Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 29 ppt

7.2 Left- and Right-Handed Limits; Sometimes the Approach Is Critical 261 lim x→0 + |x|=lim x→0 + x = 0, lim x→0 − |x|=lim x→0 − (−x) = 0, lim x→0 |x|=0. ◆ ◆ EXAMPLE 7.13 Let f(x)=|x|. (a) Sketch the graph of f and sketch the graph of f  . (b) Evaluate f  (1) and f  (−2). (c) Verify that f  (0) is undefined. SOLUTION (a) i. graph of f(x)=|x| ii. graph of f  (x) = d dx |x| 1 –1 f ′ f x x f(x) = |x| Figure 7.19 (b) For x>0f(x)=x,sof  (1)=1. For x<0f(x)=−x,sof  (−2)=−1. (c) Approaching the problem from a graphical viewpoint, we notice that the graph of f does not appear locally linear at x = 0. No matter how much the region around x = 0 is magnified the graph has a sharp corner there, so it will never look like a straight line. The slope of f is −1ifxapproaches zero from the left and +1ifxapproaches zero from the right. Analytically we can argue similarly. f  (0) = lim h→0 f(0+h) − f(0) h = lim h→0 |h|−|0| h = lim h→0 |h| h This is the limit we looked at in Example 7.11. We concluded that lim h→0 |h| h does not exist. Therefore d dx |x|    x=0 , the derivative of |x| evaluated at x = 0, does not exist. ◆ ◆ EXAMPLE 7.14 Find lim t→0 1 t 2 . SOLUTION lim t→0 + 1 t 2 =+∞;likewise, lim t→0 − 1 t 2 =+∞. 262 CHAPTER 7 The Theoretical Backbone: Limits and Continuity Therefore we say lim t→0 1 t 2 =+∞. f t f(t) = 1 t 2 Figure 7.20 ◆ The relationship between left- and right-hand limits and two-sided limits can be stated most succinctly as follows. lim x→a f(x)=Lif and only if lim x→a + f(x)= lim x→a − f(x)=L. The limit as x approaches a of f(x)is L if and only if the left- and right-hand limits are both equal to L. In this section we’ve seen that if lim x→a − f(x)=L 1 and lim x→a + f(x)=L 2 but L 1 = L 2 , then lim x→a f(x) does not exist. This is not the only situation in which a limit does not exist. Consider, for example, the function g(x) that is 1 when x is rational and −1 when x is irrational. lim x→0 g(x) does not exist since in any open interval around zero, no matter how small, there will be both rational and irrational numbers. In this case neither the left- nor right-hand limit exists. Similarly, lim x→∞ g(x) does not exist, because for any N, no matter how big N is, there will always be both rational and irrational numbers larger than it. PROBLEMS FOR SECTION 7.2 1. Let f be the first-class postage function for 1997 given in Example 2.3. The input of the function is W , where W is the weight, in ounces, of the item to be mailed and the output is the price of the postage. Return to Chapter 2 for details. Find the following. (a) lim W →1 − f(W) (b) lim W →1 + f(W) (c) lim W →1 f(W) (d) lim W →1.5 f(W) (e) lim W →2 f(W) (f) lim W →12 + f(W) (g) lim W →12 − f(W) (h) lim W →12 f(W) In Problems 2 through 7, evaluate the limits. A graph may be useful. 2. (a) lim x→2 − 1 x−2 (b) lim x→2 + 1 x−2 (c) lim x→2 1 x−2 3. (a) lim x→1 − 1 (x−1) 2 (b) lim x→1 + 1 (x−1) 2 (c) lim x→1 1 (x−1) 2 (d) lim x→−1 1 (x−1) 2 4. f(x)= |x−3| x−3 7.2 Left- and Right-Handed Limits; Sometimes the Approach Is Critical 263 (a) lim x→0 f(x) (b) lim x→4 f(x) (c) lim x→3 + f(x) (d) lim x→3 − f(x) (e) lim x→3 f(x) 5. f(x)=  3x +4, x<0 2x+4, x ≥ 0 (a) lim x→1 f(x) (b) lim x→−2 f(x) (c) lim x→0 + f(x) (d) lim x→0 − f(x) (e) lim x→0 f(x) 6. f(x)=  πx + 1, x>0 πx − 1, x ≤ 0 (a) lim x→ 1 π f(x) (b) lim x→− 1 π f(x) (c) lim x→0 f(x) 7. f(x)=  x 2 +3, x ≥ 1 2x + 2, x<1 (a) lim x→0 f(x) (b) lim x→1 f(x) (c) lim x→2 f(x) 8. Let f be the function defined by f(x)=  x +1, for x not an integer, 0, for x an integer. (a) Sketch f . (b) Find the following limits. i. lim x→1.5 f(x) ii. lim x→2 f(x) iii. lim x→0 f(x) (c) For what values of c is lim x→c f(x)=c+1? Have you excluded any values of c? If so, which ones and why? Explain. 9. Let g(x) be the function defined by g(x) =  x 2 , for x>2, x, for x ≤ 2. (a) Find the following. i. lim x→2 + g(x) ii. lim x→2 − g(x) iii. lim x→2 g(x) (b) h(x) is defined by h(x) =  x 2 , for x>2, mx, for x ≤ 2. Find the value of m that will make f(x)acontinuous function. (A graph is highly recommended.) Using the value of m you’ve found, determine lim x→2 h(x). 10. Let f be the function given by f(x)=        1 x+7 , for x ≤−5, x + 4, for −5 <x<−2, |x|, for −2 ≤ x ≤ 2, 2 −x + 1, for x>2. 264 CHAPTER 7 The Theoretical Backbone: Limits and Continuity A sketch of f is provided below. x f 1 1 –1 –1 –2–3–4–5–6–72 2 3456789 Evaluate the limits below. (a) lim x→−∞ f(x) (b) lim x→−7 f(x) (c) lim x→−5 + f(x) (d) lim x→−5 f(x) (e) lim x→−2 f(x) (f) lim x→2 + f(x) (g) lim x→∞ f(x) (h) lim x→0 f(x) 11. Let f(x)be the function defined and graphed in Problem 10. (a) Sketch a graph of f  (x). (b) Where is f  undefined? (c) Evaluate the following. i. lim x→∞ f  (x) ii. lim x→−2 − f  (x) iii. lim x→−2+ f  (x) iv. lim x→−2 f  (x) v. lim x→−7 − f  (x) vi. lim x→−7 + f  (x) 12. (a) Sketch the graph of f(x)=2|x −2|. (b) What is f  (0)? What is f  (4)? What is f  (2)? (c) Find lim h→0 + f(2+h)−f(2) h and lim h→0 − f(2+h)−f(2) h . Do your answers make sense to you? (d) On a separate set of axes graph f  (x). Each of the statements in Problems 13 through 16 is false. Produce a counterexample to show that the statement is false. 13. If f(3)=7, then lim x→3 f(x)=7. 14. If f(2)is undefined, then lim x→2 f(x)is undefined. 15. If lim x→0 + f(x)and lim x→0 − f(x)both exist, then lim x→0 f(x)exists. 16. If lim x→5 f(x)=2then f(5)=2. For Problems 17 through 21, use the graph given to determine each of the following. (a) lim x→1 + f(x) (b) lim x→1 − f(x) (c) lim x→1 f(x) 7.3 A Streetwise Approach to Limits 265 17. 18. 19. f x 1 f x 1 2 f x 1 –1 –2 20. 21. f x 1 f x 7.3 A STREETWISE APPROACH TO LIMITS Suppose you and your trusty calculator are strolling down the street one night. As you pass a dark alley the Math Mugger jumps you, throwing you a nasty limit. A quick answer is demanded. Although normally cool and level-headed, you start to panic. You barely heard the question, but the words “zero,”“one,”“infinity,” and “does not exist” are at the tip of your tongue. You try to calm yourself, asking boldly, “Could you repeat the question, please?” Let’s suppose the question thrown at you is “What is lim x →1 x 3 −1 x −1 ?” First let’s look at potholes to avoid. Then we’ll supply some calculus street-survival tactics. Finally, we’ll determine whether using these survival tactics would, in this particular instance, get you out of the bind you’re in. Potholes and Mudholes with Crocodiles In your panic you might look at lim x →1 x 3 −1 x −1 and think, “when x → 1 the numerator tends toward 0, and if the numerator is 0 then the fraction is 0, so the answer is 0.” Alternatively, you might think, “when x → 1 the denominator tends toward 0, and when the denominator goes to 0 the limit does not exist or is infinite.” If you’ve had these two thoughts in rapid succession, you might put them together and think, “so we have 0 0 , which is undefined whenever it’s not 1.” 3 There are serious problems with these lines of thought. While it is true that both the numerator and the denominator of this fraction are approaching zero, we’re interested in 3 Notice that all this panicked thinking has put you in the same place you were before you heard the question. 266 CHAPTER 7 The Theoretical Backbone: Limits and Continuity what this ratio looks like as x → 1. Think about this carefully; every derivative we calculate looks like lim x→0 f x and therefore looks like 0 0 , yet this limit could work out to be any number, depending upon the function. To assert that every limit of the form 0 0 is either 1 or does not exist is equivalent to asserting that all derivatives are 1 or undefined, an assertion you know is patently false! Another approach you might take to lim x→1 x 3 −1 x−1 is to simplify the fraction x 3 −1 x−1 . That’s a solid idea. If you can factor x 3 − 1, you’re in great shape and don’t need help. It’s entirely possible, however, that you know how to factor the difference of two perfect squares but not of two perfect cubes. Do not resort to wishful-thinking algebra. If you’re not sure of your algebra, you’ve got means at your disposal to check it. If you’re not sure of a factorization, multiply out to determine whether it works. If you think you can simplify without factoring (you can’t), do a spot-check on yourself. For example, whatever you simplify this to, when you evaluate at x = 2 you should get 2 3 −1 2−1 = 7. Suppose you’re stuck; you can’t figure out how to simplify the fraction x 3 −1 x−1 and your panic level is mounting. 4 What you need are street-survival skills. Street-Survival Tactics The survival tactics are simple and can be applied in a large variety of situations, not just limit calculations in back alleys. Investigate the situation graphically. Investigate the situation numerically. The potholes identified above are analytic and algebraic missteps. Don’t put on analytic blinders; view the problem from different angles. Graphical and Numerical Investigations We’re interested in lim x→1 x 3 −1 x−1 , so we want the graph of f(x)= x 3 −1 x−1 around x = 1. For starters we might graph on the interval [0, 2] and then zoom in around x = 1. To make sure we have the graph in the viewing window, we can begin by letting the range of y-values be from f(0)= 0 3 −1 0−1 = 1tof(2)= 2 3 −1 2−1 = 7. We can zoom in around x = 1 as many times as we like. (As pointed out in Chapter 2, although f(x) is undefined at x = 1, the graph on the calculator may not reflect this.) Use the “trace” key to trace along the curve near x = 1. 4 For help simplifying, see Section 11.2 7.3 A Streetwise Approach to Limits 267 x = .997 x = 1.002 y = 3.008 y = 2.990 tracer Figure 7.21 It is reasonable to conjecture that as x → 1, x 3 −1 x−1 → 3. Another approach is to use your calculator to calculate values of x 3 −1 x−1 for values of x approaching 1. (For instance, you could try 0.99, 0.9997, 1.0001, 1.00003, etc., or you could get your calculator to produce a table of values.) Again, your investigation would lead you to conjecture that as x → 1, x 3 −1 x−1 → 3. So if the fellow in the back alley demands an answer immediately, tell him your best guess is 3. Is this answer correct? Actually it is. x 3 − 1 can be factored into (x − 1)(x 2 + x + 1), allowing us to do the following. 5 lim x→1 x 3 − 1 x − 1 = lim x→1 (x − 1)(x 2 + x + 1) x − 1 = lim x→1 (x 2 + x + 1) = 3 (Alternatively, this limit can be interpreted as the derivative of x 3 at x = 1 and calculated as lim h→0 (1+h) 3 −1 3 h . Do this calculation as an exercise.) Cautionary Notes If you’re going to use street-survival tactics, you’re going to have to learn how to watch your back. The tactics recommended will often get you near the actual answer (not always, but frequently, if used wisely), but there is no reason to believe that they will lead you to the exact answer. For instance, in the last example at some point we simply guessed the answer was 3 (as opposed to some number very, very close to 3). If the actual answer was irrational, the calculator would probably not tell us this. There are other limitations inherent in your trusty machine; sometimes the calculator can lead you astray. Suppose, for instance, we are interested in the derivative of 2 x at x = 0. It is given by lim h→0 2 h − 1 h . 5 Chapter 11 will give you more insight into polynomials and their factorization. 268 CHAPTER 7 The Theoretical Backbone: Limits and Continuity f x (0, 1) f(x) = 2 x (h, 2 h ) tangent line at x = 0 Figure 7.22 We can approximate this derivative by 2 h −1 h for h small. The approximation should be better and better the smaller h is. If h is positive, the value obtained should be greater than the actual limit. (See Figure 7.20.) The following are data collected from a calculator. h 2 h − 1 h 10 −2 = 0.01 0.6955550057 10 −3 = 0.001 0.693387463 10 −4 = 0.0001 0.6931712 10 −5 = 0.00001 0.6931496 10 −6 = 0.000001 0.693147 10 −7 = 0.0000001 0.69315 10 −8 = 0.00000001 0.6931 10 −9 = 0.000000001 0.693 10 −10 0.69 10 −11 0.7 10 −12 1 10 −13 0 10 −14 0 Is the limit actually 0, not ≈ 0.693? The actual value of lim h→0 2 h −1 h is an irrational number. This irrational number is approximately 0.6931471806. What is going on here? The information supplied by the calculator looks reasonable up until h = 10 −6 . For the values of h smaller than this, the difference between 2 h and 1 is so tiny that the calculator is essentially losing one digit of information for each subsequent entry up through 10 −12 . By h = 10 −13 and h = 10 −14 , the difference 2 h − 1 is so miniscule that the calculator just rounds the difference off to 0. This phenomenon is not a quirk particular to this calculator, nor is it particular to the function 2 x . Similar issues arise whenever calculating a limit that can be interpreted as a derivative. So what should you do? A streetwise person is not an extremist. When investigating lim x→a f(x), try values of x close to a but not too close. And then make your guess and take your chances. You’ll win some and you’ll lose some. But frequently, thoughtful graphical and numerical investigation will put you in the vicinity of the actual answer. 7.3 A Streetwise Approach to Limits 269 Question: When is lim x→a f(x)simply equal to f(a)? In the language of the streets, when can you just plug in x =a and get the right answer even though taking the limit means x can’t really be equal to a? 6 This is a great question. Think about it for a minute. We’ll recap Examples 7.3, 7.4, and 7.5 to give you food for thought. f x 3 f x 3 2 f(x) = x 2 lim f(x) = 3 2 x→3 x→3 x→3 f(x) = f(x) = x (x–3) 2 (x–3) lim f(x) = 3 2 x 2 lim f(x) = 3 2 f x 3 for x ≠ 3 for x = –3 Figure 7.23 If f is continuous at x = a then this ‘method’ will work. This brings us back from the streets to define continuity using limits. PROBLEMS FOR SECTION 7.3 1. Is lim x→∞ (1 + 1 x ) x finite? If so, find two consecutive integers, one smaller than this limit and the other larger. We will return to this limit later in the course. In Problems 2 and 3, find the limit. 2. lim x→3 2x 3 − 8x 2 + 5x + 3 x − 3 3. lim x→0 √ 4 + x − 2 x 4. (a) Suppose you are interested in finding lim x→2 f(x), where the function f(x) is explicitly given by a formula. What approaches might you take to investigate this limit? (b) Suppose you’re now interested in finding lim x→∞ f(x), where f(x) again is explicitly given by a formula. What approaches might you take to investigate this limit? 6 The more math you know the less you’ll have to rely on street tactics. 270 CHAPTER 7 The Theoretical Backbone: Limits and Continuity 7.4 CONTINUITY AND THE INTERMEDIATE AND EXTREME VALUE THEOREMS Our intuitive notion is that a function is continuous if its graph can be drawn without lifting pencil from paper. For f to be continuous at x = a the values of f(x)must be close to f(a) for x near a.Todefine continuity we’ll begin by looking at the type of behavior we are trying to rule out. The function sketched below is discontinuous at x =−3, x =−1, x = 0, x = 1, and x = 5.2. f x – 4 –2–3 –1123456 Figure 7.24 To begin with, we might insist that if f is to be continuous at x = a, then lim x→a − f(x)= lim x→a + f(x), where this limit is finite. For the function shown in Figure 2.24, this rules out the jump discontinuities at x =−3, x = 1, and x = 5 but doesn’t take care of the removable discon- tinuities at x =−1and x = 0. The latter reminds us that f(a)must be defined and equal to the left- and right-hand limits. In other words, the following conditions must be satisfied: i. f must be defined at x = a, ii. left- and right-hand limits at x = a must be equal, iii. f(a)must be equal to lim x→a f(x). Wecan put this more succinctly. Definition The function f is continuous at x = a if lim x→a − f(x)= lim x→a + f(x)=f(a). Equivalently, f is continuous at x = a if lim x→a f(x)=f(a). f is continuous on an open interval if f is continuous at every point in the interval. 7 There are two theorems concerning continuous functions that we will use repeatedly. 7 An open interval can be of the form (a, b), (a, ∞), (−∞, b) or (−∞, ∞). . lim x→a f(x), try values of x close to a but not too close. And then make your guess and take your chances. You’ll win some and you’ll lose some. But frequently, thoughtful graphical and numerical investigation. =−3, x = 1, and x = 5 but doesn’t take care of the removable discon- tinuities at x =− 1and x = 0. The latter reminds us that f(a)must be defined and equal to the left- and right-hand limits. In. between left- and right-hand limits and two-sided limits can be stated most succinctly as follows. lim x→a f(x)=Lif and only if lim x→a + f(x)= lim x→a − f(x)=L. The limit as x approaches a of f(x)is