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CALCULUS For Business, Economics, and the Social and life Sciences Hoffmann, L.D. & Bradley, G.L. Instructor: Dr. Guangqing Lu TA: Niu Qiang Office: E409-R4 Tel: 3620635(O) 156-028-63738 (Cell) Email: guangqinglu@uic.edu.hk Uinted International College 1 They invented Calculus! Sir Isaac Newton Gottfriend Wilhelm von Leibniz (1642-1727) (1646-1716) 2 Calculus has practical applications, such as understanding the true meaning of the infinitesimals. (Image concept by Dr. Lachowska.) 3 What is Calculus all about? Calculus is the study of change, or to be more precise, changing quantities. The two key areas of Calculus are Differential Calculus and Integral Calculus. The big surprise is that these two seemingly unrelated areas are actually connected via the Fundamental Theorem of Calculus. 4 Why do I have to take Calculus? If you choose a career path that requires any level of mathematical sophistication (such as Engineering, Economics or any branch of Science), you will need Calculus. To succeed in Calculus (and Mathematics in general) you must be able to: solve problems, reason logically and understand abstract concepts. The ability to master these skills is a key indicators of success in future academic endeavors. 5 What can I do to maximize my chances for success? Top five 1. Understand, don't memorize. 2. Ask why, not how. 3. See every problem as a challenge. 4. Learn techniques, not results. 5. Make sure you understand each topic before going on to the next. 6 Some Suggestions It is recommended that you spend at least 2 hours study and review, for each hour of class time. You must pay more attention to weekly tutorials where you can work on problems and get help from your demonstrator. 7 20% Home Works Little Quizzes 50% 10% Mid-term Examination Final Examination 20% 8 Chapter 1 Functions, Graphs and Limits In this Chapter, we will encounter some important concepts. Functions Limits One-sided Limits and Continuity 9 Section 1.1 Functions A function is a rule that assigns to each object in a set A exactly one object in a set B. The set A is called the domain of the function, and the set of assigned objects in B is called the range. 10 Function, or not? f A B YES f A f B A B NO NO Section 1.1 Functions 11 To be convenient, we represent a functional relationship by an equation y f (x) In this context, x and y are called variable, furthermore, we refer to y as the dependent variable and to x as the independent variable. For instant, the function representation y f ( x) x 2 4 Noted that x and y can be substituted by other letters. For example, the above function can be represented by s t 2 4 Section 1.1 Functions 12 Function that describes tabular data Table 1.1 Average Tuition and Fees for 4-Year Private Colleges Academic Year Tuition and Ending in Period n Fees 1973 1 $1,898 1978 2 $2,700 1983 3 $4,639 1988 4 $7,048 1993 5 $10,448 1998 6 $13,785 2003 7 Section 1.1 Functions $18,273 13 So how can we use functional notation to describe this kind of data that is listed in Table 1.1? We can describe this data as a function f defined by the rule average tuition and fees at the f (n) beginning of the nth 5 - year period Thus, f (1) 1,898, f (2) 2,700,.......,f (7) 18,273 Noted that the domain of f is the set of integers A {1,2,...., } 7 Section 1.1 Functions 14 Piecewise-defined function A piecewise-defined function is such a function that is often defined using more than one formula, where each individual formula describes the function on a subset of the domain. Here is an example of such a function 1 if x 1 f ( x) x 1 3x 2 1 if x 1 Section 1.1 Functions 15 Example 1 Find f(-1/2), f(1), and f(2), where the piecewise-defined function f(x) is given at the foregoing slide. Solution: 1 Since x satisfies x<1, use the top part of the formula 2 to find 1 1 1 2 f 2 1/ 2 1 3 / 2 3 However, x=1 and x=2 satisfy x≥1, so f(1) and f(2) are both found by using the bottom part of the formula: f (1) 3(1) 2 1 4 and f (2) 3(2) 2 1 13 Section 1.1 Functions 16 Domain Convention We assume the domain of f to be the set of all numbers for which f(x) is defined (as a real number). We refer to this as the natural domain of f. In general, there are two situations where a number is not in the domain of a function: 1) division by 0 2) The even number root of a negative number Section 1.1 Functions 17 Example 2 Find the domain and range of each of these functions 1 a. f ( x) b. g (u ) 4 u 2 1 x 2 Solution: a. Since division by any number other than 0 is possible, the domain of f is the set of all numbers except -1 and 1. The range of f is the set of all numbers y except 0. b. Since negative numbers do not have real fourth roots, so the domain of g is the set of all numbers u such as u≥-2. The range of g is the set of all nonnegative numbers. Section 1.1 Functions 18 Functions Used in Economics A demand function p=D(x) is a function that relates the unit price p for a particular commodity to the number of units x demanded by consumers at that price. The total revenue is given by the product R(x)=(number of items sold)(price per item) =xp=xD(x) If C(x) is the total cost of producing the x units, then the profile is given by the function P(x)=R(x)-C(x)=xD(x)- C(x) Section 1.1 Functions 19 Example 3 Market research indicates that consumers will buy x thousand units of a particular kind of coffee maker when the unit price is p 0.27 x 51 dollars. The cost of producing the x thousand units is C ( x) 2.23 x 2 3.5 x 85 thousand dollars a. What are the revenue and profile functions, R(x) and P(x), for this production process? b. For what values of x is production of the coffee makers profitable? Section 1.1 Functions 20 Solution: a. The demand function is D( x) 0.27 x 51, so the revenue is R( x) xD( x) 0.27 x 2 51 x thousand dollars, and the profit is (thousand dollars) P( x) R( x) C ( x) 0.27 x 2 51x (2.23x 2 3.5 x 85) 2.5 x 2 47.5 x 85 b. Production is profitable when P(x)>0. We find that P ( x ) 2.5 x 2 47.5 x 85 2.5( x 2 19 x 34) 2.5( x 2)( x 17) 0 Thus, production is profitable for 2<x<17. Section 1.1 Functions 21 Composition of Functions Composition of Functions: Given functions f(u) and g(x), the composition f(g(x)) is the function of x formed by substituting u=g(x) for u in the formula for f(u). Example 4 Find the composition function f(g(x)), where f (u ) u 1 and 3 g ( x) x 1 Solution: Replace u by x+1 in the formula for f(u) to get f ( g ( x)) ( x 1)3 1 x 3 3x 2 3x 2 Question: How about g(f(x))? Note: In general, f(g(x)) and g(f(x)) will not be the same. Section 1.1 Functions 22 Example 5 An environmental study of a certain community suggests that the average daily level of carbon monoxide in the air will be c( p) 0.5 p 1 parts per million when the population is p thousand. It is estimated that t years from now the population of the community will be p(t ) 10 0.1t 2 thousand. a. Express the level of carbon monoxide in the air as a function of time. b. When will the carbon monoxide level reach 6.8 parts per million? Section 1.1 Functions 23 Solution: a. Since the level of carbon monoxide is related to the variable p by the equation c( p) 0.5 p 1 , and the variable p is related to the variable t by the equation p(t ) 10 0.1t 2 It follows that the composite function c( p(t )) c(10 0.1t 2 ) 0.5(10 0.1t 2 ) 1 6 0.05t 2 expresses the level of carbon monoxide in the air as a function of the variable t. b. Set c(p(t)) equal to 6.8 and solve for t to get 6 0.05t 2 6.8 0.05t 2 0.8 t 2 16 t4 t 4 is not a natural solution. That is, 4 years from now the level of carbon monoxide will be 6.8 parts per million. 24 Section 1.2 The Graph of a Function The graph of a function f consists of all points (x,y) where x is in the domain of f and y=f(x), that is, all points of the form (x,f(x)). Rectangular coordinate system, Horizontal axis, vertical axis. The below example shows that the function can be sketched by plotting a few points. f ( x) x x 2 2 x -3 -2 -1 0 1 2 3 4 f(x) -10 -4 0 2 2 0 -4 -10 25 Intercepts x intercepts: The points where a graph crosses the x axis. A y intercept: A point where the graph crosses the y axis. How to find the x and y intercepts: The only possible y intercept for a function is y0 f (0) , to find any x intercept of y=f(x), set y=0 and solve for x. Note: Sometimes finding x intercepts may be difficult. Following above example, the y intercept is f(0)=2. To find the x intercepts, solve the equation f(x)=0, we have x=-1 and 2. Thus, the x intercepts are (-1,0) and (2,0). Section 1.2 The Graph of a 26 Function Parabolas Parabolas: The graph of y Ax 2 Bx C as long as A≠0. All parabolas have a “U shape” and the parabola opens up if A>0 and down if A<0. The “peak” or “valley” of the parabola is called its vertex, and it always occurs where x B 2A 27 Example 6 A manufacturer determines that when x hundred units of a particular commodity are produced, they can all be sold for a unit price given by the demand function p=60-x dollars. At what level of production is revenue maximized? What is the maximum revenue? Solution: The revenue function R(x)=x(60-x) hundred dollars. Note that R(x) ≥0 only for 0≤x≤60. The revenue function can be rewritten as R ( x ) x 2 60 x which is a parabola that opens downward (Since A=-1<0) and has its high point (vertex) at x B 60 30 2A 2( 1) Thus, revenue is maximized when x=30 hundred units are produced, and the corresponding maximum revenue is R(30)=900 hundred dollars. Section 1.2 The Graph of a 28 Function Intersections of Graphs Sometimes it is necessary to determine when two functions are equal. For example, an economist may wish to compute the market price at which the consumer demand for a commodity will be equal to supply. Section 1.2 The Graph of a 29 Function Power Functions, Polynomials, and Rational Functions A Power Function: A function of the form f ( x) x , where n is n a real number. A Polynomial Function: A function of the form p ( x ) an x n an 1 x n 1 a1 x a0 where n is a nonnegative integer and a0 , a1 , , an are constants. If an 0 , the integer n is called the degree of the polynomial. p( x) A Rational Function: A quotient q( x) of two polynomials p(x) and q(x). 30 The Vertical Line Test The Vertical Line Test: A curve is the graph of a function if and only if no vertical line intersects the curve more than once. 31 Section 1.3 Linear Functions A linear function is a function that changes at a constant rate with respect to its independent variable. The graph of a linear function is a straight line. The equation of a linear function can be written in the form y mx b where m and b are constants. 32 The Slope of a Line The Slope of a Line: The slope of the non-vertical line passing through the points ( x1 , y1 ) and ( x2 , y2 ) is given by the formula change in y y y2 y1 Slope change in x x x2 x1 Sign and magnitude of slope 33 Forms of the equation of a line The Slope-Intercept Form: The equation y mx b is the equation of a line whose slope is m and whose y intercept is (0,b). The Point-Slope Form: The equation y y0 m( x x0 ) is an equation of the line that passes through the point ( x0 , y0 ) and that has slope equal to m. (0 0.5) 1 m ( 1.5 0) 3 The slope-Intercept form is 1 1 y x 3 2 The point-slope form that passes through the point (-1.5,0) is 1 y 0 ( x 1.5) Section 1.3 Linear Functions 3 34 Example 7 Table 1.2 lists the percentage of the labour force that was unemployed during the decade 1991-2000. Plot a graph with the time (years after 1991) on the x axis and percentage of unemployment on the y axis. Do the points follow a clear pattern? Based on these data, what would you expect the percentage of unemployment to be in the year 2005? Table 1.2 Percentage of Civilian Unemployment Number of Years Percentage of Year from 1991 Unemployed 1991 0 6.8 1992 1 7.5 1993 2 6.9 1994 3 6.1 1995 4 5.6 1996 5 5.4 1997 6 4.9 1998 7 4.5 1999 8 4.2 2000 9 4.0 Section 1.4 Functional Models 35 Solution: The pattern does suggest that we may be able to get useful information by finding a line that “best fits” the data in some meaningful way. One such procedure, called “least- squares approximation”, require the approximating line to be positioned so that the sum of squares of vertical distances from the data points to the line is minimized. It produces the “best-fitting line” y 0.389x 7.338 . Based on this formula, we can attempt a prediction of the unemployment rate in the year 2005: y(14) 0.389(14) 7.338 1.892 Note: Care must be taken when making predictions by extrapolating from known data, especially when the data set is as small as the one 36 in this example. Parallel and Perpendicular Lines Let m1 and m2 be the slope of the non-vertical lines L1 and L2 . Then L1and L2 are parallel if and only if m1 m2 1 L1and L2 are perpendicular if and only if m2 m1 Section 1.3 Linear Functions 37 Example 8 Let L be the line 4x+3y=3 a. Find the equation of a line L1 parallel to L through P(-1,4). b. Find the equation of a line L2 perpendicular to L through Q(2,-3). Solution: By rewriting the equation 4x+3y=3 in the slope-intercept form 4 4 y x 1, we see that L has slope mL 3 3 a. Any line parallel to L must also have slope -4/3. The required line L1 contains P(-1,4), we have y 4 4 ( x 1) y 4 x 8 3 3 3 b. A line perpendicular to L must have slope m=3/4.Since the required line L2 contains Q(2,-3), we have y 3 3 ( x 2) 4 3 9 Section 1.4 Functional Models y x 38 4 2 Section 1.4 Functional Models To analyze a real world problem, a common procedure is to make assumptions about the problem that simplify it enough to allow a mathematical description. This process is called mathematical modelling and the modified problem based on the simplifying assumptions is called a mathematical model. Real-world problem Formulation adjustments Mathematical Testing model Analysis Prediction Interpretation 39 Elimination of Variables In next example, the quantity you are seeking is expressed most naturally in term of two variables. We will have to eliminate one of these variables before you can write the quantity as a function of a single variable. Example 9 The highway department is planning to build a picnic area for motorists along a major highway. It is to be rectangular with an area of 5,000 square yards and is to be fenced off on the three sides not adjacent to the highway. Express the number of yards of fencing required as a function of the length of the unfenced side. Section 1.4 Functional Models 40 Solution: We denote x and y as the lengths of the sides of the picnic area. Expressing the number of yards F of required fencing in terms of these two variables, we get F x 2 y . Using the fact that the area is to be 5,000 square yards that is xy 5,000 y 5000 x and substitute the resulting expression for y into the formula for F to get F ( x) x 2 5000 x 10000 x x Section 1.4 Functional Models 41 Proportionality The quantity Q is said to be: directly proportional to x if Q=kx for some constant k inversely proportional to x if Q=k/x for some constant k jointly proportional to x and y if Q=kxy for some constant k For example, When environmental factors impose an upper bound on its size, population grows at a rate that is jointly proportional to its current size and the difference between its current size and the upper bound. Let p denote the size of the population, R(p) the rate of population growth, and b the upper bound. So we have where Models R(p)=kp(b-p)1.4 Functionalk is a constant. Section 42 Modelling in Business and Economics Example 10 A manufacturer can produce blank videotapes at a cost of $2 per cassette. The cassettes have been selling for $5 a piece. Consumers have been buying 4000 cassettes a month. The manufacturer is planning to raise the price of the cassettes and estimates that for each $1 increase in the price, 400 fewer cassettes will be sold each month. a: Express the manufacturer’s monthly profit as a function of the price at which the cassettes are sold. b: Sketch the graph of the profit function. What price corresponds to maximum profit? What is the maximum profit? Section 1.4 Functional Models 43 Solution: a. As we know, Profit=(number of cassettes sold)(profit per cassette) Let p denotes the price at which each cassette will be sold and let P(p) be the corresponding monthly profit. Number of cassettes sold =4000-400(number of $1 increases) =4000-400(p-5)=6000-400p Profit per cassette=p-2 The total profit is P( p) (6000 400 p)( p 2) 400 p 2 6800 p 12000 Section 1.4 Functional Models 44 b. The graph of P(p) is the downward opening parabola shown in the bottom figure. Profit is maximized at the value of p that corresponds to the vertex of the parabola. B 6800 We know p 8. 5 2A 2(400 ) Thus, profit is maximized when the manufacturer charges $8.50 for each cassette, and the maximum monthly profit is Pm ax P(8.5) 400 (8.5) 2 6800 (8.5) 12000 $16900 Section 1.4 Functional Models 45 Market Equilibrium The law of supply and demand: In a competitive market environment, supply tends to equal demand, and when this occurs, the market is said to be in equilibrium. The demand function: p=D(x) The supply function: p=S(x) The equilibrium price: pe D( xe ) S ( xe ) Shortage: D(x)>S(x) Surplus: S(x)>D(x) Example 11 Market research indicates that manufacturers will supply x units of a particular commodity to the marketplace when the price is p=S(x) dollars per unit and that the same number of units will be demanded by consumers when the price is p=D(x) dollars per unit, where the supply and demand functions are given by S ( x) x 2 14 D( x) 174 6 x a. At what level of production x and unit price p is market equilibrium achieved? b. Sketch the supply and demand curves, p=S(x) and p=D(x), on the same graph and interpret. Section 1.4 Functional Models 47 Solution: a. Market equilibrium occurs when S(x)=D(x), we have x 2 14 174 6 x ( x 10)( x 16) 0 x 10 or 16 Only positive values are meaningful, pe D(10 ) 174 6(10 ) 114 Section 1.4 Functional Models 48 Break-Even Analysis At low levels of production, the manufacturer suffers a loss. At higher levels of production, however, the total revenue curve is the higher one and the manufacturer realizes a profit. Break-even point : The total revenue equals total cost. Section 1.4 Functional Models 49 Example 12 A manufacturer can sell a certain product for $110 per unit. Total cost consists of a fixed overhead of $7500 plus production costs of $60 per unit. a. How many units must the manufacturer sell to break even? b.What is the manufacturer’s profit or loss if 100 units are sold? c.How many units must be sold for the manufacturer to realize a profit of $1250? Solution: If x is the number of units manufactured and sold, the total revenue is given by R(x)=110x and the total cost by C(x)=7500+60x Section 1.4 Functional Models 50 a. To find the break-even point, set R(x) equal to C(x) and solve 110x=7500+60x, so that x=150. It follows that the manufacturer will have to sell 150 units to break even. b. The profit P(x) is revenue minus cost. Hence, P(x)=R(x)-C(x)=110x-(7500+60x)=50x-7500 The profit from the sale of 100 units is P(100)=-2500 It follows that the manufacturer will lose $2500 if 100 units are sold. c. We set the formula for profit P(x) equal to 1250 and solve for x, we have P(x)=1250, x=175. That is 175 units must be sold to generate the desired profit. Section 1.4 Functional Models 51 Example 13 A certain car rental agency charges $25 plus 60 cents per mile. A second agency charge $30 plus 50 cents per mile. Which agency offers the better deal? Solution: Suppose a car is to be driven x miles, then the first agency will charge C1 ( x) 25 0.60 xdollars and the second will charge C2 ( x) 30 0.50 x . So that x=50. For shorter distances, the first agency offers the better deal, and for longer distances, the second agency is better. Section 1.4 Functional Models 52 Section 1.5 Limits Roughly speaking, the limit process involves examining the behavior of a function f(x) as x approaches a number c that may or may not be in the domain of f. To illustrate the limit process, consider a manager who determines that when x percent of her company’s plant capacity is being used, the total cost is 8 x 2 636 x 320 C ( x) x 2 68 x 960 hundred thousand dollars. The company has a policy of rotating maintenance in such a way that no more than 80% of capacity is ever in use at any one time. What cost should the manager expect when the plant is operating at full permissible capacity? Functional Models Section 1.4 53 It may seem that we can answer this question by simply evaluating C(80), but attempting this evaluation results in the meaningless fraction 0/0. However, it is still possible to evaluate C(x) for values of x that approach 80 from the left (x<80) and the right (x>80), as indicated in this table: x approaches 80 from the left → ←x approaches 80 from the right x 79.8 79.99 79.999 80 80.0001 80.001 80.04 C(x) 6.99782 6.99989 6.99999 7.000001 7.0000 7.00043 1 The values of C(x) displayed on the lower line of this table suggest that C(x) approaches the number 7 as x gets closer and closer to 80. The functional behavior in this example can be describe by xlim C( x) 7Limits 80 Section 1.5 54 Limit: If f(x) gets closer and closer to a number L as x gets closer and closer to c from both sides, then L is the limit of f(x) as x approaches c. The behavior is expressed by writing lim f ( x) L x c Section 1.5 Limits 55 Example 14 x 1 Use a table to estimate the limit lim x 1 x 1 Solution: x 1 Let f ( x) x 1 and compute f(x) for a succession of values of x approaching 1 from the left and from the right. x→ 1 ← x x 0.99 0.999 0.9999 1 1.00001 1.0001 1.001 f(x) 0.50126 0.50013 0.50001 0.499999 0.49999 0.49988 The numbers on the bottom line of the table suggest that f(x) approaches 0.5 as x approaches 1. That is x 1 lim Section 1.5 Limits0.5 x 1 x 1 56 It is important to remember that limits describe the behavior of a function near a particular point, not necessarily at the point itself. x) 4 Three functions for which lim f (Section 1.5 Limits 57 x 3 The figure below shows that the graph of two functions that do not have a limit as x approaches 2. Figure (a): The limit does not exist; Figure (b): The function has no finite limit as x approaches 2. Such so- discussed later. called infinite limits will be 1.5 Limits Section 58 Properties of Limits If lim f ( x) and lim g ( x) exist, then xc xc lim [ f ( x) g ( x)] lim f ( x) lim g ( x) xc xc xc lim [ f ( x) g ( x)] lim f ( x) lim g ( x) xc xc xc lim kf ( x) k lim f ( x) for any constant k xc xc lim [ f ( x) g ( x)] [ lim f ( x)][ lim g ( x)] xc xc xc f ( x) lim f ( x) xc lim [ ] if lim g ( x) 0 x c g ( x) lim g ( x) xc xc lim [ f ( x)] p [ lim f ( x)] p if [ lim f ( x)] p exists xc xc xc Section 1.5 Limits 59 For any constant k, lim k k and lim x c xc x c That is, the limit of a constant is the constant itself, and the limit of f(x)=x as x approaches c is c. Section 1.5 Limits 60 Computation of Limits Example 15 3x 3 8 Find (a) lim (3x 3 4 x 8) (b) lim x 1 x 0 x2 Solution: a. Apply the properties of limits to obtain lim (3x 4 x 8) 3 lim x 4 lim x lim 8 3(1)3 4(1) 8 9 x 1 3 x 1 3 x 1 x 1 b. Since lim ( x 2) 0, you can use the quotient rule for x 0 limits to get 3x 8 3 3 lim x 3 lim 8 0 8 lim x 0 x 0 4 x 0 x2 lim x lim 2 02 x 0 x 0 Section 1.5 Limits 61 Limits of Polynomials and Rational Functions: If p(x) and q(x) are polynomials, then lim p( x) p(c) x c and p ( x ) p (c ) lim if q(c) 0 x c q ( x ) q (c ) Example 16 x 1 Find lim x2 x2 Solution: The quotient rule for limits does not apply in this case since the limit of the denominator is 0 and the limit of the numerator is 3. So the limit of the quotient does not exist. Section 1.5 Limits 62 Indeterminate Form f ( x) lim f ( x) 0 If xc and xc lim g ( x) 0 , then is said to be lim x c g ( x) indeterminate. The term indeterminate is used since the limit may or may not exist. Example 17 x2 1 x 1 (a) Find lim 2 (b) Find lim x 1 x 3 x 2 x 1 x 1 Solution: x2 1 ( x 1)( x 1) x 1 2 a. lim 2 lim lim 2 x 1 x 3 x 2 x 1 ( x 1)( x 2) x 1 x 2 1 x 1 lim x 1 x 1 lim x 1 lim 1 1 b. lim x 1 x 1 x 1 ( x 1) x 1 x 1 ( x 1) x 1 x1 x 1 2 Section 1.5 Limits 63 Limits Involving Infinity Limits at Infinity: If the value of the function f(x) approach the number L as x increases without bound, we write lim f ( x) L x Similarly, we write lim f ( x) M x when the functional values f(x) approach the number M as x decreases without bound. Section 1.5 Limits 64 Reciprocal Power Rules: For constants A and k, with k>0 A A lim 0 and lim 0 x x k x x k Example 18 x2 Find lim x 1 x 2 x 2 Solution: x2 x2 / x2 lim 1 1 lim lim x 0.5 x 1 x 2 x 2 x 1 / x x / x 2 x / x 2 2 2 2 lim 1 / x lim 1 / x lim 2 0 0 2 2 x x x Section 1.5 Limits 65 Procedure for Evaluating a Limit at Infinity of f(x)=p(x)/q(x) Step 1. Divide each term in f(x) by the highest power xk that appears in the denominator polynomial q(x). Step 2. Compute xlim f ( x) or lim f ( x) using algebraic x properties of limits and the reciprocal rules. Exercise 3x 4 8 x 2 2 x lim sin( x) lim (Optional Question!) x 5x 4 1 x x Example 19 If a crop is planted in soil where the nitrogen level is N, then the crop yield Y can be modeled by the Michaelis- Menten function Y ( N ) AN N 0 BN where A and B are positive constants. What happens to crop yield as the nitrogen level is increased indefinitely? Section 1.5 Limits 66 Solution: We wish to compute AN lim Y ( N ) lim N N B N AN / N lim N B / N N / N A A lim N B / N 1 0 1 A Thus, the crop yield tends toward the constant value A as the nitrogen level N increases indefinitely. For this reason, A is called the maximum attainable yield. Section 1.5 Limits 67 Infinite Limits: If f(x) increases or decreases without bound as x→c, we have lim f ( x) or lim f ( x) xc xc x For example lim x2 ( x 2) 2 From the figure, we can guest that x lim x 2 ( x 2) 2 Section 1.5 Limits 68 Section 1.6 One-sided Limits and Continuity If f(x) approaches L as x tends toward c from the left (x<c), we write lim f ( x) L xc where L is called the limit from the left (or left- hand limit) Likewise if f(x) approaches M as x tends toward c from the right (x>c), then lim f ( x) M x c M is called the limit from the right (or right-hand limit.) 69 Example 20 For the function 1 x 2 if x 2 f ( x) 2 x 1 if x 2 evaluate the one-sided limits lim f ( x) and lim f ( x) x 2 x 2 Solution: Since f ( x) 1 x 2 for x<2, we have lim f ( x) lim (1 x 2 ) 3 x 2 x 2 Similarly, f(x)=2x+1 if x≥2, so lim f ( x) lim (2 x 1) 5 x 2 x 2 Section 1.6 One-sided Limits and 70 Continuity Existence of a Limit: The two-sided limit lim2 f ( x) x exists if and only if the two one-sided limits lim f ( x) x 2 and lim f ( x) exist and are equal, and then x 2 lim f ( x) lim f ( x) lim f ( x) x2 x2 x2 Notice that the limit of the piecewise-define function f(x) in example 20 does not exist, that is lim f ( x) does not exist, since lim f ( x) lim f ( x) x2 x 2 x 2 Section 1.6 One-sided Limits and 71 Continuity lim f ( x) does not exist! x1 Since the left and right hand limits are not equal. At x=1: lim f x 0 Left-hand limit x 1 lim f x 1 Right-hand limit x 1 f 1 1 value of the function Section 1.6 One-sided Limits and 72 Continuity lim f ( x) does exist! x2 Since the left and right hand limits are equal, However, the limit is not equal to the value of function. At x=2: lim f x 1 Left-hand limit x 2 lim f x 1 Right-hand limit x 2 f 2 2 value of the function Section 1.6 One-sided Limits and 73 Continuity lim f ( x) does exist! x3 Since the left and right hand limits are equal, and the limit is equal to the value of function. At x=3: x3 f x 2 lim Left-hand limit lim f x 2 Right-hand limit x 3 f 3 2 value of the function Section 1.6 One-sided Limits and 74 Continuity Nonexistent One-sided Limits A simple example is provide by the function f ( x) sin(1 / x) As x approaches 0 from either the left or the right, f(x) oscillates between -1 and 1 infinitely often. Thus neither one-sided limit at 0 exists. Section 1.6 One-sided Limits and 75 Continuity