Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 28 ppt

7.1 Investigating Limits—Methods of Inquiry and a Definition 251 beginning of this section, we will not be applying this definition much in practice; it is the meaning behind the definition that is of primary importance to us. f x L + L – L 3–δ 3+δ δ δδ δ 3 f x L 3–δ 3+δ3 For all x within of 3, f(x) is within of L. For all x within of 3, f(x) is well within of L. (i) (ii) ∈ L + ∈ ∈ L – ∈ ∈ ∈ ∈∈ Figure 7.7 ◆ EXAMPLE 7.4 Consider the function f(x)=  x 2 , for x = 3, undefined, for x = 3. 3 f x f(x) Figure 7.8 We can write this function compactly as f(x)= x(x−3) 2(x−3) , because x(x−3) 2(x−3) = x 2 for x = 3 and is undefined at x = 3. Find lim x→3 x(x−3) 2(x−3) . SOLUTION lim x→3 x(x − 3) 2(x − 3) = lim x→3 x 2 (provided x = 3) = 3 2 The argument given in Example 7.3 holds without alteration since we always worked with the condition x = 3. A single hole in the graph makes no difference in the limit. In fact, inserting any finite number of holes in the graph of a function has no effect on the computation of limits. ◆ ◆ EXAMPLE 7.5 Let f(x)=  x 2 , for x = 3, 2, for x = 3. Find lim x→3 f(x). 252 CHAPTER 7 The Theoretical Backbone: Limits and Continuity 3 f x f(x) Figure 7.9 SOLUTION Again lim x→3 f(x)= 3 2 since f(x)= x 2 for x = 3 and the arguments given above, both formal and informal, hold here as well. ◆ Notice that we’ve established that lim x→3 f(x)= 3 2 for each of the functions f(x) drawn below. The limit tells us about the behavior of f near x = 3 but not at x = 3. 3 f x 3 f x 3 f x Figure 7.10 ◆ EXAMPLE 7.6 Find lim x→0 4x+x 2 x . We’ re interested in the behavior of f(x)= 4x+x 2 x = (4+x)x x for x near zero but not at x = 0. For x = 0, (4+x)x x = 4 + x,solim x→0 (4+x)x x = lim x→0 4 + x = 4. More formally, we must show that for any positive , |f(x)−4|<provided |x| is small enough (and nonzero). For x = 0, we have: |f(x)−4|=     (4+x)x x − 4     =|4+x−4|=|x|, so if 0 < |x| <,then |f(x)−4|<as well. 4 f x f(x) = = (4 + x) x 4 + x for x ≠ 0 undefined for x = 0 x Figure 7.11 ◆ The limit in Example 7.6 is the limit that we would compute if we were computing the derivative of g(x) = x 2 at x = 2. 7.1 Investigating Limits—Methods of Inquiry and a Definition 253 g  (2) = lim h→0 g(2 + h) − g(2) h = lim h→0 (2 + h) 2 − 4 h = lim h→0 4 + 4h + h 2 − 4 h = lim h→0 (4 + h)h h This is the limit given in Example 7.6. = lim h→0 4 + h = 4 (2, 4) (2 + h, (2 + h) 2 ) g(x) = x 2 g x Figure 7.12 EXERCISE 7.1 Show that lim x→3 2x 2 − 6x x − 3 = 6 using the formal definition of a limit. Try this first on your own. The answer is provided below. Answer Let f(x)= 2x 2 −6x x−3 . We must show that for any positive , |f(x)−6|<provided |x − 3| is small enough but nonzero. If |x − 3| = 0 then |f(x)−6|=     2x(x − 3) x − 3 − 6     =|2x−6|=2|x−3|. Therefore, |f(x)−6|<provided 0 < |x − 3| <  2 . ◆ EXAMPLE 7.7 Interpret lim x→3 x 2 −9 x−3 as the derivative of some function K(x) evaluated at x = a. (a) Determine K(x) and the value of a. (b) Evaluate the limit. 254 CHAPTER 7 The Theoretical Backbone: Limits and Continuity SOLUTION (a) A derivative is the limit of the slope of the secant lines, so we want to interpret x 2 −9 x−3 as K x , the slope of the line through the points (x, x 2 ) and (3, 9). We see that the limit is being taken as x approaches 3, so the point (x, x 2 ) is approaching (3, 9) along the curve K(x) = x 2 . (3, 9) (x, x 2 ) K K(x) = x 2 x Figure 7.13 K(x) = x 2 and a = 3. That is, the derivative of x 2 at x = 3islim x→3 x 2 −9 x−3 . (b) There are several ways of evaluating this limit. Method (i): In part (a) we showed that lim x→3 x 2 −9 x−3 is the derivative of x 2 at x = 3. In Chapter 6 we showed that the derivative of x 2 is 2x. Evaluating this at x = 3gives lim x→3 x 2 −9 x−3 = 6. Method (ii): Following the reasoning of the previous few examples, we can observe that x 2 − 9 x − 3 = (x − 3)(x + 3) x − 3 =  x + 3, for x = 3, undefined, for x = 3. So lim x→3 x 2 − 9 x − 3 = lim x→3 (x − 3)(x + 3) x − 3 = lim x→3 x + 3 since x − 3 x − 3 = 1 for x = 3 = 6. Alternative ways of approaching the limit: Suppose you’re staring at this limit and feel at a loss, or for some reason are feeling queasy about your answer. Or maybe you just want to double-check your work. You can investigate the limit graphically or investigate it numerically. 6 3 3 x 2 – 9 x – 3 x 2 – 9 x – 3 y y = x x = x + 3 for x ≠ 3 undefined for x = 3 2.99 2.999 3.001 3.01 5.99 5.999 6.001 6.01 Figure 7.14 7.1 Investigating Limits—Methods of Inquiry and a Definition 255 These methods lead you to conjecture that the limit is 6. ◆ ◆ EXAMPLE 7.8 Find lim x→∞ x 2 . SOLUTION We are being asked about the behavior of x 2 as x grows without bound. As x grows without bound, x 2 grows without bound; about these facts there should be no controversy. The problem is how we should answer the question. Reasonable people may differ. Some mathematicians don’t like to write lim x→∞ x 2 =∞. After all, infinity is not something that f(x)=x 2 can snuggle up to; f(x) can’t get arbitrarily close to infinity. On the other hand, “as x →∞, x 2 →∞” is a reasonable shorthand for “as x grows without bound, x 2 grows without bound.” Therefore in this text we use the convention that lim x→∞ x 2 =∞ is a shorthand for “as x grows without bound, x 2 grows without bound.” Similarly, lim x→a f(x)=∞is shorthand for “as x approaches a, f(x)grows without bound”. ◆ Before moving on from the definition of a limit, let’s take a moment to debunk a common misconception with the next example. ◆ EXAMPLE 7.9 Find lim x→0 x x . The graph of x x is given below. x x = 1 for x = 0 and is undefined for x = 0. lim x→0 x x = lim x→0 1 = 1 1 x x y y = x = 1 for x ≠ 0 undefined for x = 0 Figure 7.15 ◆ From this example we see that lim x→a f(x)=Ldoes not mean “as x gets closer and closer to a, f(x) gets closer and closer to L but never reaches L”; in Example 7.9 f(x) does reach L = 1asxapproaches 0. It is possible (but not required) that f will hit the value L any number of times. EXERCISE 7.2 Show, using the methods of Example 7.3, that lim x→3 k = k for any constant k. Conclude that lim x→c k = k for c and k any constants. EXERCISE 7.3 Show, using the methods of Example 7.3, that lim x→c x = c, where c is any constant. 256 CHAPTER 7 The Theoretical Backbone: Limits and Continuity PROBLEMS FOR SECTION 7.1 1. Evaluate the following limits; then discuss lim x→∞ b x for b>0. (a) lim x→∞ (1.1) x (b) lim x→∞ (0.9) x (c) lim x→0 (1.1) x (d) lim x→−∞ (1.1) x (e) lim x→−∞ (0.9) x 2. We can define  − 1 2  n for any positive integer n, but not for every real number. For instance,  − 1 2  1/2 =  − 1 2 , which is not defined in real numbers. We’ll write lim n→∞  − 1 2  n = L if  − 1 2  n can be made arbitrarily close to L for all positive integers n sufficiently large. (a) Find lim n→∞  − 1 2  n , where n takes on only positive integer values. (b) Which two of the following statements are true? Explain. i. lim n→∞ (−2) n =∞ ii. lim n→∞ (−2) n =−∞ iii. lim n→∞ (−2) n does not exist iv. lim n→∞ −2 n =∞ v. lim n→∞ −2 n =−∞ 3. Each of the following limits are of the form lim x→a f(x).Evaluate the limit and sketch a graph of f on some interval including a. Make it clear from your sketch whether or not f is defined at a. (a) lim h→−2 (h − 3)(h + 2) h + 2 (b) lim x→5 x 2 − 25 x + 5 (c) lim x→−5 x 2 − 25 x + 5 (d) lim t→0 t 2 + πt t (e) lim h→0 hk + h 2 h , where k is a constant (f) lim w→2 (w − 3)(w + 1)(w − 2) 3w − 6 4. Discussion Question. Consider lim x→∞ (1 + 1 x ) x . This is a very important limit. (a) What is your guess for this limit? You are not expected to guess the right answer. Once you complete part (b) you will see that the problem is subtle. (b) Use a calculator or computer to investigate the limit, graphically and numerically. Give a revised estimate of this limit. We will return to this limit in Chapter 15. In Problems 5 through 13, graph f and evaluate the limit(s). 5. f(x)= x 2 + 3; lim x→2 f(x) 6. f(x)=πx − 4; (a) lim x→0 f(x) (b) lim x→1 f(x) (c) lim x→∞ f(x) 7.1 Investigating Limits—Methods of Inquiry and a Definition 257 7. f(x)=|x−2|; (a) lim x→0 f(x) (b) lim x→2 f(x) 8. f(x)= x 2 −2x x − 2 ; (a) lim x→0 f(x) (b) lim x→2 f(x) 9. f(x)=  5x +1, x = 2 7, x = 2 ; (a) lim x→1 f(x) (b) lim x→2 f(x) 10. f(x)= (x + 2)(x 2 − x) x(x −1) ; (a) lim x→0 f(x) (b) lim x→1 f(x) 11. f(x)= x 2 −3x −4 x + 1 ; (a) lim x→1 f(x) (b) lim x→−1 f(x) 12. f(x)=  |x|, x =3 0, x = 3 ; (a) lim x→0 f(x) (b) lim x→3 f(x) (c) lim x→−∞ f(x) 13. f(x)= x 2 −9 x + 3 ; (a) lim x→3 f(x) (b) lim x→−3 f(x) In Problems 14 through 18, evaluate the limit by interpreting it as the derivative of some function f evaluated at x = a. Specify f and a, then calculate the limit. 14. lim x→2 e x − e 2 x − 2 15. lim h→0 √ 9 + h − 3 h 16. lim h→0 e 1+h − e h 17. lim h→0 √ 7 + h − √ 7 h 18. lim h→0 e h − 1 h In Problems 19 and 20, give an example of a function having the set of characteristics specified. 258 CHAPTER 7 The Theoretical Backbone: Limits and Continuity 19. (a) lim x→5 f(x)=7; f(5)=7 (b) lim x→5 g(x) = 7; g(5) = 8 20. (a) lim x→∞ f(x)=∞; lim x→−∞ f(x)=−∞; lim x→0 f(x)=1 (b) lim x→∞ g(x) =∞; lim x→−∞ g(x) =∞; lim x→0 g(x) =−1 21. Look back at Example 7.6. When approximating the slope of x 2 at x = 2, we end up with the expression (4h + h 2 )/h. If we assume h = 0, then we can cancel the h’s, arriving at 4 + h. In this problem, we will investigate a function like (4h + h 2 )/h. Consider the following pay scale for employees at the nepotistic Nelson Nattle Company. Let D be the date of hire (we use the start-up date of the Nattle Company as our benchmark time, D = 0, D measured in years) and let S be the starting annual salary. S(D) =      (15000 + 200D)D(D − 1) D(D − 1) 50, 0000 for D = 0 35, 000 for D = 1 (a) Sketch the graph of S(D). The graph looks a little weird until you find out that Nelson Nattle is the only person hired at D = 0 and that his son, Nelson Nattle, Jr., is due back from college exactly one year after the Nattle Company’s start- up date. The expression (15000+200D)D(D−1) D(D−1) is equal to 15000 + 200D as long as D = 0 and D = 1. For D = 0 and D = 1, (15000+200D)D(D−1) D(D−1) is undefined. (b) Nelson has just received a letter from his brother Nathaniel asking for a position in the company. Nathaniel’s projected date of hire is D = 1.5. Nelson is thinking of offering his brother a starting salary of $40,000. Adjust S(D) to define S(1.5) appropriately. 22. Let f(t)= t(3+ t) t . (a) Sketch the graph of f(t). (b) What are the domain and range of f(t)? 7.2 LEFT- AND RIGHT-HANDED LIMITS; SOMETIMES THE APPROACH IS CRITICAL ◆ EXAMPLE 7.10 How does f(x)= 1 x behave as x approaches zero? SOLUTION Let’s begin by investigating this graphically and numerically. 7.2 Left- and Right-Handed Limits; Sometimes the Approach Is Critical 259 f f(x) = x x 1 x x 1 .01 .001 .0001 –.0001 –.001 –.01 100 1000 10000 –10000 –1000 –100 Figure 7.16 As x approaches zero from the right (i.e., through numbers greater than zero), 1 x grows without bound. We can express this by writing lim x→0 + 1 x =∞, where “x → 0 + ” indicates that only numbers greater than zero are being considered. We read “lim x→0 + ” as “the limit as x approaches 0 from the right.” As x approaches zero from the left (i.e., through numbers less than zero), 1 x decreases without bound. We can express this by writing lim x→0 − 1 x =−∞, where “x → 0 − ” indicates that only numbers less than zero are being considered. ◆ We can define finite one-sided limits in general as follows. Definitions If the values of f(x) stay arbitrarily close to L provided x is sufficiently close to a but greater than a, we write lim x→a + f(x)=L and read this as “the limit of f(x)as x approaches a from the right is L.” Analogously, if the values of f(x) can be made to stay arbitrarily close to L provided x is sufficiently close to a but less than a, then we write lim x→a − f(x)=L. These definitions can be made more precise in exactly the same manner the definition of lim x→a f(x)=L wasmade more precise. From that latter definition we see that in order to have lim x→a f(x)=L it is necessary that both the left- and right-hand limits be equal to L as well. 260 CHAPTER 7 The Theoretical Backbone: Limits and Continuity If lim x→a − f(x)= lim x→a + f(x), then lim x→a f(x)does not exist. (We will make a stronger statement about the relationship between one-sided and two-sided limits later in this section.) It follows that lim x→0 1 x does not exist. ◆ EXAMPLE 7.11 Evaluate lim x→0 + |x| x , lim x→0 − |x| x , and lim x→0 |x| x . SOLUTION The basic strategy for dealing with absolute values is to split the problem into two cases, one where the quantity inside the absolute value is negative and the other where this quantity is nonnegative.                If x>0, |x| x = x x = 1; if x<0, |x| x = −x x =−1; if x = 0, |x| x is undefined. 1 –1 |x| x y y = x Figure 7.17 lim x→0 + |x| x = lim x→0 + 1 = 1, lim x→0 − |x| x = lim x→0 − −1 =−1, lim x→0 |x| x does not exist because the left- and right-hand limits are unequal. ◆ ◆ EXAMPLE 7.12 Evaluate lim x→0 + |x|, lim x→0 − |x|, and lim x→0 |x|. SOLUTION  If x ≥ 0, |x|=x; if x<0, |x|=−x. y y = |x | x Figure 7.18 . Sketch the graph of f(t). (b) What are the domain and range of f(t)? 7.2 LEFT- AND RIGHT-HANDED LIMITS; SOMETIMES THE APPROACH IS CRITICAL ◆ EXAMPLE 7.10 How does f(x)= 1 x behave as x approaches zero? SOLUTION. hit the value L any number of times. EXERCISE 7.2 Show, using the methods of Example 7.3, that lim x→3 k = k for any constant k. Conclude that lim x→c k = k for c and k any constants. EXERCISE. limit of the slope of the secant lines, so we want to interpret x 2 −9 x−3 as K x , the slope of the line through the points (x, x 2 ) and (3, 9). We see that the limit is being taken as x approaches