.1 Limits of Sequences. CHAPTER.1.0. a) True. If converges, then there is an M > 0 such that M. Choose by Archimedes an N N such that N > M/ε. Then n N implies /n M/n M/N < ε. b) False. = n does not converge, but /n = 1/ n 0 c) False. = 1 converges and y n = ( 1) n is bounded, but y n = ( 1) n does not converge. d) False. = 1/n converges to 0 and y n = n > 0, but y n = n does not converge..1.1. a) By the Archimedean Principle, given ε > 0 there is an N N such that N > 1/ε. Thus n N implies ( 1/n) 1/n 1/N < ε. b) By the Archimedean Principle, given ε > 0 there is an N N such that N > π /ε. Thus n N implies 1 + π/ n 1 π/ n π/ N < ε. c) By the Archimedean Principle, given ε > 0 there is an N N such that N > 3/ε. Thus n N implies 3(1 + 1/n) 3 3/n 3/N < ε. d) By the Archimedean Principle, given ε > 0 there is an N N such that N > 1/ 3ε. Thus n N implies (n + 1)/(3n ) /3 1/(3n ) 1/(3N ) < ε..1.. a) By hypothesis, given ε > 0 there is an N N such that n N implies 1 < ε/. Thus n N implies 1 + 3 1 < ε. b) By hypothesis, given ε > 0 there is an N N such that n N implies > 1/ and 1 < ε/4. In particular, 1/ <. Thus n N implies (π )/ (π ) ( 1)/ < 4 1 < ε. c) By hypothesis, given ε > 0 there is an N N such that n N implies > 1/ and 1 < ε/(1 + e). Thus n N and the triangle inequality imply (x n e)/ (1 e) 1 1 + e ( 1 1 + e ) < 1 (1 + e) < ε..1.3. a) If n k = k, then 3 ( 1) n k converges to ; if n k = k + 1, then 3 ( 1) n k 4 converges to 4. b) If n k = k, then ( 1) 3nk + ( 1) 6k + = 1 + = 3 converges to 3; if n k = k + 1, then ( 1) 3nk + ( 1) 6k+3 + = 1 + = 1 converges to 1. c) If n k = k, then (n k ( 1) n k n k 1)/n k 1/(k) converges to 0; if n k = k+1, then (n k ( 1) n k n k 1)/n k (n k 1)/n k = (4k + 1)/(k + 1) converges to..1.4. Suppose is bounded. By Definition.7, there are numbers M and m such that m M for all n N. Set C := max{1, M, m }. Then C > 0, M C, and m C. Therefore, C C, i.e., < C for all n N. Conversely, if < C for all n N, then is bounded above by C and below by C..1.5. If C = 0, there is nothing to prove. Otherwise, given ε > 0 use Definition.1 to choose an N N such that n N implies b n b n < ε/ C. Hence by hypothesis, n N implies By definition, a a C b n < ε..1.6. If = a for all n, then a = 0 is less than any positive ε for all n N. Thus, by definition, a 10
.1.7. a) Let a be the common limit point. Given ε > 0, choose N N such that n N implies a and y n a are both < ε/. By the Triangle Inequality, n N implies y n a + y n a < ε. By definition, y n 0 b) If n converges to some a, then given ε = 1/, 1 = (n + 1) n < (n + 1) a + n a < 1 for n sufficiently large, a contradiction. c) Let = n and y n = n + 1/n. Then y n = 1/n 0 as n, but neither nor y n converges..1.8. By Theorem.6, if a then k a. Conversely, if k a for every subsequence, then it converges for the subsequence.. Limit Theorems...0. a) False. Let = n and y n = n and note by Exercise..a that + y n b) True. Let ε > 0. If as n, then choose N N such that n N implies < 1/ε. Then < 0 so = > 0. Multiply < 1/ε by ε/( ) which is positive. We obtain ε < 1/, i.e., 1/ = 1/ < ε. c) False. Let = ( 1) n /n. Then 1/ = ( 1) n n has no limit d) True. Since ( x x) = x log 1 > 1 for all x, i.e., x x is increasing on [, ). In particular, x x > 0, i.e., x > x for x. Thus, since as n, we have xn > for n large, hence xn < 1 0..1. a) 1/n 0 as n and we can apply the Squeeze Theorem. b) n/(n + π) = (/n)/(1 + π/n ) 0/(1 + 0) = 0 by Theorem.1. c) ( n + 1)/(n + ) = (( / n) + (1/n))/(1 + ( /n)) 0/(1 + 0) = 0 by Exercise..5 and Theorem.1. d) An easy induction argument shows that n + 1 < n for n = 3, 4,.... We will use this to prove that n n for n = 4, 5,.... It s surely true for n = 4. If it s true for some n 4, then the inductive hypothesis and the fact that n + 1 < n imply (n + 1) = n + n + 1 n + n + 1 < n + n = n+1 so the second inequality has been proved. Now the second inequality implies n/ n < 1/n for n 4. Hence by the Squeeze Theorem, n/ n 0... a) Let M R and choose by Archimedes an N N such that N > max{m, }. Then n N implies n n = n(n 1) N(N 1) > M( 1) = M. b) Let M R and choose by Archimedes an N N such that N > M/. Notice that n 1 implies 3n 3 so 1 3n. Thus n N implies n 3n = n(1 3n) n N < M. c) Let M R and choose by Archimedes an N N such that N > M. Then n N implies (n + 1)/n = n + 1/n > N + 0 > M. d) Let M R satisfy M 0. Then + sin θ 1 = 1 implies n ( + sin(n 3 + n + 1)) n 1 > 0 M for all n N. On the other hand, if M > 0, then choose by Archimedes an N N such that N > M. Then n N implies n ( + sin(n 3 + n + 1)) n 1 N > M...3. a) Following Example.13, + 3n 4n 1 n + 3n = (/n ) + (3/n) 4 (1/n ) (/n) + 3 4 3 b) Following Example.13, n 3 + n n 3 + n = 1 + (1/n ) (/n 3 ) + (1/n ) (/n 3 ) 1 11
c) Rationalizing the expression, we obtain ( 3n + n)( 3n + + n) n + 3n + n = = 3n + + n 3n + + n as n by the method of Example.13. (Multiply top and bottom by 1/ n.) d) Multiply top and bottom by 1/ n to obtain 4n + 1 n 4 + 1/n 1 1/n = 1 9n + 1 n + 9 + 1/n 1 + /n 3 1 = 1...4. a) Clearly, Thus Since y 0, y n y / for large n. Thus x y n y = y xy n = y xy + xy xy n. yy n yy n x y n y 1 y n x + x yy n y n y. x y n y y x + x y y n y 0 as n by Theorem.1i and ii. Hence by the Squeeze Theorem, /y n x/y b) By symmetry, we may suppose that x = y =. Since y n implies y n > 0 for n large, we can apply Theorem.15 directly to obtain the conclusions when α > 0. For the case α < 0, > M implies α < αm. Since any M 0 R can be written as αm for some M R, we see by definition that..5. Case 1. x = 0. Let ɛ > 0 and choose N so large that n N implies < ɛ. By (8) in 1.1, < ɛ for n N, i.e., 0 Case. x > 0. Then xn x = ( ( xn + ) x x) xn + = x x xn + x. Since 0, it follows that x x x. This last quotient converges to 0 by Theorem.1. Hence it follows from the Squeeze Theorem that x..6. By the Density of Rationals, there is an r n between x + 1/n and x for each n N. Since x r n < 1/n, it follows from the Squeeze Theorem that r n x..7. a) By Theorem.9 we may suppose that x =. By symmetry, we may suppose that x =. By definition, given M R, there is an N N such that n N implies > M. Since w n, it follows that w n > M for all n N. By definition, then, w n b) If x and y are finite, then the result follows from Theorem.17. If x = y = ± or x = y =, there is nothing to prove. It remains to consider the case x = and y =. But by Definition.14 (with M = 0), > 0 > y n for n sufficiently large, which contradicts the hypothesis y n...8. a) Take the limit of +1 = 1 1, We obtain x = 1 1 x, i.e., x x = 0. Therefore, x = 0, 1. b) Take the limit of +1 = + We obtain x = + x, i.e., x 5x + 6 = 0. Therefore, x =, 3. But x 1 > 3 and induction shows that +1 = + > + 3 = 3, so the limit must be x = 3. c) Take the limit of +1 = + We obtain x = + x, i.e., x x = 0. Therefore, x =, 1. But +1 = + 0 by definition (all square roots are nonnegative), so the limit must be x =. This proof doesn t change if x 1 >, so the limit is again x =. 1
..9. a) Let E = {k Z : k 0 and k 10 n+1 y}. Since 10 n+1 y < 10, E {0, 1,..., 9}. Hence w := sup E E. It follows that w 10 n+1 y, i.e., w/10 n+1 y. On the other hand, since w + 1 is not the supremum of E, w + 1 > 10 n+1 y. Therefore, y < w/10 n+1 + 1/10 n+1. b) Apply a) for n = 0 to choose x 1 = w such that x 1 /10 x < x 1 /10 + 1/10. Suppose s n := n n x k 10 k x < x k 10 k + 1 10 n. Then 0 < x s n < 1/10 n, so by a) choose +1 such that +1 /10 n+1 x s n < +1 /10 n+1 + 1/10 n+1, i.e., n+1 n+1 x k 10 k x < x k 10 k + 1 10 n+1. c) Combine b) with the Squeeze Theorem. d) Since an easy induction proves that 9 n > n for all n N, we have 9 n < 1/n. Hence the Squeeze Theorem implies that 9 n 0 Hence, it follows from Exercise 1.4.4c and definition that Similarly,.4999 = 4 10 + lim n k=.999 = lim.3 The Bolzano Weierstrass Theorem. 9 10 k = 4 ( 10 + lim 1 1 1 ) 10 9 n = 4 10 + 1 10 = 0.5. n ( 9 10 k = lim 1 1 ) 9 n = 1..3.0. a) False. = 1/4 + 1/(n + 4) is strictly decreasing and 1/4 + 1/5 < 1/, but 1/4 as n. b) True. Since (n 1)/(n 1) 1/ as n, this factor is bounded. Since cos(n + n + 1) 1, it follows that { } is bounded. Hence it has a convergent subsequence by the Bolzano Weierstrass Theorem. c) False. = 1/ 1/n is strictly increasing and 1/ < 1 + 1/n, but 1/ d) False. = (1 + ( 1) n )n satisfies = 0 for n odd and = n for n even. Thus x k+1 0 as k, but is NOT bounded..3.1. Suppose that 1 < 1 < 0 for some n 0. Then 0 < 1 + 1 < 1 so 0 < 1 + 1 < 1 + 1 and it follows that 1 < 1 + 1 1 =. Moreover, 1 + 1 1 1 1 = 0. Hence by induction, is increasing and bounded above by 0. It follows from the Monotone Convergence Theorem that a Taking the limit of 1 + 1 1 = we see that a + a = 0, i.e., a = 1, 0. Since increases from x 0 > 1, the limit is 0. If x 0 = 1, then = 1 for all n. If x 0 = 0, then = 0 for all n. Finally, it is easy to verify that if x 0 = l for l = 1 or 0, then = l for all n, hence l.3.. If x 1 = 0 then = 0 for all n, hence converges to 0. If 0 < x 1 < 1, then by 1.4.1c, is decreasing and bounded below. Thus the limit, a, exists by the Monotone Convergence Theorem. Taking the limit of +1 = 1 1, as n, we have a = 1 1 a, i.e., a = 0, 1. Since x 1 < 1, the limit must be zero. Finally, +1 = 1 1 xn = 1 (1 ) (1 + 1 ) 1 1 + 1 = 1..3.3. Case 1. x 0 =. Then = for all n, so the limit is. Case. < x 0 < 3. Suppose that < 1 3 for some n 1. Then 0 < 1 1 so 1 1, i.e., = + 1 1. Moreover, = + 1 +1 = 3. Hence by induction, is increasing and bounded above by 3. It follows from the Monotone Convergence Theorem that a Taking the limit of + 1 = we see that a 5a + 6 = 0, i.e., a =, 3. Since increases from x 0 >, the limit is 3. Case 3. x 0 3. Suppose that 1 3 for some n 1. Then 1 1 so 1 1, i.e., = + 1 1. Moreover, = + 1 + 1 = 3. Hence by induction, is decreasing 13
and bounded above by 3. By repeating the steps in Case, we conclude that decreases from x 0 3 to the limit 3..3.4. Case 1. x 0 < 1. Suppose 1 < 1. Then 1 = 1 < 1 + 1 = < = 1. Thus { } is increasing and bounded above, so x. Taking the limit of = (1 + 1 )/ as n, we see that x = (1 + x)/, i.e., x = 1. Case. x 0 1. If 1 1 then 1 = 1 + 1 = 1 = 1. Thus { } is decreasing and bounded below. Repeating the argument in Case 1, we conclude that 1 as n..3.5. The result is obvious when x = 0. If x > 0 then by Example. and Theorem.6, lim x1/(n 1) = lim m x1/m = 1. If x < 0 then since n 1 is odd, we have by the previous case that x 1/(n 1) = ( x) 1/(n 1) 1.3.6. a) Suppose that { } is increasing. If { } is bounded above, then there is an x R such that x (by the Monotone Convergence Theorem). Otherwise, given any M > 0 there is an N N such that x N > M. Since { } is increasing, n N implies x N > M. Hence b) If { } is decreasing, then is increasing, so part a) applies..3.7. Choose by the Approximation Property an x 1 E such that sup E 1 < x 1 sup E. Since sup E / E, we also have x 1 < sup E. Suppose x 1 < x < < in E have been chosen so that sup E 1/n < < sup E. Choose by the Approximation Property an +1 E such that max{, sup E 1/(n + 1)} < +1 sup E. Then sup E 1/(n + 1) < +1 < sup E and < +1. Thus by induction, x 1 < x <... and by the Squeeze Theorem, sup E.3.8. a) This follows immediately from Exercise 1..6. b) By a), +1 = ( + y n )/ < / =. Thus y n+1 < +1 < < x 1. Similarly, y n = y n < y n = y n+1 implies +1 > y n+1 > y n > y 1. Thus { } is decreasing and bounded below by y 1 and {y n } is increasing and bounded above by x 1. c) By b), +1 y n+1 = + y n y n < + y n y n = y n. Hence by induction and a), 0 < +1 y n+1 < (x 1 y 1 )/ n. d) By b), there exist x, y R such that x and y n y By c), x y (x 1 y 1 ) 0 = 0. Hence x = y..3.9. Since x 0 = 1 and y 0 = 0, x n+1 y n+1 = ( + y n ) ( + y n ) = x n + y n = = ( 1) n (x 0 y 0 ) = ( 1) n. Notice that x 1 = 1 = y 1. If y n 1 n 1 and 1 1 then y n = 1 + y n 1 1 + (n 1) = n and = 1 + y n 1 1. Thus 1/y n 0 as n and 1 for all n N. Since x n yn = x n yn = 1 0 as n, it follows that /y n ± Since, y n > 0, the limit must be. y n 14 y n
.3.10. a) Notice x 0 > y 0 > 1. If 1 > y n 1 > 1 then y n 1 1 y n 1 = y n 1 (y n 1 1 ) > 0 so y n 1 (y n 1 + 1 ) < 1 y n 1. In particular, = 1y n 1 1 + y n 1 > y n 1. It follows that > y n 1 > 1, so > y n 1 = y n > 1 1 = 1. Hence by induction, > y n > 1 for all n N. Now y n < implies y n < + y n. Thus +1 = y n + y n <. Hence, { } is decreasing and bounded below (by 1). Thus by the Monotone Convergence Theorem, x for some x R. On the other hand, y n+1 is the geometric mean of +1 and y n, so by Exercise 1..6, y n+1 y n. Since y n is bounded above (by x 0 ), we conclude that y n y as n for some y R. b) Let n in the identity y n+1 = +1 y n. We obtain, from part a), y = xy, i.e., x = y. A direct calculation yields y 6 > 3.141557494 and x 7 < 3.1416101..4 Cauchy sequences..4.0. a) False. a n = 1 is Cauchy and b n = ( 1) n is bounded, but a n b n = ( 1) n does not converge, hence cannot be Cauchy by Theorem.9. b) False. a n = 1 and b n = 1/n are Cauchy, but a n /b n = n does not converge, hence cannot be Cauchy by Theorem.9. c) True. If (a n + b n ) 1 converged to 0, then given any M R, M 0, there is an N N such that n N implies a n + b n 1 < 1/ M. It follows that n N implies a n + b n > M > 0 > M. In particular, a n + b n diverges to. But if a n and b n are Cauchy, then by Theorem.9, a n +b n x where x R. Thus a n +b n x, NOT. d) False. If x k = log k and = 0 for n k, then x k x k 1 = log(k/(k 1)) 0 as k, but x k does not converge, hence cannot be Cauchy by Theorem.9..4.1. Since (n + 3)/(n 3 + 5n + 3n + 1) 0 as n, it follows from the Squeeze Theorem that 0 Hence by Theorem.9, is Cauchy..4.. If is Cauchy, then there is an N N such that n N implies x N < 1. Since x N Z, it follows that = x N for all n N. Thus set a := x N..4.3. Suppose and y n are Cauchy and let ε > 0. a) If α = 0, then α = 0 for all n N, hence is Cauchy. If α 0, then there is an N N such that n, m N implies x m < ε/ α. Hence α αx m α x m < ε for n, m N. b) There is an N N such that n, m N implies x m and y n y m are < ε/. Hence + y n (x m + y m ) x m + y n y m < ε for n, m N. c) By repeating the proof of Theorem.8, we can show that every Cauchy sequence is bounded. Thus choose M > 0 such that and y n are both M for all n N. There is an N N such that n, m N implies x m and y n y m are both < ε/(m). Hence for n, m N. y n (x m y m ) x m y m + y n y m < ε.4.4. Let s n = n 1 x k for n =, 3,.... If m > n then s m+1 s n = m k=n x k. Therefore, s n is Cauchy by hypothesis. Hence s n converges by Theorem.9. 15
.4.5. Let = n ( 1)k /k for n N. Suppose n and m are even and m > n. Then S := m ( 1) k k=n k 1 ( 1 n n + 1 1 ) ( 1 n + m 1 1 ). m Each term in parentheses is positive, so the absolute value of S is dominated by 1/n. Similar arguments prevail for all integers n and m. Since 1/n 0 as n, it follows that satisfies the hypotheses of Exercise.4.4. Hence must converge to a finite real number..4.6. By Exercise 1.4.4c, if m n then x m+1 = m (x k+1 x k ) k=n m k=n ( 1 a k = 1 1 a m (1 1 ) 1 a n ) a 1. Thus x m+1 (1/a n 1/a m )/(a 1) 0 as n, m since a > 1. Hence { } is Cauchy and must converge by Theorem.9..4.7. a) Suppose a is a cluster point for some set E and let r > 0. Since E (a r, a + r) contains infinitely many points, so does E (a r, a + r) \ {a}. Hence this set is nonempty. Conversely, if E (a s, a + s) \ {a} is always nonempty for all s > 0 and r > 0 is given, choose x 1 E (a r, a + r). If distinct points x 1,..., x k have been chosen so that x k E (a r, a + r) and s := min{ x 1 a,..., x k a }, then by hypothesis there is an x k+1 E (a s, a + s). By construction, x k+1 does not equal any x j for 1 j k. Hence x 1,..., x k+1 are distinct points in E (a r, a + r). By induction, there are infinitely many points in E (a r, a + r). b) If E is a bounded infinite set, then it contains distinct points x 1, x,.... Since { } E, it is bounded. It follows from the Bolzano Weierstrass Theorem that contains a convergent subsequence, i.e., there is an a R such that given r > 0 there is an N N such that k N implies k a < r. Since there are infinitely many k s and they all belong to E, a is by definition a cluster point of E..4.8. a) To show E := [a, b] is sequentially compact, let E. By the Bolzano Weierstrass Theorem, has a convergent subsequence, i.e., there is an x 0 R and integers n k such that k x 0 as k. Moreover, by the Comparison Theorem, E implies x 0 E. Thus E is sequentially compact by definition. b) (0, 1) is bounded and 1/n (0, 1) has no convergent subsequence with limit in (0, 1). c) [0, ) is closed and n [0, ) is a sequence which has no convergent subsequence..5 Limits supremum and infimum..5.1. a) Since 3 ( 1) n = when n is even and 4 when n is odd, lim sup = 4 and lim inf =. b) Since cos(nπ/) = 0 if n is odd, 1 if n = 4m and 1 if n = 4m +, lim sup = 1 and lim inf = 1. c) Since ( 1) n+1 + ( 1) n /n = 1 + 1/n when n is even and 1 1/n when n is odd, lim sup = 1 and lim inf = 1. d) Since 1/ as n, lim sup = lim inf = 1/ by Theorem.36. e) Since y n M, y n /n M/n 0 Therefore, lim sup = lim inf = 0 by Theorem.36. f) Since n(1 + ( 1) n ) + n 1 (( 1) n 1) = n when n is even and /n when n is odd, lim sup = and lim inf = 0. g) Clearly Therefore, lim sup = lim inf = by Theorem.36..5.. By Theorem 1.0, lim inf ( ) := lim ( inf ( x k)) = lim A similar argument establishes the second identity. (sup x k ) = lim sup..5.3. a) Since lim (sup x k ) < r, there is an N N such that sup k N x k < r, i.e., x k < r for all k N. b) Since lim (sup x k ) > r, there is an N N such that sup k N x k > r, i.e., there is a k 1 N such that x k1 > r. Suppose k ν N have been chosen so that k 1 < k < < k j and x kν > r for ν = 1,,..., j. Choose N > k j such that sup k N x k > r. Then there is a k j+1 > N > k j such that x kj+1 > r. Hence by induction, there are distinct natural numbers k 1, k,... such that x kj > r for all j N. 16
.5.4. a) Since inf x k + inf y k is a lower bound of x j + y j for any j n, we have inf x k + inf y k inf j n (x j + y j ). Taking the limit of this inequality as n, we obtain lim inf + lim inf y n lim inf ( + y n ). Note, we used Corollary 1.16 and the fact that the sum on the left is not of the form. Similarly, for each j n, inf (x k + y k ) x j + y j sup x k + y j. Taking the infimum of this inequality over all j n, we obtain inf (x k +y k ) sup x k +inf j n y j. Therefore, lim inf ( + y n ) lim sup + lim inf y n. The remaining two inequalities follow from Exercise.5.. For example, lim sup + lim inf y n = lim inf ( ) lim sup( y n ) lim inf ( y n ) = lim sup( + y n ). b) It suffices to prove the first identity. By Theorem.36 and a), lim + lim inf y n lim inf ( + y n ). To obtain the reverse inequality, notice by the Approximation Property that for each n N there is a j n > n such that inf (x k + y k ) > x jn 1/n + y jn. Hence inf (x k + y k ) > x jn 1 n + inf y k for all n N. Taking the limit of this inequality as n, we obtain lim inf ( + y n ) lim + lim inf y n. c) Let = ( 1) n and y n = ( 1) n+1. Then the limits infimum are both 1, the limits supremum are both 1, but + y n = 0 0 If = ( 1) n and y n = 0 then lim inf ( + y n ) = 1 < 1 = lim sup + lim inf y n..5.5. a) For any j n, x j sup x k and y j sup y k. Multiplying these inequalities, we have x j y j (sup x k )(sup y k ), i.e., sup x j y j (sup x k )(sup y k ). j n Taking the limit of this inequality as n establishes a). The inequality can be strict because if { 0 n even = 1 y n = 1 n odd then lim sup ( y n ) = 0 < 1 = (lim sup )(lim sup y n ). b) By a), lim inf (y n ) = lim sup( y n ) lim sup( ) lim sup y n = lim inf lim sup y n..5.6. Case 1. x =. By hypothesis, C := lim sup y n > 0. Let M > 0 and choose N N such that n N implies M/C and sup n N y n > C/. Then sup k N (x k y k ) y n (M/C)y n for any n N and sup k N (x k y k ) (M/C) sup n N y n > M. Therefore, lim sup ( y n ) =. 17
Case. 0 x <. By Exercise.5.6a and Theorem.36, lim sup ( y n ) (lim sup )(lim sup y n ) = x lim sup y n. On the other hand, given ɛ > 0 choose n N so that x k > x ɛ for k n. Then x k y k (x ɛ)y k for each k n, i.e., sup (x k y k ) (x ɛ) sup y k. Taking the limit of this inequality as n and as ɛ 0, we obtain lim sup( y n ) x lim sup y n..5.7. It suffices to prove the first identity. Let s = inf n N (sup x k ). Case 1. s =. Then sup x k = for all n N so by definition, lim sup = lim (sup x k ) = = s. Case. s =. Let M > 0 and choose N N such that sup k N x k M. Then sup x k sup k N x k M for all n N, i.e., lim sup =. Case 3. < s <. Let ɛ > 0 and use the Approximation Property to choose N N such that sup k N x k < s + ɛ. Since sup x k sup k N x k < s + ɛ for all n N, it follows that for n N, i.e., lim sup = s. s ɛ < s sup x k < s + ɛ.5.8. It suffices to establish the first identity. Let s = lim inf. Case 1. s = 0. Then by Theorem.35 there is a subsequence k j such that x kj 0, i.e., 1/x kj as j. In particular, sup (1/x k ) = for all n N, i.e., lim sup (1/ ) = = 1/s. Case. s =. Then x k, i.e., 1/x k 0, as k. Thus by Theorem.36, lim sup (1/ ) = 0 = 1/s. Case 3. 0 < s <. Fix j n. Since 1/ inf x k 1/x j implies 1/ inf x k sup j n (1/x j ), it is clear that 1/s lim sup (1/ ). On the other hand, given ɛ > 0 and n N, choose j > N such that inf x k + ɛ > x j, i.e., 1/(inf x k + ɛ) < 1/x j sup (1/x k ). Taking the limit of this inequality as n and as ɛ 0, we conclude that 1/s lim sup (1/ )..5.9. If 0, then 0. Thus by Theorem.36, lim sup = 0. Conversely, if lim sup 0, then 0 lim inf lim sup 0, implies that the limits supremum and infimum of are equal (to zero). Hence by Theorem.36, the limit exists and equals zero. 18