Chapter 1. Limits and Continuity. 1.1 Limits

Transcription

1 Chapter Limits and Continuit. Limits The its is the fundamental notion of calculus. This underling concept is the thread that binds together virtuall all of the calculus ou are about to stud. In this section, we develop the notion of it using some common language and illustrate the idea with some simple eamples. As a start, consider the function f() = 2 and g() = 2 2 Notice that both functions are undefined at =. But we can eamine their behavior for close to, as in the following table. f() = 2 g() = , , , , 000, Notice that as ou move down the first column of the table, the -values get closer to, but all less than. We use the notation to indicate that approaches from the left side. Notice that the table suggest that as gets closer and closer to (with < ), f() is getting closer and closer to 2. In view of this, we sa that the it of f() as approaches from the left is 2, written f() = 2. On the other hand, the table indicate that as gets closer and closer to (with < ), g() increases without bound. In this case, g() is said to increase without bound as approaches from the left, written g() = +

2 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 2 Likewise, we need to consider what happens to the function values for close to but larger than. Here, we use the notation + to indicate that approaches from the right side. We compute some of these values in the following table. f() = 2 g() = , , , Again, the table suggest that as gets closer and closer to (with > ), f() is getting closer and closer to 2. In view of this, we sa that the it of f() as approaches from the right is 2, written +f() = 2. Finall, the table suggest that g() decreases without bound as approaches from the right. We sa that +g() = We call f() and +f() one-sided its. Since the two one-sided its of f() are the same, we summarize our results b saing that the it of f() as approaches is 2, written f() = 2 On the other hand, g() do not approach a real number as approaches either from the left or from the right, and In this case, we sa that g() = + and +g() =. g() does not eist Hence a it eists if and onl if both one-sided its eist and are equal. That is, f() = L, for some number L, if and onl if f() = +f() = L Moreover, we can write 2 f() = = ( )(+) = (+) = 2 where we can cancel the factors ( ) since in the it as, is close to, but, so that 0.

3 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 3 Eample.. Determine whether or not 0 eists. We first graph = and compute some function values for close to 0. 0 = / ±0. ±0 ±0.0 ±00 ±0.00 ±000 ±0.000 ±0, 000 ± ±00, 000 Notice that, as 0 +, increases without bound. Thus, Likewise, we can sa that Therefore, Eample.2. Evaluate 0 sin = +. 0 =. does not eist. We graph f() = sin and compute some function values. 3 2 = sin (sin)/ ± ± ± ± ±

4 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 4 The graph and the tables of value lead us to Thus, Eample.3. sin 0 + = and 0 sin = sin 0 = For the function whose graph is given, state the value of the given quantit, if it does not eist, eplain wh. (a) f() = (c) 2 f() = (e) 5 +f() = (b) +f() = (d) 5 f() = (f) f() = 5

5 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 5 Eercises.. = f() For the function whose graph is given, state the value of the given quantit, if it does not eist, eplain wh. 2. (a) 0 f() (c) 0 f() (e) 2 +f() (g) 3 f() (b) 0 +f() (d) 2 f() (f) f() = f() -3-4 For the function whose graph is given, state the value of the given quantit, if it does not eist, eplain wh. (a) f() (c) 3 +f() (e) f(3) (g) 2 +f() (i) f( 2) (b) 3 f() (d) f() 3 (f) 2 f() (h) f() 2

6 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana = f() For the function whose graph is given, state the value of the given quantit, if it does not eist, eplain wh. (a) 3 f() (c) f( ) (e) f() (g) +f() (b) f( 3) (d) f() (f) f() (h) f() Answer to Eercises.. (a) 2 (b) 2 (c) Does not eist (d) (e) 3 (f) Does not eist (g) 2 2. (a) 3 (b) 2 (c) 2 (d) Does not eist (e) (f) (g) (h) (i) 3 3. (a) 2 (b) (c) 2 (d) 5 2 (e) 2 (f) 2 (g) (h) Does not eist.2 Computing Limits In this section, we present some basic rules for dealing with common it problem. Theorem.. For an constant c and an real number a, c = c. Theorem.2. For an real number a, = a.

7 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 7 Theorem.3. Suppose that f() and g() both eist and let c be an constant. The following then appl:. [ cf() ] = c f() 2. [ f()±g() ] = f()± g() 3. [ f() g() ] = f() g() f() f() 4. g() = g() ( ) if g() 0 Corollar.. Suppose that f() eists. Then, [ ] [ ] 2 2. f() = f() Likewise, for an positive integer n, [ ] n f() = [ f() ] n. Corollar.2. For an integer n > 0 and for real number a, n = a n. Eample.4. Appl the rules of its to evaluate the following. (a) 2 ( ) (b)

8 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 8 Theorem.4. For an polnomial and an real number a, p() = c n n +c n n + +c +c 0 p() = c na n +c n a n + +c a+c 0 = p(a) Theorem.5. Let f() = p() q() be the rational function, and let a be an real number. (a) If q(a) 0, then f() = f(a). (b) If q(a) = 0 but p(a) 0, then f() does not eist. Eample.5. Evaluate Eample.6. ( Evaluate 2 ). 2

9 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 9 Eample Evaluate Eample.8. 2/3 4 /3 Evaluate. 2 2 Theorem.6. Suppose that f() = L and n is an positive integer. Then, where for n even, we assume that L > 0. n f() = n f() = n L,

10 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 0 Eample.9. Evaluate Eample.0. Evaluate Eample Evaluate. 4 2

11 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Eample Evaluate. 2 2 For functions that are defined piecewise, a two-sided it at a point where the formula for the function changes is best obtained b first finding the one-sided its at the point. Eample.3. Evaluate 2 f(), where f is defined b f() = for 2 for > 2

12 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 2 Theorem.7 (Squeeze Theorem). Supposethatg() f() h()forallinsomeinterval(c,d), ecept possibl at the point a (c,d) and that g() = h() = L for some number L. Then, it follows that L f() = L. h g 0 a f Remark. Squeeze Theorem also applied to one-sided its. Eample.4. Evaluate f() if f() 4+5 for all R. Eample.5. [ Determine the value of cos 0 ( 50π )].

13 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 3. Given that Eercise.2 f() = 3, g() = 0, h() = 8 find the its that eist. If the it does not eist, eplain wh. [ ] [ ] 2 (a) f()+h() (b) f() (c) 3 h() (e) f() h() (g) f() g() 2. Evaluate the following its. (a) 0 ( 2 3+) (c) (e) (g) (i) (+h) 4 (k) h 0 h t (m) t t t 2 +t 6 (o) t 2 t (q) 2 (s) 2 [ 2 2 (3+h) 3 (u) h 0 h (w) () Evaluate 2 f() where f() = ] (d) f() (f) g() f() 2f() (h) h() f() (b) ( 3 +2)( 2 5) 3 ( ) (d) (f) (h) 2 (j) (l) h 0 (2+h) 3 8 h (n) 3 t 2 t 2 (p) t 0 t 2 8 (r) 9 3 [ (t) t 0 t +t ] t (v) 2 t 9 9 t 2 2 () 0 +3 e 2+ (z) { if < if 2

14 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 4 { 2 if < 2 4. Evaluate f() where f() = 2 2 if 2 { 5. Evaluate f() where f() = 2 + if < 0 3+ if 6. Evaluate f() where f() = 2+ if < 3 if < 2+ if 7. Find the it, if it eists. If the it does not eist, eplain wh. (a) (c) (b) 4 +4 ( (d) 0 + ) 8. Use the Squeeze Theorem to find the following its. (a) 0 2 sin(/) (b) 0 2 cos(20π) (c) sin π Answer to Eercise.2. (a) 5 (b) 9 (c) 2 (d) 3 (e) 3 8 (f) 0 (g) (h) 6 2. (a) (b) 74 (c) 3 8 (d) 4 9 (e) 7 (f) 0 (g) 7 (h) 5 (i) 3 (j) 3 (k) 4 (l) 2 (m) 2 (n) 6 (o) 5 (p) 2 (q) 32 (r) 08 (s) (t) (u) (v) (w) () 2 () (z) Does not eist 7. (a) 0 (b) (c) Does not eist (d) 0 8. (a) 0 (b) 0 (c) 0

15 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 5.3 Limits at Infinit Let s begin b investigating the behavior of the function f defined b as becomes large. f() = = f() = ±0 ± 0 ± ± ± ± ± ± ± ± As grows larger and larger we can see from the graph and the table of values that the values of f() get closer and closer to. That is, In general, we use the notation =. f() = L indicates that the values of f() becomes closer and closer to L as becomes larger and larger. Definition.. Let f be a function defined on some interval (a, ). Then f() = L mean that the value of f() can be made arbitraril close to L b taking sufficientl large. Referring back to the above Figure, we see that for numericall large negative values of, the values of f() are close to, that is, The general definition is as follows =.

16 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 6 Definition.2. Let f be a function defined on some interval (,a). Then f() = L mean that the value of f() can be made arbitraril close to L b taking sufficientl large negative. Definition.3. The line = L is called a horizontal asmptote of the curve = f() if either f() = L or f() = L For instance, the curve illustrated in Figure above has the line = as a horizontal asmptote because =. Theorem.8. For an rational number t > 0, ± t = 0 where for the case where, we assume that t = p q where q is odd. Theorem.9. For a polnomial of degree n > 0, p n () = a n n +a n n + +a +a 0, we have {, if p an > 0 n() =, if a n < 0

17 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 7 Eample.6. Evaluate Eample.7. Evaluate Eample.8. ( ) Evaluate 2 +3.

18 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 8 Eample Evaluate Evaluate the it, if it eists. Eercise ( ) ( ( ) sin2 2. ln(2) ) ( )(2+) 2. (+2)(2 3) sin ( (3 5 2 ) 8. e e 3 cos (ln2) )

19 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Find the horizontal asmptotes of each curve, if it eists. 25. f() = f() = f() = f() = f() = f() = Answer to Eercise Does not eist = 26. =, = 27. = No horizontal asmptote 29. No horizontal asmptote 30. =, =.4 Limit of Trigonometric Functions Lemma.. sin = 0 0 Lemma.2. sin 0 = Corollar.3. For an real number k 0, sink 0 = k Eample.20. Evaluate 0 cos2 2.

20 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 20 Eample.2. Evaluate sin tan2. Eample.22. Evaluate 0 2cot 2 csc. Eample.23. Evaluate 0 cos sin.

21 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 2 0 Evaluate the its. sin cosθ 3. θ 0 sinθ tan π/4 4 sin cos cos2 +sin 9. 0 tan cos. 0 sin Eercise.4 sin8t 2. t 0 sin9t sin cot csc 2cot csc 0. 0 sin(cos) sec Answer to Eercise sin. 2.5 Continuit It is helpful for us to first tr to see what it is about the function whose graphs are shown below that makes them discontinuous at the point = a. a a a This suggests the following definition of continuit at a point. Definition.4. A function f is continuous at = a when (i) f(a) is defined, (ii) f() eists and (iii) f() = f(a). Otherwise, f is said to be discontinuous at = a.

22 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 22 Eample.24. From the graph of f, state the numbers at which f is discontinuous and eplain wh Eample.25. Where are each of the following functions discontinuous? (a) f() = if 3 (b) g() = 3 6 if = 3

23 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 23 Eample.26. For what value of a is continuous at ever. f() = { 2 if < 3 2a if 3 Eample.27. Let ( 2) 2 +2k if > f() = h if = 2 2+k if < 2 Find the values h and k so that f is continuous at = 2.

24 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 24 Definition.5. A function f is continuous from the right at a number a if +f() = f(a) and f is continuous from the left at a if f() = f(a). Definition.6. Afunctionf isissaidtocontinuous on a closed interval [a,b]ifthefollowing conditions are satisfied:. f is continuous on (a,b). 2. f is continuous from the right at a. 3. f is continuous from the left at b. Eample.28. Show that the function f() = 6 2 is continuous on the interval [ 4, 4].

25 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 25 Theorem.0. If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: (a) f +g (b) f g (c) cf (d) f g (e) f g if g(a) 0 The following tpes of functions are continuous at ever number in their domains: polnomials rational functions root functions trigonometric functions inverse trigonometric functions eponential functions logarithmic functions Eample.29. Where is the function h() = continuous? Theorem.. If g() = L and f is continuous at L, then f( g() ) ( ) = f g() = f(l) Corollar.4. If g is continuous at a and f is continuous at g(a), then the composite function f g given b (f g)() = f ( g() ) is continuous at a.

26 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 26 Theorem.2 (Intermediate Value Theorem). Suppose that f is continuous on the closed interval [a,b] and let N be an number between f(a) and f(b). Then there eists a number c (a,b) such that f(c) = N. f(b) f(b) N = f() N f(a) a = f() c b f(a) ac c 2 c 3 b (a) (b) Note that the value N can be taken on once [as in part (a)] or more that once [as in part (b)]. Corollar.5. Suppose that f is continuous on [a,b] and f(a) and f(b) have opposite signs. Then, there is at least one number c (a,b) for which f(c) = 0. Eample.30. Show that there is a root of the equation 3 = 0 between and 2.

27 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 27 Eercise.5. Use the given graph to identif all discontinuilit of the functions. (a) (b) (c) 2. Use the definition of continuit and the properties of its to show that the function is continuous at the given number. (a) f() = 2 + 7, = 4 (b) g() = (+2 3 ) 4, = (c) h() = , = 4 3. Use the definition of continuit and the properties of its to show that the function is continuous on the given interval. (a) f() = +, (, ) (b) f() = 2 4, ( 2,2) (c) g(t) = 9 4t 2, [ 3 2, 3 2 ]

28 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana 28 (d) h(z) = (z )(3 z), [,3] 4. Eplain wh the function is discontinuous at the given number. (a) f() = ; = (b) f() = sin ; = 0 (c) f() = e / ; = 0 (d) f() = ln 2 ; = 2 { (e) f() = if 2 if = ; = (f) f() = 2 + ; = if 4 (g) f() = 4 ; = 4 3 if = 4 { if 2 (h) f() = 2 2 if > 2 ; = 2 2 if < 2 (i) f() = 3 if = 2 ; = if > 2 (j) f() = 2 if < 0 if 0 if > ; = 5. Determine the intervals on which f() is continuous. (a) f() = 2+ 3 (b) f() = +3 (c) f() = 2 + (e) f() = (g) f() = 3 2 (i) f() = sin 2 (d) f() = (f) f() = 3 (h) f() = 4 2

29 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Find the constant c that makes f() continuous on (, ). { +c if < 0 (a) f() = 4 2 if 0 { c (b) f() = 2 2 if < 0 2( c) 2 if 0 { c+ if 3 (c) f() = c 2 if > 3 { (d) f() = 2 c 2 if < 4 c+20 if 4 7. For what value of the constant k is the function ; 2 < < 3 3 f() = k ; 3 + continuous on ( 2, ). 8. For what value of the constant k is the function if < 3 f() = +3 ( 2 k) 3 if 3 continuous on (, ). 9. Find the values a and b so that f is continuous on (, ). + if < f() = a+b if < 2 3+ if 2 0. Let Show that f() = { 2, 0 4, = 0 and g() = 2. f( g() ) ( ) f g() Use the Intermediate Value Theorem to verif that f() has a zero in the given interval. (a) f() = 2 7; [2,3] (b) f() = 3 4 2; [,0] (c) f() = cos ; [0,]

30 MA: Prepared b Asst.Prof.Dr. Archara Pacheenburawana Use the Intermediate Value Theorem to show that = 0 has a real root between 0 and. 3. Use the Intermediate Value Theorem to show that (cost)t 3 +6sin 5 t 3 = 0 has a real root between 0 and 2π. 4. Show that the equation = 0 has at least one real root. Answer to Eercise.5. (a) = 2,2 (b) = 2,,4 (c) = 2,2,4 4. (a) f() is not defined (b) f(0) is not defined (c) f(0) is not defined (d) f(2) is not defined (e) f() is not defined (f) f( ) is not defined (g) f() f(4) (h) f() is not defined (i) f() f(2) (j) f() is not defined 5. (a) R (b) (, 3) ( 3, ) (c) R (d) (,5) (5, ) (e) (,2) (2, ) (f) (,) (, ) (g) (,0) (0,) (, ) (h) (, 2) ( 2,2) (2, ) (i) (,0) (0, ) 6. (a) 4 (b) 0 (c) 3 (d) a = 5, b = 3