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# Calculus 1 Week One Notes (Improved) MATH 2231 (IAI M1 900-1/MTH901)

College of DuPage

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This 7 page Class Notes was uploaded by Tiffany Montgomery on Saturday February 13, 2016. The Class Notes belongs to MATH 2231 (IAI M1 900-1/MTH901) at College of DuPage taught by Dr. O'Leary in Spring 2016. Since its upload, it has received 52 views. For similar materials see Calculus and Analytic Geometry I in Mathematics (M) at College of DuPage.

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Date Created: 02/13/16

CalculusandAnalyticGeometryI SPRINGSEMESTER2016 INSTRUCTOR:DR.M.O’LEARY 25 January 2016 1.1: A Preview of Calculus We will be solving two problems: ● the tangent line problem (How do you find the slope of a curve?) ● and the area problem (How do you find the area under a curve?) 1.2: Finding Limits Graphically and Numerically Let f(x) = 2x + 1. What value if any does f(x) approach (get closer and closer to) as x approaches 4? If we look at the graph, we can take a quick guess: (picture here) f(4) = 2(4) + 1 = 9 We can also start by filling out a table: 3.9 8.8 3.99 8.98 3.99 8.998 9 4.001 9.00 2 4.01 9.02 4.1 9.2 The process of finding a number that you are approaching is called finding a limit. When finding limits the domain of x does not include the number you are approaching. A limit is written like this: x→ cf(x) = L 1 And is read like this: f(x) approaches L as x approaches c, where x = / c . Examples 1. Let h(x) = . If possible, find limh(x) . x x→0 a. The limit does not exist because as the graph of 1 approaches zero it does not approach any one x number on either side. 2. f(x) = |x−3| x−3 a. ** Quick notes on the absolute value function: i. |x| = distance ii. on a number line from zero to x: 1. x = x if x >= 0 2. x = -x if x = 0 iii. It is useful to get rid of the absolute value ASAP using this piecewise function. a. the limit at 3 DNE, but not just because you can’t divide by three… It does not exist because it approaches too different numbers (-1 and 1) on both sides. b. Conclusion: a limit has two sides. Both sides must agree and approach a single number c. **We will talk about one-sided limits and limits that approach infinity next week but for now limit have to match on both sides*** 1.3 Evaluating Limits Analytically Properties List: Let: ● f,g be functions ● b, c be in the set of all real numbers ● n be a positive integer. 1. x→cx = c 2. limb = b(Because x is the thing that is changing not b) x→c 3. limx = x→c n n−1 n n−1 4. lim a xn + a n−1x + ... + a1 + a0 = n L + an−1L + ... + a1 L + a0 x→c n n 5. lx→c x√ = c√ Suppose limf(x) = L and limg(x) = M x→c x→c 2 4. x→cf(x) ± g(x) = L ± M 5. limf(x)g(x) = LM x→c 6. lim f(x)= L if M / 0 x→c g(x) M 7. lim b ∙ f(x) = b ∙ L x→c 8. lim (f g°(x) = f(M) if , M ∈ dom f(x) x→c 9. limcos(x) = cos(c) (etc. for all trig functions) x→c Examples starting on page 67 3x +5x−2 44. lim x+2 x→−2 (3x − 1)(x+2) ● lix→−2 x+2 ● lim 3x − 1 x→−2 ● lim 3(− 7) − 1(You can just plug in the x that the limit is approaching here) x→−2 ● = 7 The functions in the first step and the steps that follow are not the same! They differ at one point − 2, f(− 2). However, because the limit does not take into account what is happening at the point it is approaching the limits will almost always coincide. 50. lim 5 − x x→5x − 25 Because the limits of the numerator and the denominator both are zero you can not use your limit properties to simplify the problem. We must simplify by factoring. ● lim (x−5)(x+5) x→5 ● lim −1 x→5 x+5 −1 ● = 10 Almost Agreement Theorem: Let f, g be functions such that there exists a real number c such that for all x in dom(f) and dom(g), f(x) = g(x) except f(c) =/= g(c). For any d: limf(x) = limg(x), provided that both limits exist x→d x→d 58. lim 1 − / x x→0x+4 4 3 This limit also has an indeterminate form so you can not use limit properties here. x+4 −4 ● lim x x→0 4(x+4) ● lim x x→0 −x ● lim 4(x+4 x→0 x −1 ● lim 4(x+4) x→0 −1 ● = 16 Almost agreement theorem is used to transition from line 4 to 5 54 . lim √x+1−2 x→3 x−3 ● lim √x+1−2 ∙√x+1+2 x→3 x−3 √x+1+2 (x+1)−4 ● lim x→3 (x−3)( x+1 +2) 1 ● lim √x+1 + 2 x→3 1 ● = 4 Note: You can not draw any conclusions from an indeterminate form other than you can not use your limit properties. Each of the following have indeterminate form: x 2 1. lim = 0 x→0 x 2 2. lim x = 1 x→0 x x 3. lim 2= DNE x→0 x Famous Limit approaching!! (get it…?) lim six(x= 1… But why? x→ 0 Squeeze Theorem: Let f, g, h be functions. Suppose g(x) ≤ f(x) ≤ h(x)for all x in an open interval around c and limg(x) = limh(x) = L: limf(x) = L. x→c x→c x→c 4 5 Examples: sin(2x) 24. lim sin(3x) x→0 sin(2x)3x 2 3x ● limx→0 sin(3x)2x ∙3 Here we multiply by 2x in order to get both the numerator and sin(x) denominator into the form. We need to multiply by 2 so that we will still be x 3 multiplying by one. 2 sin(2x)3x ● 3 lim sin(3x)2x see limit property #2 x→0 ● 2 lim sin(2x∙ lim 3x see limit property #5 3 x→0 2x x→0 sin(3x) 2 ● = 3 Both of the previous limits are equal to 1. 1 − tan(x) 72. lim πsin(x)−cos(x) x→ 4 1 − tan(x)cos(x) ● lim πsin(x)−cos(xcos(x) x→ 4 cos(x) − sin(x) ● lim πsin(x)−cos(x)(cos(x)) x→ 4 −1 ● lim πcos(x) x→ 4 −1 ● = √2 2 ● = − 2 √ 1.4 Continuous Functions When is a function continuous at a point? You need three conditions to satisfy this: 1. limf(x)exists x→c 2. f(c) exists 3. limf(x) = f(c) x→c In layman’s terms…. there are no gaps, no holes, and all of the points are on the same line. 6 Let I be an open interval (end points are excluded). f(x)is continuous on I if: f(x)is continuous at x = c for all c in I . Example: Let f(x) = tan(x). π ● f(x) is not continuous at x = 2 ● f(x) is continuous at x = 0 ● f(x) is continuous on ( , ) −2 2 More examples: 79. f(x) = 4 This function is not continuous at x = 6 because there is a hole here. x−6 Continuity Property List: Let f, g be continuous at x = c. Let a ∈ ℜ . The following are continuous at x = c . 1. f ± g 2. f ∙ g f 3. g except where g(x) = 0 4. a ∙ f 5. f g° when f(x) is continuous at f(g(x)) 7

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