1. Introduction2. Administration3. Precalculus Review3.1. Exercise 1.2.213.2. Exercise 1.2.233.3. Exercise 1.2.243.4. Exercise 1.4.553.5. Exercise 1.4.563.6. Exercise 1.4.573.7. Exercise 1.4.58. Perpendicular Lines3.8. Exercise 1.4.594. Basic Limits4.1. Basic Limit Laws4.2. Exercise 2.3.224.3. Exercise 2.3.294.4. Exercise 2.3.304.5. Exercise 2.3.314.6. Exercise 2.3.324.7. Exercise 2.3.38Math 31A 2010.01.07MATH 31A DISCUSSIONJED YANG1. IntroductionLecture 1• Instructor: Steve Butler.• Location: HAINES 39.Sections 1A a nd 1B• Email: mailto:[email protected].• Office: MS 6617A.• Office hours: R 12:30–1 3:30.• Discussion Location: MS 5117 (T) and 5138 (R).• Website: http://www.math.ucla.edu/~jedyang/31a.1.10w/.• SMC: Jan. 6–Mar. 11, M–R 09:00–15:00, MS 3974, T 12:00–13:00.2. Administration• HW due Fridays in lec tur e , can turn in ear ly to me, and I will hand backin section.• Textbook: Rogawski, Single Variable Calculus, 2008.• Confirm office hour.3. Precalculus Review3.1. Exercise 1.2.21. Find the equation of the perpendicular bise c tor of the seg-ment joining (1, 2) and (5, 4).Solution. Slope of segment is m1=4−25−1=12. Slope of perpendicular bisector ism2= −1/m1= −2. Mid point is (1+52,2+42). So the equation can be written asy −3 = −2(x − 3). 3.2. Exercise 1.2.23. Find the equation of the line w ith x-intercept x = 4 andy-intercept y = 3.Solution. Equation of the line is y = mx+b, where b is the y-intercept, hence b = 3.The x-intercept x = 4 will yield y = 0 (by definition), s o substituting, we may solvefor m. We get 0 = 4m + 3, hence m = −34. So the equation can b e written asy = −34x + 3. 3.3. Exercise 1.2.24. A line of slope m = 2 passes through (1, 4). Find y suchthat (3, y) lies on the line.Math 31A Yang 2Solution. One way is to wr ite down an equation of the line in point-slope form:y = 2(x − 1) + 4. Then we see c learly that if x = 3, then y = 8. Alternatively,the slope m is the change of y ove r the change of x. Symbolically, m =∆y∆x, or∆y = m∆x. This co nce pt will be useful later when we deal with differentialsdy = m dx. Since the change in x is ∆x = 3 −1 = 2, we get that the change in y is∆y = y − 4 = 2 · 2 = 4, hence y = 8. This method seems longer, but conceptuallyit is easier to do in o ne’s head, and will lead to intution for calculus later . 3.4. Exercise 1.4.55. Use the addition formulae for sine and cosine to provetan(a + b ) =tan a + tan b1 − tan a tan b(1)cot(a − b) =cot a cot b + 1cot b − cot a. (2)Proof. Recall thatsin(a + b ) = sin a cos b + cos a sin b (3)cos(a + b) = cos a cos b − sin a sin b. (4)Nowtan(a + b) =sin(a + b )cos(a + b)(5)=sin a co s b + cos a sin bcos a cos b − sin a sin b(6)=sin acos a+sin bcos b1 −sin acos asin bcos b(7)=tan a + tan b1 − tan a tan b, (8)where we get from (6) to (7) by dividing top and bottom by cos a cos b.The case for cotangent is completely analogous. Remember cot x =cos xsin xandthat sin(−b) = −sin(b) and cos(−b) = cos(b). Work out the details and convinceyourself. 3.5. Exercise 1.4.56. Let θ be the angle between the line y = mx + b and thex-axis. Prove that m = tan θ.Proof. This is trivial. 3.6. Exercise 1.4.57. Let L1and L2be the lines of slope m1and m2, respectively.Show that the angle θ between L1and L2satisfies cot θ =m2m1+1m2−m1.Proof. This is immediate by using Exercises 55 and 56. 3.7. Exercise 1.4.58. Perpendicular Lines. Use Exercise 57 to prove that twolines with nonzero slopes m1and m2are perpendicular if and only if m2= −1/m1.Proof. What is cot(π/2)? Math 31A Yang 33.8. Exercise 1.4.59. Apply the double-angle formula to prove:(a) cosπ8=12p2 +√2.(b) cosπ16=12q2 +p2 +√2.Guess the values of cosπ32and of cosπ2nfor all n.Proof. Recall cos2t =1+cos(2t)2. For the general case, let a0= 0 and define in-ductively an=√2 + an−1. We claim that for n ≥ 1, we have cosπ2n=12an−1.The base case is trivial. By induction, assume cosπ2n=12an−1. By the half-angleformula, we get cosπ2n+1=q12(1 +12an−1) =q14(2 + an−1) =12an. 4. Basic Limits4.1. Basic Limit Laws. Assume that limx→cf(x) and limx→cg(x) exist. Then:(a) Sum Law:limx→c(f(x) + g(x)) = limx→cf(x) + limx→cg(x).(b) Constant Multiple Law: For any number k ∈ R,limx→ckf(x) = k limx→cf(x).(c) Product Law:limx→c(f(x)g(x)) =limx→cf(x)limx→cg(x).(d) Quotient Law: If limx→cg(x) 6= 0, thenlimx→cf(x)g(x)=limx→cf(x)limx→cg(x).4.2. Exercise 2.3.22. Evaluate the limit limz→1z−1+zz+1.Solution. Recall that limz→1z = 1 and limz→11 = 1. By the Quotient Law,limz→1z−1=limz →11limz →1z=11= 1. By the Sum Law, limz→1z−1+ z = limz→1z−1+limz→1z = 1 + 1 = 2. By the Sum Law, limz→1z + 1 = 2. So by the Quotient Law,limz→1z−1+zz+1=limz →1z−1+zlimz →1z+1=22= 1. 4.3. Exercise 2.3.29. Can the Quotient Law b e applied to evaluate limx→0sin xx?Solution. The Quotient Law requires the limit of the denominator, namely, limx→0x,to exist and be nonzero. This is not the case, so we cannot apply dire ctly. 4.4. Exercise 2.3.30. Show that the Product Law cannot be use d to evaluatelimx→π/2(x − π/2) ta n x.Solution. The Product Law req uires the limit of each factor to exist. However,limx→π/2tan x doe s not exist. 4.5. Exercise 2.3.31. Give an example where limx→0(f(x) + g(x)) exists but nei-ther limx→0f(x) nor limx→0g(x) exists.Math 31A Yang 4Solution. Let f(x) be any function defined on a neighborhood of 0 (but not nec-essarily at 0) such that limx→0f(x) does not exist (e.g., f(x) = 1/x). Let g(x) =−f(x). Then of course limx→0g(x) also does not exist (otherwise by the ConstantMultiple Law, limx→0f(x) also exists). But notice f(x) + g(x) is identicaly zeroin a neighborhood of 0 (but not necessarily at 0). So limx→0(f(x) + g(x)) = 0exists. 4.6. Exercise 2.3.32. Assume that the limit La= limx→0ax−1xexists and thatlimx→0ax= 1 for all a > 0. Prove that Lab= La+ Lbfor a, b > 0. [Hint:(ab)x− 1 = ax(bx− 1) + (ax− 1).]Solution. By definition, Lab= limx→0(ab)x−1x= limx→0axbx−1x+ax−1x. Sincelimx→0ax= 1 by assumption and limx→0bx−1x= Lbexists by assumption, theProduct Law states limx→0axbx−1x= 1·Lb. Now limx→0ax−1x= Laby assumption,so the Sum Law yields limx→0axbx−1x+ax−1x= Lb+ La. 4.7. Exercise 2.3.38. Assuming that limx→0f(x)x= 1, which of the fo llowingstatements is necessarily true?(a) f(0) = 0.(b) limx→0f(x) = 0.Solution.