (= + ) if, for every L > 0 there exists a δ > 0 such that 0 < x a < δ = f(x) > L. One defines lim x a+ f(x) = (= + ) and lim x a f(x) = (= + ) in a similar manner. We write lim x a f(x) = if, for every L > 0 there exists a δ > 0 such that 0 < x a < δ = f(x) < L. One defines lim x a+ f(x) = and lim x a f(x) = in a similar manner. Note that, + and are symbols and not numbers that can be manipulated in the usual manner. Example 39. (a) compute lim x 1 (b) compute lim x 1 x (x 2 1) ; 2 x (x 2 1) ; 2 x x It suffices to prove (a) since lim x 1 (x 2 1) = lim 2 x 1 (x 2 1). Let us concentrate on (a). 2 Choose L >> 0. Let δ = min( 2/L/5, 1/2). If x 1 < δ 1/2 = 1/2 < x < 3/2 = 0 < x + 1 < x x 5/2. Then (x 2 1) = 2 (x+1) 2 (x 1) > (1/2)(4/25)(1/δ 2 ) > (2/25)(1/( 2/L/5) 2 ) = (2/25)(25L/2) = L. 2 Lemma 40. Let f be defined near a R but not necessarily at a. Then lim x a f(x) = = lim x a 1/f(x) = 0. Conversely, f(x) > 0 for x near a and lim x a 1/f(x) = 0 = lim x a f(x) = +. Similarly, f(x) < 0 for x near a and lim x a 1/f(x) = 0 = lim x a f(x) =. Of course, similar statements hold for and lim x a+ f(x) and lim x a f(x). Suppose that lim x a f(x) = +, let ϵ > 0 and let L = 1/ϵ > 0. Choose δ > 0 so that 0 < x a < δ = f(x) > L = 0 < 1/f(x) 0 = 1/f(x) < 1/L = ϵ = lim x a 1/f(x) = 0. Conversely, suppose that f(x) > 0 for x near a, that lim x a 1/f(x) = 0, L > 0 and ϵ = 1/L. Choose δ > 0 such that 0 < x a < δ = 0 < 1/f(x) = 1/f(x) 0 < ϵ = f(x) > 1/ϵ = L. Apply this lemma to give a very simple analysis of the previous two examples. Next, an aid in graphing. Definition 41. If lim x a f(x) = ± or lim x a+ f(x) = ± or lim x a f(x) = ±, then the line x = a is said to be a vertical asymptote. Example 42. (a) vertical asymptotes of g(x) = (x 2) (x 1) 2 (x 3) ; 22

24 2 5x (b) compute lim x 3+ x 3 ; 2 5x (c) compute lim x 3 x 3 ; x (d) compute lim 3 +5x 2 6x x 4+ x 3 4x ; 2 (e) vertical asymptotes of f(x) = x2 4x+3 x 2 1 ; (f) compute lim θ 0+ cot(θ); (g) compute lim θ 0 cot(θ) Exercise 6. Look over the end of Section 1.5 on infinite limits and vertical asymptotes. You should should be able to do #29 - #40, p. 61. Graded: #1 problem #36, p. 61; #2 problem #37, p. 61; #3 problem #40, part (a) only, p. 61; These homeworks are due at the beginning of class on Wednesday, September 19, Remember that homeworks make up circa 40% of your grade. 23

25 Limits at Infinity. Definition 43. lim x + f(x) = lim x 0+ f(1/x) [, + ] and lim x f(x) = lim x 0 f(1/x) [, + ]. Example 44. (a) compute lim x ( x ); 2 (b) compute lim x + (5 + sin(x) x ); (c) compute lim x + 1/x n for n 1; (d) compute lim x ± (3x 4 6x 2 + x 10); (e) compute lim x ± ( 2x 3 + 3x 2 12); 3x+2 (f) compute lim x ± x 2 1 ; 40x (g) compute lim 4 +4x 2 1 x ± 10x 4 +8x 2 +1 ; 2x (h) compute lim 2 +6x 2 x ± x+1 ; 10x (i) compute lim 3 3x 2 +8 x ± ; 25x6 +x 4 +2 (j) compute lim x + ( x + 1 x) Exercise 7. Graded: #1 problem #14, p. 242; #2 problem #22, p. 242; #3 problem #23, p. 242; #4 problem #29, p. 242; #5 problem #32, p. 242 Look over the problems at the end of section 3.4 and let us know if any seem of interest or are particularly difficult. These homeworks are due at the beginning of class on Wednesday, September 19, Remember that homeworks make up circa 40% of your grade. 24

26 Section 1.8. Continuity. Smooth graphs without jumps or wild oscillations (a la sin(1/x) near 0). Definition 45. Let f be defined in an open interval about a. Then f is continuous at a if and only if lim x a f(x) = f(a). Use a similar definition at the end points of a closed interval using one-sided limits if f is only defined to the right or left of the point a. f is continuous on an interval if it is continuous at every point of the interval. Theorem 46. The following functions with the obvious domains are continuous: (a) k on R; (b) x n if n 1 on R; (c) x on R; (d) x q on [0, + if q Q, q > 0 and x q on 0, + if q Q; (e) sin(x) on R, cos(x) on R, tan(x) on R {nπ + π/2 n Z}, csc(x) on R {mπ n Z}, sec(x) on R {nπ + π/2 n Z}, cot(x) on R {mπ m Z}; (f) p(x) a polynomial; (g) r(x) = p(x)/q(x), a rational function, on {x q(x) 0}; (h) p(cos(θ), sin(θ)) where p(x, y) is a polynomial in x and y; (i) p(cos(θ), sin(θ))/q(cos(θ), sin(θ)), where p(x, y) and q(x, y) are polynomials in x and y, on R {θ q(cos(θ), sin(θ)) = 0}. Theorem 47. Let λ R and let f and g be continuous at a. Then λf, f + g, fg and f/g if g(a) 0 are continuous at a. Therefore if f and g are continuous, then λf, f + g, fg and f/g (on {x R g(x) 0}) are continuous. Theorem 48. Let f be continuous at b, let g be continuous at a and let g(a) = b. Then f g is continuous at a. Therefore the composition of two continuous functions is continuous. x a = g(x) g(a) = f(g(x)) f(g(a)). Using this theorem we can create very, very complicated continuous functions through algebraic operations and mutiple compositions. Let f(x) = x on the open interval I = 0, 1. Then f does not attain a maximum or minimum on I, so a continuous function on an open interval need not attain a maximum or minismum. Let f(0) = f(1) = 1/2 and let f(x) = x on 0, 1. Then f is not continuous and does not attain a maximum or minimum on the closed interval [0, 1]. However the following theorem is true but its proof lies beyond the scope of this course. We will be using it in the future. Theorem 49. Let f : [a, b] R be a continuous function on the closed bounded interval [a, b]. Then there exists x max, x min [a, b] such that f(x min ) f(x) f(x max ) for all x [a, b]. x max and x min of course are not necessarily unique. 25

27 The following important theorem is also true but its proof also lies beyond the scope of this course. We will be using it in the future. Theorem 50 (Intermediate Value Theorem). Let = I R be an interval, f : I R be continuous and let p, q I. If y lies between f(p) and f(q), then there is an x I that lies between p and q such that f(x) = y. x of course is not necessarily unique. Exercise 8. Graded: #1 let f(0) = 0 and f(x) = x sin(1/x) if x 0. Is f continuous on R? Look over the problems at the end of section 1.8 and let us know if any seem of interest or are particularly difficult. These homeworks are due at the beginning of class on Wednesday, September 26, Remember that homeworks make up circa 40% of your grade. 26

28 Derivatives and Rates of Change. Slope of and equation for the tangent line to the graph of a function f. Example 51. f(x) = 16x x at (1, 80). Definition 52. Let f be defined on an open interval containing the point a. Then the derivative of f at a is defined to be f(x) f(a) f(a+h) f(a) lim x a x a = lim h 0 h if this limit exists in R. If this limit exists it is called the derivative of f at a and is variously denoted f (a), (Df)(a) or df dx (a). Similar definitions apply at the end points of a closed interval. Example 53. f(x) = x. Compute f (x) and find the equation for the tangent line to the graph of f at (4, 2). Example 54. f(x) = 1/x 2. Compute f (x). Example 55. f(x) = x and a = 0. Then f (0) does not exist. Theorem 56. If f (a) exists then f is continuous at a. The convrse is false. f(a + h) = f(a+h) f(a) h h + f(a) Df(a) 0 + f(a) = f(a) as h 0. The converse is false - consider f(x) = x at x = 0. Exercise 9. Graded: #1 problem #35, p. 114; #2 problem #36, p. 114; Look over the problems at the end of section 2.1 and let us know if any seem of interest or are particularly difficult. These homeworks are due at the beginning of class on Wednesday, September 26, Remember that homeworks make up circa 40% of your grade. 27

29 Differentiation Formulas. Theorem 57. Let λ R, a R and suppose that f (a) and g (a) exist. Then: (i) λ = 0; (ii) (f + g) (a) = f (a) + g (a); (iii) (fg) (a) = f (a)g(a) + f(a)g (a) - in particular, (λf) (a) = λf (a); (iv) ( f g ) (a) = f (a)g(a) f(a)g (a) g 2 (a) if g(a) 0 - in particular, ( 1 g ) (a) = g (a) g 2 (a) if g(a) 0; (v) D(x) = 1; (vi) D(x n ) = nx n 1 if n 1 (here, and only here, we let 0 0 = 1 for convenience); (vii) D(x m ) = mx m 1 if m Z and x 0; (viii) D(x q ) = qx q 1 if x > 0 and q Q. (i). Let f(x) = λ for all x R. Then f(a+h) f(a) h = λ λ h = 0/h = 0 0 = f (x) = 0. (ii). Dg(a). (f+g)(a+h) (f+g)(a) h = f(a+h) f(a) h + g(a+h) g(a) h Df(a) + Dg(a) = D(f + g)(a) = Df(a) + (fg)(a+h) (fg)(a) (iii). h = f(a+h) f(a) h g(a + h) + f(a) g(a+h) g(a) h Df(a) g(a) + f(a) Dg(a) = (fg) = f g + fg. (λf) (a) = λ f(a) + λf (a) = 0 f(a) + λf (a) = λf (a). (iv). (f/g)(a + h) (f/g)(a) h f(a + h)g(a) f(a)g(a + h) = (1/h) g(a + h)g(a) f(a + h) f(a) = (1/g(a + h)g(a)) [ h (1/g 2 (a))[df(a) g(a) f(a) Dg(a)] g(a) f(a) = (f/g) = (f g fg )/g 2. In particular, (1/g) = g /g 2. (v). Let f(x) = x for all x R. Then f(a+h) f(a) h = (a+h) a h = 1 1 = D(x) = 1. g(a + h) g(a) ] h (vi). Use induction. The formula is true if n = 1. If the formula is true for n 1, then (x n+1 ) = (x n x) = (x n ) x + x n x = nx n 1 x + x n 1 = (n + 1)x n = (n + 1)x (n+1) 1. (vii). The formula is true if m 0. If m = n < 0, then (x m ) = (x n ) = (1/x n ) = 1 x n 1 (x n ) (x n ) 2 = ( n)x n 1 x 2n = ( n)x n 1 = mx m 1. (viii). The result is true if q = 0. For n 1 ( n x n a)/(x a) = 1/ 1 l n 1/na (n 1)/n = (1/n)a (1/n) 1 x (n l)/n a (l 1)/n 28

30 = D(x 1/n ) = (1/n)x (1/n) 1. D(x 1/n ) = D(1/x 1/n ) = D(x 1/n )/(x 1/n ) 2 = ( 1/n) x (1/n) 1 /x 2/n = ( 1/n)x ( 1/n) 1. Therefore D(x q ) = qx q 1 if q = 1/n, n 0, n Z. Next, let 0 n Z and let S = {m N D(x m n ) = ( m n )x m n 1 }. 1 S. If 1 m S, then D(x m+1 n ) = D(x m n x 1 n ) = ( m n )x m n 1 x 1 n + x m n ( 1 n )x 1 n 1 = ( m m+1 n )x n 1 + ( 1 m+1 n )x n 1 = ( m+1 m+1 n )x n 1 = m + 1 S = S = N. This proves (vii). The n-th derivative of f is denoted D n f. Definition 58. Define 0! = 1, 1! = 1, 2! = 2 1, 3! = = 3 2! and recursively n! = n (n 1)! for n 1. Lemma 59. For n 1 we have D n (x n ) = n!. Proof by induction. D 1 (x 1 ) = D(x) = 1 = 1!. D n+1 (x n+1 ) = D n (D(x n+1 )) = D n ((n + 1)x n ) = (n + 1)D n (x n ) = (n + 1)n! = (n + 1)!. Proposition 60. Given n distinct objects, the number of distinct words of length 1 k n that use each object at most once (i.e., the number of distinct k-tuples) is n(n 1)(n 2) (n k + 1) = n!/(n k)!. Lemma 61. If 1 k n then D k (x n ) = n!/(n k)!x n k. Definition 62. Let 0 k n. Then ( n k) = n!/k!(n k)!, a binomial coefficient, called n choose k. Corollary 63. Given n distinct objects, the number of subsets of size 0 k n is ( n k). Lemma ( 64. n ( k) = n ( n k). n ( 0) = n ( n) = 1 and n ) ( 1 = n n 1) = n. Lemma 65. If 1 k n then ( ( n k 1) + n ) ( k = n+1 ) k. ( ) n + k 1 ( ) n k = = = = = n! (k 1)!(n (k 1))! + n! k!(n k)! n! (k 1)!(n k)! [ 1 n + 1 k + 1 k ] n! (k 1)!(n k)! n + 1 k(n + 1 k) (n + 1)! k!(n + 1 k)! ( ) n + 1 k 29

31 This is also obvious by a simple combinatorial argument using a fixed distinguished element out of n + 1 distinct objects. Theorem 66 (Binomial Theorem). If A and B are commuting variables and n 1, then (A + B) n = 0 l n ( n l) A l B n l. Induct. If n = 1 then ( 1 ) 0 l 1 l A l B 1 l = ( ) 1 0 A 0 B ( 1 1) A 1 B 1 1 = B + A = (A + B) 1. If the theorem is true for n, then (A + B) n+1 = (A + B)(A + B) n Corollary 67. If n 1 then D(x n ) = nx n 1. = (A + B) 0 l n ( n l ) A l B n l = ( ) n A l+1 B n l + ( ) n A l B n+1 l l l 0 l n 0 l n ( ) n = A l B n+1 l + ( ) n A l B n+1 l l 1 l 1 l n+1 0 l n = A n+1 + ( ) n A l B n+1 l + l 1 1 l n 1 l n = A n+1 + ( ) ( ) n n [ + ]A l B n+1 l + B n+1 l 1 l 1 l n = A n+1 + ( ) n + 1 A l B n+1 l + B n+1 l 1 l n ( ) n + 1 = A l B n+1 l. l 0 l n+1 ( ) n A l B n+1 l + B n+1 l O.k. if n = 1. If n 2 then (a + h) n a n h = 1 ( ) n a l h n l h l 0 l n 1 = na n 1 + ( ) n a l h n l 1 = na n 1 + h na n 1 0 l n 2 0 l n 2 l ( n l ) a l h n l 2 30

32 as h 0. Exercise 10. Look over Section 2.2. You should should be able to do #19 - #29, pp Graded: #27. Look over Section 2.3. You should should be able to do #1 - #45, #51 - #52, #55 - #63, #65 - #67, #69 - #72, #75 - #83, #87 - #92, #98 - #107, pp Graded: #18, #27, #57, #76, #82, #101, #106, pp These homeworks are due at the beginning of class on Wednesday, October 3, Remember that homeworks make up circa 40% of your grade. 31

33 Extra Section Proofs by induction. A very important basic observation. If = A N, then A has a smallest element. Give two intuitive proofs based on what we think the integers are. Principle of Induction: Given a sequence of statements S 1, S 2,..., suppose that: (1) S 1 is true; (2) if S n is true, then S n+1 is true. Conclusion: all of the S n s are true. Give an inutitive proof and one based on the very important basic observation. Note that we can start with any integer. Note that both (1) and (2) must be carefully checked. For example, let S n be n = n 2. Then (1) is true, but (2) is not, for if n = n 2, then n + 1 = (n + 1) 2 = n 2 + 2n + 1 = 2n = 0 = n = 0, a contradiction. For another example, let S n be n = n+1. Then (1) is false but (2) is true, for n = n+1 = n+1 = (n+1)+1. Examples. (i) S n is the statement n = n(n+1) 2. Give proof by induction and Gauss proof. (ii) if x 1 then (1 + x) n 1 + nx. We have strict inequality if x 0. Give proof by induction. This is called Bernoulli s inequality. (iii) Let S n be the statement (2n 1) = n 2. Extra Exercise 1. Let S n be the statement n 2 = n(n+1)(2n+1) 6. Prove by induction. Extra Exercise 2. Let S n be the statement n 3 = ( n) 2. Extra Exercise 3. This one is tricky. For what integers n is 3 2n n+2 divisible by 7? Extra Exercise 4. For which integers n is 9 n 8n 1 divisible by 8? The Σ convention for sums, factorials, binomial coefficients, the Binomial Theorem. 32

34 Derivatives of the Trigonometric Functions. Theorem 68. (1) sin (0) = 1; (2) cos (0) = 0; (3) sin (x) = cos(x); (4) cos (x) = sin(x); (5) tan (x) = sec 2 (x); (6) sec (x) = sec(x) tan(x); (7) cot (x) = csc 2 (x); (8) csc (x) = csc(x) cot(x). (1) and (2) follow from Example 37. as h 0. This proves (3). sin(x + h) sin(x) h sin(x) cos(h) + cos(x) sin(h) sin(x) = h = sin(x) cos(h) 1 + cos(x) sin(h) h h sin(x) 0 + cos(x) 1 = cos(x) cos(x + h) cos(x) h cos(x) cos(h) sin(x) sin(h) cos(x) = h = cos(x) cos(h) 1 sin(x) sin(h) h h cos(x) 0 sin(x) 1 sin(x) as h 0. This proves (4). (5), (6), (7) and (8) follow from (3) and (4) using the quotient rule for derivatives. Go over examples. Exercise 11. Look over Section 2.4. You should should be able to do (no graphing) #1 - #52, #56, #57, pp Graded: #16, #32, #37, #47, #52. #56, pp These homeworks are due at the beginning of class on Wednesday, October 3, Remember that homeworks make up circa 40% of your grade. 33

35 The Chain Rule. Theorem 69 (Chain Rule). Let I R be an interval, g : I R, a I, Dg(a) exist, f : J R where J is an interval with g(i) J, and (Df)(g(a)) exist. If h(t) = (f g)(t) = f(g(t)), h : I R, then h is differentiable at a and (Dh)(a) = (Df)(g(a)) (Dg)(a). g is continuous at a since Dg(a) exists. Let y = g(a). By the definition of the derivative, we have that g(t) g(a) = (t a)[(dg)(a) + u(t)] and f(s) f(y) = (s y)[(df)(y) + v(s)], where t I, s J, u(t) 0 = u(a) as t a and v(s) 0 = v(y) as s y. Let s = g(t). Then h(t) h(a) = f(g(t)) f(g(a)) = [g(t) g(a)] [(Df)(y) + v(s)] = (t x) [(Dg)(a) + u(t)] [(Df)(y) + v(s)], or, if t a, h(t) h(a) t a = [(Df)(y) + v(s)] [(Dg)(a) + u(t)]. Let t a = s y since g is continuous at a and so [(Dg)(a) + u(t)] [(Df)(y) + v(s)] (Df)(y)(Dg)(a) = (Df)(g(a))(Dg)(a). Corollary 70. Suppose that n 1, I R is an interval, g : I R, a I, Dg(a) exists and h(x) = g n (x). Then h (a) = ng n 1 (a)g (a). The same formula holds if n Z and g(a) 0. Let f(w) = w n = f (w) = nw n 1 (of course except at 0 if n < 0). Then h(x) = (f g)(x) = h (a) = f (g(a))g (a) = ng n 1 (a)g (a). Corollary 71. Suppose that q Q, I R is an interval, g : I 0, +, a I, Dg(a) exists and h(x) = g q (x). Then h (a) = qg q 1 (a)g (a). Let f(w) = w q = f (w) = qw q 1. Then h(x) = (f g)(x) = h (a) = f (g(a))g (a) = qg q 1 (a)g (a). Corollary 72. Let λ R and h(x) = f(λx). Then h (x) = λf (λx). Let g(x) = λx. Then h(x) = f(g(x)) = (f g)(x) = h (x) = f (g(x))g (x) = f (λx)(λx) = λf (λx). Example 73. (sin(x 2 )) = sin (x 2 )(x 2 ) = 2x cos(x 2 ). (sin 17 (x)) = 17 sin 16 (x) cos(x). Go over more examples from textbook. 34

36 Exercise 12. Look over Section 2.5. You should should be able to do #1 - #54, #59 - #64, #69 - #80, #83, #85, pp Graded: #42, #46, #60, #47, #52. #56, pp These homeworks are due at the beginning of class on Wednesday, October 3, Remember that homeworks make up circa 40% of your grade. 35

37 Sample Questions for Test #1 on Wednesday, October 4, Problem 1. Find f g and g f if f(x) = x and g(x) = 3 x 1. Problem 2. For each n 1 let S n be the statement 2n 2 n. Use induction to prove that each S n is true. Problem 3. For each n 1 let S n be the statement 1 + 2n 3 n. Use induction to prove that each S n is true. Problem 4. Let 1 λ R. For each n 1 let S n be the statement 1 + λ + λ 2 + λ λ n = 1 λn+1 1 λ. Use induction to prove that each S n is true. Problem 5. Compute lim x 0+ x sin(1/ 3 x) if the limit exists. Problem 6. Compute lim x 1 x 7 +1 x+1 if the limit exists. Problem 7. Compute lim x 1 3 x 1 x 1 if the limit exists. Problem 8. Compute lim x 16 4 x 2 x 16 if the limit exists. Problem 9. x 1 Compute lim x 1 x 1 if the limit exists. Problem 10. x 1 Compute lim x 1 4x+5 3 if the limit exists. Problem 11. Compute lim x 1 3(x 4) x+5 3 x+5 if the limit exists. Problem 12. x Compute lim x 0 cx+1 1 if the limit exists. Problem 13. Compute lim x π/4 Problem 14. Compute lim x 3 1 x 3 Problem 15. Compute lim p 1 p 5 1 p 1 cos(x) cos(x) sin(x) ( 1 x if the limit exists. if the limit exists. ) if the limit exists. 36

38 Problem 16. Compute lim x 0 2/ tan(x) if the limit exists. Problem 17. Compute lim u 0+ u 1 sin(u) if the limit exists. Problem 18. Compute lim x + 25x2 +8 x+2 if the limit exists. Problem 19. Compute lim r + 1 cos(r)+1 Problem 20. Compute lim x ± x+1 4x 2 +1 Problem 21. Compute lim x ± 12x 2 14x4 +7 Problem x+1 Compute lim x ± x 3 if the limit exists. if the limit exists. if the limit exists. if the limit exists. Problem 23. Find all vertical and horizontal asymptotes for f(x) = x2 x x 2 1. Problem 24. Find all vertical and horizontal asymptotes for f(x) = 2x2 +6 2x 2 +3x 2. Problem 25. Let f(x) = x2 + x+sin(x)+1 x+ x. Compute f (x). Problem 26. Let g(x) = x2 16 x 4 Problem 27. Compute lim x 0 sin(3x) 2x. Problem 28. Compute lim x 0 x 2 sec(x) 1. Problem 29. Compute lim x 0 x 2 1+cot 2 (3x). if x = 4 and g(4) = 8. Is g a continuous function? Problem 30. Give a δ, ϵ proof that lim x 2 x 2 = 4. Problem 31. Compute lim x 0 x 2 2x sin(3x). 37

39 Problem 32. Compute lim x 0 1 cos(4x) 9x 2. Problem 33. Let f(0) = 0 and f(x) = x 2 sin(1/x 2 ) if x 0. Compute f (0) if it exists. Problem 34. Let f(x) = (ax + b)(cx + d), where a, b, c and d are constants. Compute f (x) if it exists. Problem 35. Let f(x) = 6x2 1 x 4 +5x+1. Compute f (x) if it exists. Problem 36. Find the equations for the tangent and normal lines to the graph of f(x) = (2, 2). Problem 37. Let f(x) = (1 + 1 x )(1 + 1 x 2 ). Compute f (x). Problem 38. Find a function f such that f (x) = 4x 3 2x + 4. Problem 39. Let f(x) = sin 2 (x) tan(x). Compute f (x). Problem 40. Find a function f such that f (x) = 2x 2 3x 1/x 2. Problem 41. Let f(x) = x3 +x 2 +x 1 x 3 x 2 +x+1. Find f (x). Problem 42. Let f(x) = 3x x 4. Compute (D 4 f)(x). 3x 1 2x Problem 43. Let f(x) = x 3 if x 0 and f(x) = 0 if x < 0. Compute (Df)(0) and (D 2 f)(0) if they exist. at the point (2, f(2)) = Problem 44. Let f(t) = 2t t+3 denotes the position of a particle on the real axis. Find the position, velocity, acceleration and speed of the particle at the time t 0 = 3. Problem 45. Let f(x) = x3 1+cos(x) cos(1) x 1 Problem 46. Let f(x) = x+ x+1 x x 1. Find f (x). Problem 47. Let f(x) = 2 x x x+1. Find f (x). if x 1. Compute lim x 1 f(x). 38

40 Problem 48. Let A, B and ω R and let y(t) = A cos(ωt) + B sin(ωt). Prove that y (t) + ω 2 y(t) = 0. Problem 49. Compute lim x 0 sin(13x)/ tan(17x). Problem 50. Prove that cot (x) = csc 2 (x). Problem 51. Prove that csc (x) = csc(x) cot(x). Problem 52. Find a function f such that f (x) = 2 cos(x) 3 sin(x). Problem 53. Find a function f such that f (x) = sec 2 (x) csc 2 (x). Problem 54. Let f(0) = 0 and f(x) = x 2 sin(1/x) if x 0. Show that f (x) exists for each x R and compute f (x) for each x, including for x = 0. Is f continuous at 0? Problem 55. Compute lim x 0 sin(5x) 3x. Problem 56. Compute lim x 0 sin(3x) tan(4x). Problem 57. If f(x) = sin(x)/x, compute (D 2 f)(x). Problem 58. Evaluate lim x 0 sin(x) x 3. Problem 59. Let f(x) = sec 3 (5x). Compute f (x). Problem 60. Let f(x) = 3 2x. Compute f (x). Problem 61. Let f(x) = sin( x 1 x ). Compute f (x). Problem 62. Compute lim x x 2 x Problem 63. Let f(x) = tan( x x+1 ). Compute f (x). 39