In order to be successful in AP Calculus, you are expected to KNOW everything that came before. All topics from Algebra I, II, Geometry and of course Precalculus are expected to be mastered before you enter this class in September. Because the AP test is in early May, and the southern schools begin a month before us, we need to grab a head start. Therefore, you are expected to cover sections 1 and of chapter. These sections introduce the concept of limit. Attached you will find lessons on these sections, practice problems, and a written copy of the first two sections of the book, with assigned problems. Please complete this packet before the first day of school. It will be gone over quickly, and a quiz on the material given within the first week or so. Advanced Placement Calculus is a high level course, and you must be motivated to work hard for the ENTIRE school year, work independently on AP materials and put in extra time. Students must be committed to putting in the time and effort, or they will not succeed. You must decide if this is what you want- it is a challenge, even for students who have never found any high school course challenging before. On the school website, you may find AP packets from previous years. You are encouraged to look at it, and make sure that you can do it, but that packet is not assigned for this school year. THIS PACKET REPLACES ANYTHING THAT COMES BEFORE IT. The questions/problems in this booklet have been compiled by past and present CHS mathematics teachers based on open resources from publishers and math websites such as NCTM.
Section.1 Rates of change and Limits Objective: To define the concept of a limit and establish its properties. Example: consider the function f( x) x 4 x We can see that f() is undefined. (Plugging x = into the function results in a 0 in the denominator) But what happens as our x value gets really, really close to? x x 4 f( x) x 1.9 (1.9 4)/(1.9 ) = 3.9 1.99 (1.99 4)/(1.99 ) = 3.99 1.999 (1.999 4)/(1.999 ) = 3.999 1.99999999 (1.99999999 4)/(1.99999999 ) = 3.99999999 Undefined.00000001 (.00000001 4)/(.00000001 ) = 4.00000001 We see that as x gets really, really close to, f(x) (or y), gets really, really close to 4. We say the limit as x approaches on the function f( x) x 4 x is 4 and we write: lim x 4 x x = 4 In general, lim f(x) = L means as x gets really, really close to c, f(x) (or y) gets really, really close to L. x c
Properties of Limits: Let: lim f(x) = L lim g(x) = M, and k is any real number. x c x c Then: 1) Sum and difference property: lim (f(x) + g(x)) = L + M and lim (f(x) g(x)) = L M x c x c Example: Given: lim x = 9 and lim x 3 = 7 x 3 x 3 Then lim (x + x 3 ) = 9 + 7 = 36 and lim (x x 3 ) = 9 7 = 18 x 3 x 3 ) Product Property: lim (f(x)g(x)) = LM x c Example: Given lim x =9 and lim (x+5) = 8, then lim (x (x+5)) = 9*8 = 7 x 3 x 3 x 3 3) Constant Multiples: lim k(f(x)) = kl x c Example: Given lim x = 9, then lim 5x = 5*9 = 45 x 3 x 3 4) Division Property : lim x c f( x) L gx ( ) M AS LONG AS M 0 Example: : Given lim x =9 and lim (x+5) = 8, then lim x 3 x 3 x 3 x 9 x 5 8 r r r s s s 5) Exponent Property: lim ( f( x)) L Provided L (is real). x c Example: lim (x+5) = 8, then lim x 3 x 3 3 3 ( x 5) 8 4
Methods for taking limits 1) Limits by substitution (always the first thing to try) If f(x) is continuous, then lim f(x) = f(c) x c Example: lim 3x 5x+ = 3( ) 5() + = 4 x Theorem: n f( x) a x a x... a x a then n n 1 n 1 1 0 n n 1 If lim f(x) = ac a c... ac a x c n n 1 1 0 (limits by substitution always works in polynomials) Theorem: If f(x) and g(x) are polynomials, then lim f ( x) f( c), provided g(c) 0 x c gx ( ) gc ( ) (limits by substitution always works in rational functions, as long as the denominator doesn t go to 0). 0 0 0 ) Indeterminate forms (,,0,,0,1, ). These 7 forms are indeterminate, that 0 is, they could be anything. We can not leave any answer in this form. We must deal with it using : Algebraic methods such as factoring, finding a common denominator, rationalization Trigonometric methods using trig identities or special trig limits Logarithmic methods the last 3 indeterminate forms (which we will not cover this year), need logarithms to simplify. For example, look at the first limit we saw: lim x x 4 x When we try to substitute x = into the function, we get 0/0. But we can factor: lim x x 4 x = lim ( x )( x ) x x Now, AS LONG AS x (it isn t, it s just really really close), we can cancel the (x ) s and get lim (x+) (try substitution again) = + = 4. x Read pages 58 66, Do p. 66 # s 1 14, 19 3 and the following worksheet.
Name Calculus AP Limits hw day1 Determine the limit, if it exists. If a limit fails to exist, indicate with. 1) lim x 1/ 3x (x 1) ) lim x 4 (x + 3) 014 3) lim x 1 (x 3 + 3x x 17) 4) lim x + 4x+ 3 x 3 x 3 5) lim x (cos(πx) +4) 6) lim x 1 x 1 x 1 7) lim x 3x + x x 4 8) lim 5x 3 + 8x x 0 3x 4 16x 9) 1 1 lim x x x 0 10) lim (x + ) 3 8 x 0 x
One Sided Limits Definition Piecewise defined function a function whose definition changes over different intervals of x. x+5 x<3 For example: f(x) = x 1 x 3 We would like to know what value the function approaches as x approaches 3, but we have to look at both sides. Def Left hand limit lim f(x) the limit as x approaches c from the left (from x < c) x c Right hand limit lim f(x) the limit as x approaches c from the right (from x > c) x c + Theorem If a limit exists, then it is unique. (If a limit is not unique, then it doesn t exist) Example: In the piecewise defined function above, lim f(x) = lim (x+5) = 11 x 3 x 3 and lim f(x) = lim (x 1) = 8 x 3 + x 3 + Since the left and right hand limits are not equal, lim f(x) Does Not Exist ( ) x 3 Thm lim f(x) = L iff lim f(x) = lim f(x) = L x c x c x c + (A limit only exists if the left and right hand limits are equal, and it equals that value) Example: 3x 5 x f(x) = consider the limit as x x 1 x > lim f(x) = lim (3x 5) = 1 lim f(x) = lim (x 1) = 1 x x x + x + Since the left and right hand limits are both 1, lim f(x) = 1 x Do p. 66, 67 # s 37 44, 51 54
The Sandwich Theorem THM The Sandwich Theorem if g(x) f(x) h(x) for all x c in some interval about c, and lim g(x) = lim h(x) = L, then lim f(x) = L x c x c x c Example: Show that lim x sin(1/x) = 0 x 0 (note: substitution won t work here, since sin(1/x) is bouncing up and down faster and faster as x 0) We know: 1 sin (1/x) 1 (sine of any angle is always between 1 and 1) Multiply by x : x x sin(1/x) x take limits: lim x lim x sin(1/x) lim x x 0 x 0 x 0 Evaluate : 0 lim x sin(1/x) 0 x 0 Conclusion: Therefore, by the Sandwich Theorem, lim x sin(1/x) = 0 x 0 Using the Sandwich Theorem, we can derive an important limit: lim sinθ = 1 θ 0 θ Please watch the following Khan Academy video (note: They call the Sandwich Theorem the Squeeze Theorem, it s the same thing). Here is the link. https://www.khanacademy.org/math/differential calculus/limits topic/squeeze theorem/v/proof lim sin x x In this video, we established that lim θ 0 sinθ = 1, Notice that the angle in the numerator must match the quantity in the denominator θ Example: lim sin 3x x 0 x We need the angle to match the denominator, but we can t touch the angle. So we multiply numerator and denominator by 3 and we get: lim 3 sin 3x x 0 3x, Which is 3*1 = 3. Do page 66 # s 4 8, page 68 # s 59 6 and the following worksheet.
Name Date Calculus.1 Show all work neatly. Evaluate the limit if it exists. If a limit does not exist, indicate with. Trig limits 1) lim sin(9 x ) 5) lim x 0 9x x 0 cos 4x x ) lim sin 3x 6) lim x 0 x x 0 sin 9x x x 5 3) lim 8x sin5x 7) lim x 0 4x x 0 sin 6x sin 7x 4) lim x 0 sin 7 7x x Given: f(x) = x sin3x x sin 5x x x<0 0 < x < 1 cosx x > Find: 8) lim f(x) 11) lim f(x) x 0 x 9) lim f(x) 1) lim f(x) x 0 + x π + 10) lim f(x) 13) lim f(x) x 0 x π
Section. Limits involving Infinity It is important to remember that infinity ( ) is not a number, but represents the idea that something is growing without bound. Def Horizontal Asymptote the line y = b is a horizontal asymptote of a function y = f(x) if either lim f(x) = b or lim f(x) = b x x Looking for horizontal asymptotes taking limits as x Although is not a number, it is useful to think of it as the reciprocal of 0. THM lim 1 = 0 x x So, when evaluating limits as x, we want to get all of our x s in denominators. 4 3 Example: 3x 5x x 1 lim 4 x 7x 4x 9 In order to move all the x s to denominators, we multiply every term by 1/x 4, since x 4 is the highest power of x in the problem. 4 3 1 (3x 5x x 1) 4 lim x this results in: 5 1 3 3 4 x 4 1 (7x 4x 9) lim x x x 4 x x 4 9 7 4 x x Using the theorem above, together with our sum/difference, constant multiple and power rules for limits, makes any term with an x in the denominator go to 0 as x. And so the limit is 3, and the 7 equation of the horizontal asymptote to the function is y = 3 7. Notice, that as x, only the highest power in a func on ma ers. If the highest power is in the denominator, the limit of the function as x = 0, and the x axis (y = 0) is a horizontal asymptote of that function. If the highest power is in the numerator, the limit of the function as x = (which is not a number, and so the limit does not exist), and the function has no horizontal asymptotes. If the highest power exists both in the numerator and the denominator (it s a tie), then the, the limit of the function as x will equal the ra o of the coefficient of the highest power term in the numerator over the highest power term in the denominator, and y = (that ratio) is the equation of the horizontal asymptote of the function, (as it was in the example above).
But what about sine and cosine functions? The Sandwich Theorem Revisited: find lim x x 5cosx 3x 6x 1 We know: 1 cos x 1 Mult by 5: 5 5 cos x 5 Add x : x 5 x + 5 cos x x + 5 Divide by 3x + 6x +1 : x 5 x + 5 cos x x + 5 3x + 6x +1 3x + 6x +1 3x + 6x +1 Take limits: : lim x 5 lim x + 5 cos x lim x + 5 x 3x + 6x +1 x 3x + 6x +1 x 3x + 6x +1 In the first and third limits, the highest power is a tie, and so they are both /3 lim x + 5 cos x 3 x 3x + 6x +1 3 Conclusion: Therefore, by the Sandwich Theorem, lim x + 5 cos x = /3 x 3x + 6x +1 So basically, we can treat sines and cosines as 1 s when x is approaching infinity. Read pages 70 75 Do p. 76 # s 3 6 and the following worksheet
Name Calculus AP Limits x For part a of each problem, determine the limit, if it exists. If a limit fails to exist, indicate with. For part b of each problem, give the equation of the horizontal asymptote of the function. If there is no horizontal asymptote, write none for part b 1) lim x 3 a) x 5x 7 b) x 1 ) lim 3 a) x 3 x b) 3) lim 1 1x a) x 4x 3x 1 b) 4) lim x 3 + 4x+ 3 a) x 5x 3 b) 5) lim 5 4 10x x 31 a) x 6 x b) 6) lim (x 3) a) x 5x x 1 b) 7) lim 5x sin(3x) + a) x 3x 4x + 7 b) 8) lim 5x 3 + 8x 4 cosx a) x 3x 4 16x 3 b)