Chapter 11: Sequences; Indeterminate Forms; Improper Integrals

Transcription

1 Chapter 11: Sequences; Indeterminate Forms; Improper Integrals Section 11.1 The Least Upper Bound Axiom a. Least Upper Bound Axiom b. Examples c. Theorem d. Example e. Greatest Lower Bound f. Theorem Section 11.2 Sequences of Real Numbers a. Definition b. Laws of Formation c. Types of Sequences d. Range of a Sequence e. Increasing, Decreasing, Monotonic Sequences f. Example Section 11.3 Limit of A Sequence a. Definition b. Example c. Uniqueness of Limit Theorem d. Convergent and Divergent Sequences e. Convergence Theorem f. Example g. Theorem h. Pinching Theorem for Sequences i. Corollary j. Continuous Functions Applied to Convergent Sequences Section 11.4 Some Important Limits a. Common Limits b. Limit Properties c. More Properties Section 11.5 The Indeterminate Form (0/0) a. L Hôpitals Rule (0/0) b. Example c. The Cauchy Mean-Value Theorem Section 11.6 The Indeterminate Form ( / ); Other Indeterminate Forms a. L Hôpitals Rule ( / ) b. Limit Properties c. Example d. Indeterminates 0 0, 1, 0 e. Example Section 11.7 Improper Integrals a. Integrals Over Unbounded Intervals b. Examples c. Limit Property d. Comparison Test e. Integrals of Unbounded Functions f. Example

2 The Least Upper Bound Axiom

3 The Least Upper Bound Axiom We indicate the least upper bound of a set S by writing lub S. As you will see from the examples below, the least upper bound idea has wide applicability. (1) lub (, 0) = 0, lub(, 0] = 0. (2) lub ( 4, 1) = 1, lub( 4, 1] = 1. (3) lub {1/2, 2/3, 3/4,..., n/(n + 1),... } = 1. (4) lub { 1/2, 1/8, 1/27,..., 1/n 3,... } = 0. (5) lub {x : x 2 < 3} = lub{x : 3< x < 3} = 3 (6) For each decimal fraction we have b = 0.b 1 b 2 b 3,..., b = lub {0.b 1, 0.b 1 b 2, 0.b 1 b 2 b 3,... }. (7) If S consists of the lengths of all polygonal paths inscribed in a semicircle of radius 1, then lub S = π (half the circumference of the unit circle).

4 The Least Upper Bound Axiom

5 The Least Upper Bound Axiom Example (a) Let S = {1/2, 2/3, 3/4,..., n/(n + 1),... } and take ε = Since 1 is the least upper bound of S, there must be a number s S such that < s < 1. There is: take, for example, 99,999 s =. 100,000 (b) Let S = {0, 1, 2, 3} and take ε = It is clear that 3 is the least upper bound of S. Therefore, there must be a number s S such that There is: s = < s 3.

6 The Least Upper Bound Axiom

7 The Least Upper Bound Axiom

8 Sequences of Real Numbers

9 Sequences of Real Numbers Suppose we have a sequence of real numbers a 1, a 2, a 3,..., a n,... By convention, a 1 is called the first term of the sequence, a 2 the second term, and so on. More generally, a n, the term with index n, is called the nth term. Sequences can be defined by giving the law of formation. For example: b n 1 an = n n = n c = n n is the sequence 1,,,, is the sequence,,,,... is the sequence 1, 4, 9, 16,... It s like defining f by giving f (x).

10 Sequences of Real Numbers Sequences can be multiplied by constants; they can be added, subtracted, and multiplied. From we can form a 1, a 2, a 3,..., a n,... and b 1, b 2, b 3,..., b n,... the scalar product sequence : αa 1, αa 2, αa 3,..., αa n,..., the sum sequence : a 1 + b 1, a 2 + b 2, a 3 + b 3,..., a n + b n,..., the difference sequence : a 1 b 1, a 2 b 2, a 3 b 3,..., a n b n,..., the product sequence : a 1 b 1, a 2 b 2, a 3 b 3,..., a n b n,.... If the b i s are all different from zero, we can form the reciprocal sequence : the quotient sequence : ,,,...,,... b1 b2 b3 b n a a a a b b b b 1, 2, 3,..., n, n

11 Sequences of Real Numbers The range of a sequence is the set of values taken on by the sequence. While there can be repetition in a sequence, there can be no repetition in the statement of its range. The range is a set, and in a set there is no repetition. A number is either in a particular set or it s not. It can t be there more than once. The sequences 0, 1, 0, 1, 0, 1, 0, 1,... and 0, 0, 1, 1, 0, 0, 1, 1,... both have the same range: the set {0, 1}. The range of the sequence is the set of integers. 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4,...

12 Sequences of Real Numbers Many of the sequences we work with have some regularity. They either have an upward tendency or they have a downward tendency. The following terminology is standard. The sequence with terms a n is said to be increasing if a n < a n+1 for all n, nondecreasing if a n a n+1 for all n, decreasing if a n > a n+1 for all n, nonincreasing if a n a n+1 for all n. A sequence that satisfies any of these conditions is called monotonic. The sequences are monotonic. The sequence is not monotonic. 1, ½, 1 3, ¼,..., 1 n,... 2, 4, 8, 16,..., 2 n,... 2, 2, 4, 4, 6, 6,..., 2n, 2n,... 1, ½, 1, 1 3, 1, ¼, 1,...

13 Sequences of Real Numbers Example The sequence a n n = n + 1 is increasing. It is bounded below by ½ (the greatest lower bound) and above by 1 (the least upper bound). Proof Since 2 a ( n+ 1/ 1 ) ( n+ 2 n+ ) n+ 1 n+ 1 n + 2n+ 1 = = = > 2 1 a n/ n+ 1 n+ 2 n n + 2n n ( ) we have a n < a n+1. This confirms that the sequence is increasing. The sequence can be displayed as ,,,,...,,, It is clear that ½ is the greatest lower bound and 1 is the least upper bound.

14 Limit of A Sequence

15 Example Since 2 n 2 = n n Limit of A Sequence it is intuitively clear that To verify that this statement conforms to the definition of limit of a sequence, we must show that for each ε > 0 there exists a positive integer K such that if n K, then 2 n 2 <ε n + 1 To do this, we fix ε > 0 and note that 2 2 n lim = 2 n n + 1 ( n ) n 2 n = = = < n+ 1 n+ 1 n + 1 n + 1 n We now choose K sufficiently large that 2 K < ε. If n K, then 2 n 2 K < ε and consequently 2 n 2 2 < <ε n + 1 n

16 Limit of A Sequence

17 Limit of A Sequence

18 Limit of A Sequence

19 Limit of A Sequence Example We shall show that the sequence is convergent. Since 3 = (3 n ) 1/n < (3 n + 4 n ) 1/n < (4 n + 4 n ) 1/n = (2 4 n ) 1/n = 2 1/n 4 8, the sequence is bounded. Note that Since we have n n ( 3 4 ) 1/ a = + (3 n + 4 n)(n+1)/n = (3 n + 4 n ) 1/n (3 n + 4 n ) = (3 n + 4 n ) 1/n 3 n + (3 n + 4 n ) 1/n 4 n. n (3 n + 4 n ) 1/n > (3 n ) 1/n = 3 and (3 n + 4 n ) 1/n > (4 n ) 1/n = 4, (3 n + 4 n ) (n+1)/n > 3 (3 n ) + 4 (4 n ) = 3 n n+1. Taking the (n + 1)st root of the left and right sides of this inequality, we obtain (3 n + 4 n ) 1/n > (3 n n+1 ) 1/(n+1) The sequence is decreasing. Being also bounded, it must be convergent. n

20 Limit of A Sequence

21 Limit of A Sequence

22 Limit of A Sequence The following is an obvious corollary to the pinching theorem. A limit to remember

23 Limit of A Sequence Continuous Functions Applied to Convergent Sequences

24 Some Important Limits

25 Some Important Limits

26 Some Important Limits

27 The Indeterminate Form (0/0)

28 The Indeterminate Form (0/0) Example Find lim + x 0 sin x x Solution As x 0 +, both numerator and denominator tend to 0 and ( ) f x 1 2 x 0 = = = 0 g ( x) ( cos x)( 1/ 2 x ) cos x 1 It follows from L Hôpital s rule that x lim 0 + x 0 sin x For short, we can write x 2 x lim lim = sin x cos x x 0 x 0

29 The Indeterminate Form (0/0)

30 Other Indeterminate Forms

31 Other Indeterminate Forms

32 Other Indeterminate Forms Example 2 n Determine the behavior of a n = as n. 2 n Solution x 2 To use the methods of calculus, we investigate lim x 2 x Since both numerator and denominator tend to with x, we try L Hôpital s rule: ( ) 2 x x x 2 2 ln 2 2 ln 2 lim lim lim x 2 x x 2x x 2 = Therefore the sequence must also diverge to.

33 Other Indeterminate Forms The Indeterminates 0 0, 1, 0 Such indeterminates are usually handled by first applying the logarithm function: y = [ f (x)] g(x) gives ln y = g(x) ln f (x).

34 Other Indeterminate Forms Example (1 ) Find lim 1 x 0 + ( + x) 1/ x Solution Here we are dealing with an indeterminate of the form 1 : as x 0 +, 1 + x 1 and 1/x. Taking the logarithm and then applying L Hôpital s rule, we have ( + x) 1/ x ln 1 1 lim ln ( 1+ x) = lim lim = x 1+ x x 0 x 0 x 0 As x 0 +, ln (1 + x) 1/x 1 and (1 + x) 1/x = e ln(1+x)1/x e 1 = e. Set x = 1/n and we have the familiar result: as n, [1 + (1/n)] n e.

35 Improper Integrals Integrals over Unbounded Intervals We begin with a function f which is continuous on an unbounded interval [a, ). For each number b > a we can form the definite integral ( ) If, as b tends to, this integral tends to a finite limit L, b a f x dx lim b a b ( ) f x dx = L then we write and say that a ( ) f x dx = L the improper integral Otherwise, we say that a f ( ) x dx converges to L. the improper integral a f ( ) x dx diverges.

36 Improper Integrals As a tends to, sin πa oscillates between 1 and 1. Therefore the integral oscillates between 1/π and 1/π and does not converge.

37 Improper Integrals

38 Improper Integrals

39 Improper Integrals Integrals of Unbounded Functions Improper integrals can arise on bounded intervals. Suppose that f is continuous on the half-open interval [a, b) but is unbounded there. For each number c < b, we can form the definite integral ( ) If, as c b, the integral tends to a finite limit L, namely, if c a f x dx lim c b c a ( ) f x dx = L then we write and say that b ( ) f x dx = L a the improper integral b ( ) a f x dx converges to L. Otherwise, we say that the improper integral diverges.

40 Example Test ( ) for convergence. Improper Integrals ( x ) Solution The integrand has an infinite discontinuity at x = 2. For integral ( ) to converge both 2 dx 4 dx and 1 2 ( ) 2 2 x 2 ( x 2) must converge. Neither does. For instance, as c 2, c c dx 1 1 = 1 1 = x 2 x 2 c 2 ( ) 2 4 dx dx This tells us that 1 2 diverges and shows that ( ) diverges. ( x 2) If we overlook the infinite discontinuity at x = 2, we can be led to the incorrect conclusion that 4 4 dx 1 3 = 1 2 = x 2 x 2 2 ( ) 2 1 1