Solutions to Section 1

Transcription

1 Solutions to Section Exercise. Show tht nd. This follows from the fct tht mx{, } nd mx{, } Exercise. Show tht = { if 0 if < 0 Tht is, the bsolute vlue function is piecewise defined function. Grph this function in the rectngulr coordinte system. If 0 then 0 so tht = mx{, } =. If < 0 then > 0 so tht = mx{, } =. The grph is shown in Figure. Figure Exercise.3 Show tht 0 with = 0 if nd only if = 0. We see from the grph of tht 0 nd = 0 if nd only if = 0 Exercise.4 Show tht if = b then = ±b. Suppose first tht 0. Then =. If b 0 then b = b. In this cse, = b implies tht = b.

2 If b < 0 then b = b. In this cse, = b implies tht = b. Suppose now tht < 0. Then =. If b 0 then b = b. In this cse, = b implies tht = b which is the sme s = b. If b < 0 then b = b. In this cse, = b implies tht = b which is equivlent to = b Exercise.5 Solve the eqution 3x = 5x + 4. We hve either 3x = 5x + 4 or 3x = (5x + 4). Solving the first eqution we find x = 3. Solving the second eqution, we find x = 4 Exercise.6 Show tht =. If 0 then < 0. Thus, = nd = ( ) =. Hence, =. If < 0 then > 0. In this cse, = nd =. Tht is, = Exercise.7 Show tht b = b. Suppose first tht 0. Then =. If b 0 then b = b. Moreover, b 0. In this cse, b = b = b. If b < 0 then b = b. Moreover, b 0. In this cse, b = b = ( b) = b. Suppose now tht < 0. Then =. If b 0 then b = b. Moreover, b 0. In this cse, b = b = ( )b = b. If b < 0 then b = b. Moreover, b > 0. In this cse, b = b = b Exercise.8 Show tht =, where 0.

3 If > 0 then > 0. Thus, = =. If < 0 then = = = Exercise.9 Show tht b = where b 0. b < 0. Thus, Using both Exercise.7 nd Exercise.8 we cn write the following b = b = b = = b b Exercise.0 Show tht for ny two rel numbers nd b we hve b b. From Exercise., we hve b b = b, where we used Exercise.7 Exercise. Recll tht number b 0 is the squre root of number, written = b, if nd only if = b. Show tht =. Since (±) = we cn write = if < 0 or = if 0. But this is equivlent to writing = by Exercise. Exercise. suppose tht A nd B re points on coordinte line tht hve coordintes nd b, respectively. Show tht b is the distnce between the points A nd B. Thus, if b = 0, mesures the distnce from the number to the origin. Let d be the distnce between A nd B. If < b (See Figure ()) then d = b nd b = ( b) = b = d. If > b (See Figure (b)) then d = b nd b = b = d 3

4 Figure Exercise.3 Grph the portion of the rel line given by the inequlity () x < δ (b) 0 < x < δ where δ > 0. Represent ech grph in intervl nottion. () The grph is given in Figure 3(). In intervl nottion, we hve ( δ, + δ). (b) The grph is given in Figure 3(b). In intervl nottion, we hve ( δ, ) (δ, + δ) Figure 3 Exercise.4 Show tht x < k if nd only if k < x < + k, where k > 0. From the previous exercise, we cn write x < k k < x < + k Exercise.5 Show tht x > k if nd only if x < k or x > + k, where k 0. Using Figure 4, we see tht x > k x < k or x > k. This is equivlent to x < k or x > + k Figure 4 4

5 Exercise.6 Solve ech of the following inequlities: () x 3 < 5 nd (b) x + 4 >. () Using Exercise.4 we cn write x 3 < 5 5 < x 3 < < x < < x < 4. (b) Using Exercise.5 we cn write x + 4 > x + 4 < or x + 4 > which is equivlent to x < 6 or x > Exercise.7 (Tringle inequlity) Use Exercise., Exercise.7, nd the expnsion of ( + b ) to estblish the inequlity + b + b, where nd b re rbitrry rel numbers. We hve ( + b ) = ( + b)( + b) = + b + b + b + b = ( + b ). Now, the result follows by tking the squre root of both sides Exercise.8 Show tht for ny rel numbers nd b we hve b b. Hint: Notice tht = ( b) + b. Using Exercise.7 we cn write = ( b)+b b + b. Subtrcting b from both sides to obtin the desired inequltiy Exercise.9 Let R. Show tht mx{, 0} = ( + ) nd min{, 0} = ( ). The results re cler if = 0. If > 0 then = nd mx{, 0} = = ( + ). If < 0 then = mx{, 0} = 0 = ( + ). Likewise for the minimum Exercise.0 Show tht + b = + b if nd only if b 0. 5

6 Suppose first tht + b = + b. Squring both sides we find + b + b = + b + b or equivlently b = b = b. But this is true only when b 0. Conversely, suppose tht b 0. If = 0 we hve 0 + b = b = 0 + b. Likewise when b = 0. So ssume tht b > 0. Suppose tht > 0 nd b > 0. Then +b > 0 nd in this cse +b = +b = + b. Similr rgument when < 0 nd b < 0 Exercise. Suppose 0 < x <. Simplify x+3 x +5x 3. We hve x + 3 x + 5x 3 = x + 3 (x )(x + 3) = x = x Exercise. Write the function f(x) = x + + x 4 s piecewise defined function. Sketch its grph. We hve Likewise, Combining we find { x + if x x + = (x + ) if x < x 4 = { x 4 if x 4 (x 4) if x < 4 x + if x < f(x) = 6 if x < 4 x if x 4 The grph is given below 6

7 Exercise.3 Prove tht b b for ny rel numbers nd b. We hve b b = b so tht b b b by Exercise.8. Now using Exercise.4, the result follows Exercise.4 Solve the eqution 4 x 3 3 x 3 =. Let u = x 3. Then 4u 3u = 0 implies u = or u =. Since u 0 4 we must hve x 3 = u =. Hence, x 3 = or x 3 =. Thus, x = or x = 4 Exercise.5 Wht is the rnge of the function f(x) = x x for ll x 0? If x > 0 then x x {, } = x x =. If x < 0 then x x = x x =. Thus, the rnge is Exercise.6 Solve 3 x 7. Write your nswer in intervl nottion. Solving the inequlity x 7 we find 5 x 9. Solving the inequlity x 3 we find x or x 5. Thus, the common intervls re [ 5, ] [5, 9] 7

8 Exercise.7 Simplify x x. The nswer is x x = x x = Exercise.8 Solve the inequlity x+ < 3. Write your nswer in intervl nottion. x We hve 3 < x+ x+ x+7 < 3. The inequlity < 3 implies < 0. Solving x x x this inequlity we find x < or x > 7 x+. Likewise, The inequlity > 3 x implies 4x 5 > 0. Solving this inequlity we find x < 5 or x >. Hence, the x 4 common intervl is (, 5) ( 7, ) 4 Exercise.9 Suppose x nd y re rel numbers such tht x y < x. Show tht xy > 0. Since x y < x we hve x < x y < x. Multiplying through by nd dding x we obtin x x < y < x + x. If x = 0 then 0 < y < 0 which is impossible. Therefore either x > 0 or x < 0. If x > 0 then 0 < y < x. Hence xy > 0. If x < 0 then x < y < 0 nd so xy > 0 8

9 Solutions to Section Exercise. Prove tht A is bounded if nd only if there is positive constnt C such tht x C for ll x A. Suppose tht A is bounded. Then there exist rel numbers m nd M such tht m x M for ll x A. Let C = { m, M + } > 0. Then C m m x M M + C. Tht is, x C for ll x A. Conversely, suppose tht x C for ll x A nd for some C > 0. Then, by Exercise.4, we hve C x C for ll x A. Let m = C nd M = C Exercise. Let A = [0, ]. () Find n upper bound of A. How mny upper bounds re there? (b) Find lower bound of A. How mny lower bounds re there? () Any number greter thn is n upper bound. (b) Any number less thn 0 is lower bound Exercise.3 Consider the set A = { n : n N}. () Show tht A is bounded from bove. Find the supremum. Is this supremum mximum of A? (b) Show tht A is bounded from below. Find the infimum. Is this infimum minimum of A? () The supremum is which is lso the mximum of A. (b) The infimum is 0 which is not minimum of A Exercise.4 Consider the set A = { : n N}. n () Show tht is n upper bound of A. (b) Suppose L < is nother upper bound of A. Let n be positive integer such tht n >. Such number n exist by the Archimedin property L which we will discuss below. Show tht this leds to contrdiction. Thus, L. This shows tht is the lest upper bound of A nd hence sup A =. 9

10 () Since > 0 we hve < for ll n N. Thus, is n upper bound n n of A. (b) Since n >, we find L <. But this contrdicts the fct tht L is L n n upper bound of A. Hence, we must hve < L. This shows, tht is the smllest upper bound nd so is the supremum of A Exercise.5 Let, b R with > 0. () Suppose tht n b for ll n N. Show tht the set A = {n : n N} hs supremum. Cll it c. (b) Show tht n c for ll n N. Tht is, c is n upper bound of A. Hint: n + N for ll n N. (c) Conclude from (b) tht there must be positive integer n such tht n > b. () Since n b for ll n N, the set A is bounded from bove. By the completeness xiom of R, A hs supremum, denote it by c = sup{a}. (b) Let n N. Then n + N so tht (n + ) c or n c. (c) From (b), we hve tht c is n upper bound of A. By the definition of supremum, we must hve c c which is impossible. This shows tht A cnnot hve n upper bound. Tht is, there must be positive integer n such tht n > b Exercise.6 Let nd b be two rel numbers such tht < b. () Let [] denote the gretest integer less thn or equl to. Show tht [] < < [] +. (b) Let n be positive integer such tht n >. Show tht n + < nb. b (c) Let m = [n] +. Show tht n < m < nb. Thus, < m < b. We see n tht between ny two distinct rel numbers there is rtionl number. () Since [] <, we hve < [] < which is equivlent to [] < < [] +. (b) Since n > nd b > 0 we cn hve n(b ) > or n + < nb. b (c) We hve m = [n] n < [n] + = m < n + < nb. Thus, n < m < nb. Dividing through by n > 0 we obtin < m < b n 0

11 Exercise.7 Consider the set A = { ( )n : n N}. n () Show tht A is bounded from bove. Find the supremum. Is this supremum mximum of A? (b) Show tht A is bounded from below. Find the infimum. Is this infimum minimum of A? () The supremum is which is lso mximum. (b) The infimum is which is lso minimum Exercise.8 Consider the set A = {x R : < x < }. () Show tht A is bounded from bove. Find the supremum. Is this supremum mximum of A? (b) Show tht A is bounded from below. Find the infimum. Is this infimum minimum of A? () is supremum tht is not mximum. (b) is n infimum tht is not minimum Exercise.9 Consider the set A = {x R : x > 4}. () Show x A nd x < leds to contrdiction. Hence, we must hve tht x for ll x A. Tht is, is lower bound of A. (b) Let L be lower bound of A such tht L >. Let y = L+. Show tht < y < L. (c) Use () to show tht y A nd L y. Show tht this leds to contrdiction. Hence, we must hve L which mens tht is the infimum of A. () If x A nd x < then x < 4 which contrdicts the fct tht x A. Thus, for ll x A we hve x. This shows tht is lower bound of A. (b) Since L > we hve L + > 4 nd this implies y = L+ >. Also, y = L+ < L+L = L. (c) Since y > we hve y > 4 so tht y A. But L is lower bound of A so we must hve L y. But this contrdicts y < L from (b). It follows tht is the lest lower bound of A

12 Exercise.0 Show tht for ny rel number x there is positive integer n such tht n > x. Let = nd b = x in the Archimeden property Exercise. Let nd b be ny two rel numbers such tht < b. () Let w be fixed positive irrtionl number. Show tht there is rtionl number r such tht < wr < b. (e) Show tht wr is irrtionl. Hence, between ny two distinct rel numbers there is n irrtionl number. () Since < b, we hve < b. By Exercise??, there is rtionl number w w r such tht < r < b or < rw < b. w w (e) If rw = s with s rtionl then w = s which is rtionl, contrdiction. r Hence, rw is irrtionl Exercise. Suppose tht α = sup A <. Let ɛ > 0 be given. Prove tht there is n x A such tht α ɛ < x. Suppose the contrry. Tht is, α ɛ x for ll x A. In this cse, α ɛ is n upper bound of A. Thus, we must hve α α ɛ which is impossible Exercise.3 Suppose tht β = inf A <. Let ɛ > 0 be given. Prove tht there is n x A such tht β + ɛ > x. Suppose the contrry. Tht is, β + ɛ x for ll x A. In this cse, β + ɛ is lower bound of A. Thus, we must hve β + ɛ β which is impossible Exercise.4 For ech of the following sets S find sup{s} nd inf{s} if they exist. () S = {x R : x < 5}. (b) S = {x R : x > 7}. (c) S = { n : n N}.

13 () sup{s} = 5 nd inf{s} = 5. (b) sup{s} = nd inf{s} =. (c) sup{s} = 0 nd inf{s} = 3

14 Solutions to Section 3 Exercise 3. Find simple expression for the generl term of ech sequence. (),, 3, 4, (b), 3, 4 3, 5 4, (c), 3, 5, 7, 9, (d),,,,,, () n = ( ) n n. (b) n = n+ n. (c) n = n (d) n = ( ) n Exercise 3. Show tht the sequence { n} converges to 0. n= Let ɛ > 0 be given. We wnt to find positive integer N ɛ such tht if n N ɛ then 0 < ɛ. But this lst inequlity implies tht n >. Let N n ɛ ɛ be positive integer greter thn. If n N ɛ ɛ > then 0 ɛ n = < ɛ. This n shows tht lim n n = 0 Exercise 3.3 Show tht the sequence { } + C converges to, where C 0 is constnt. n n= Let ɛ > 0. We wnt to find positive integer N ɛ such tht + C n < ɛ for ll n N ɛ. But + C n < ɛ implies n > C. By choosing N ɛ ɛ to be positive integer greter thn C the result follows ɛ Exercise 3.4 Is there number L with the property tht ( ) n L < for ll n N, where N is some positive integer? Hint: Consider the inequlity with n even integer greter thn N nd n odd integer greter thn N. 4

15 If n e N is n even integer thn ( ) ne L = L <. If n o N is n odd integer thn ( ) no L = L <. This shows tht L is within one unit of both nd which is impossible. Thus, L does not exist Exercise 3.5 Use the previous exercise to show tht the sequence {( ) n } n= is divergent. Assume the contrry. Tht is, suppose there is n L such tht lim n ( ) n = L. Let ɛ =. Then, there is positive integer N such tht n N implies ( ) n L <. But by the previous exercise, this is impossible. Hence, the given sequence is divergent Exercise 3.6 Suppose tht lim n n = nd lim n n = b with < b. Show tht by choosing ɛ = b > 0 we end up with the impossible inequlity b < b. A similr result holds if b <. Thus, we must hve = b. Hint: Exercise.6 nd Exercise.7. Let ɛ = b > 0. Since the sequence converges to, we cn find positive integer N such tht n N = n < b. Similrly, since the sequence converges to b we cn find positive integer N such tht n N = n b < b. Let N = mx{n, N }. Then for n N we hve n N nd n N. Moreover, by using Exercise.6 nd Exercise.7 we hve b = b = (b n ) + ( n ) b n + n = n b + n < b + b = b Thus, we conclude tht b < b which is impossible. Likewise, if b < we end up with b < b which is impossible. Tht is, either < b or b < leds to contrdiction. Hence, = b 5

16 Exercise 3.7 Show tht ech of the following sequences is bounded. Identify M in ech cse. () n = ( ) n. (b) n = n ln (n+). () We hve n = ( ) n = so tht M =. (b) n n n. Also, ln (n + ) ln ln (n+). ln Hence, n ln Exercise 3.8 Let { n } n= be sequence such tht n K for ll n N. Show tht this sequence is bounded. Identify your M. Let M = N + K. Then n M for ll n. Tht is, the sequence is bounded Exercise 3.9 Show tht convergent sequence is bounded. convergence with ɛ =. Hint: use the definition of Let { n } n= be convergent sequence with limit L. Let ɛ =. There is positive integer N such tht n L < for ll n N. By Exercise.8, we obtin n L < or n < + L for ll n N. By the previous problem with K = + L, the sequence is bounded Exercise 3.0 Give n exmple of bounded sequence tht is divergent. Let n = ( ) n. We know tht this sequence is bounded (Exercise 3.7()). We lso know tht this sequence is divergent (Exercise ) be three sequences with the following condi- Exercise 3. Let { n } n=, {b n} n=, {c n} n= tions: 6

17 () b n n c n for ll n K, where K is some positive integer. () lim n b n = lim n c n = L. Show tht lim n n = L. Hint: Use the definition of convergence long with Exercise.4. Let ɛ > 0. By hypothesis, there exist positive integers N nd N such tht b n L < ɛ for ll n N or equivlently L ɛ < b n < L + ɛ for ll n N nd c n L < ɛ for ll n N or equivlently L ɛ < c n < L + ɛ for ll n N. Let N = N + N + K. Suppose n N. Then L ɛ < b n n c n < L + ɛ. Tht is L ɛ < n < L + ɛ for ll n N. This implies tht lim n n = L Exercise 3. An expnsion of (+b) n, where n is positive integer is given by the Binomil formul n ( + b) n = C(n, k) k b n k k=0 where C(n, k) = n! k!(n k)!. () Use the Binomil formul to estblish the inequlity ( + x) n + x n, x 0 (b) Show tht if then lim n n =. Hint: Use Exercise 3.3. () Using the Binomil formul with = nd b = x n we find or ( + x n ) n = + n x+(other positive terms) + x. n + x ( + x n) n. Tking the n th root of both sides we find ( + x) n + x n. 7

18 (b) If then n. By letting x = 0 in () we find n + n. By Exercise 3.3 we know tht lim n + =. Thus, by the squeeze rule n we obtin lim n =, n Exercise 3.3 Prove tht the sequence {cos (nπ)} n= is divergent. Note tht {cos (nπ)} n= divergent = {( ) n } n= nd by Exercise, this sequence is Exercise 3.4 Let { n } n= be the sequence defined by n = n for ll n N. Explin why the sequence { n } n= does not converge to ny limit. The sequence is unbounded Exercise 3.5 () Show tht for ll n N we hve n! n n n. (b) Show tht the sequence { n } n= where n = n! n n its limit. is convergent nd find () We know tht n i n! for ll 0 i n. Thus, n n n n n n. n n n n n n n! (b) By the Squeeze rule we find tht lim n = 0 n n Exercise 3.6 Using only the definition of convergence show tht lim n 3 n n 00 =. = n(n )(n ) n n n = 8

19 Let ɛ > 0. We wnt to find positive integer N such tht if n N then 3 n n 00 < ɛ or n 00 < ɛ. Let n > Then 3 n 00 > 0 so tht the previous inequlity becomes Solving this for n we find n 00 < ɛ. ( n > + 00). ɛ Let N be positive integer greter thn ( ) 3 ɛ. Then for n N we hve 3 n n 00 < ɛ Exercise 3.7 Consider the sequence defined recursively by = nd n+ = + n for ll n N. Show tht n for ll n N. The proof is by induction on n. For n = we hve =. Suppose tht n. Then n+ = + n + = Exercise 3.8 Clculte lim n (n +) cos n n 3. We hve n + (n + ) cos n n +. n 3 n 3 n 3 By the Squeeze rule we conclude tht the limit is 0 9

20 Exercise 3.9 Clculte lim n ( ) n+3 n. We hve n ( )n+3 n n. By the Squeeze rule the limit is 0 Exercise 3.0 Suppose tht lim n n = L with L > 0. Show tht there is positive integer N such tht x N > x. Let ɛ = L. Then there is positive integer N such tht if n N we hve n L < L. Thus, N L < L or L < N L. Hence, N > L or N > L Exercise 3. Let R nd n N. Clerly, < +. n () Show tht there is Q such tht < < +. Hint: Exercise n.6(c). (b) Show tht there is Q such tht < <. (c) Continuiung the bove process we cn find sequence { n } n= such tht < n < + for ll n N. Show tht this sequence converges to. n We hve proved tht if is rel number then there is sequence of rtionl numbers converging. We sy tht the set Q is dense in R. () This follows from Exercise.6(c). (b) Similr to (). (c) Applying the Squeeze rule, we obtin lim n n = 0

21 Solutions to Section 4 Exercise 4. Suppose tht lim n n = A nd lim n b n = B. Show tht lim n ± b n = A ± B. n We will prove the result for ddition. The difference cse is similr. Let ɛ > 0. Since the two sequences re convergent, there exist positive integers N nd N such tht nd n A < ɛ for ll n N b n B < ɛ for ll n N. Let N = N + N. Then for ll n N we hve n N nd n N. Hence, ( n +b n ) (A+B) = ( n A)+(b n B) n A + b n B < ɛ + ɛ = ɛ. This estblishes the desired result Exercise 4. Suppose tht lim n n = A nd lim n b n = B. () Show tht b n M for ll n N, where M is positive constnt. (b) Show tht n b n AB = ( n A)b n + A(b n B). (c) Let ɛ > 0 be rbitrry nd K = M + A. Show tht there exists positive integer N such tht n A < ɛ K for ll n N. (d) Let ɛ > 0 nd K be s in (c). Show tht there exists positive integer N such tht b n B < ɛ K for ll n N. (e) Show tht lim n n b n = AB. () Since {b n } n= is convergent, the sequence is bounded. Thus, there is positive constnt M such tht b n M for ll n. (b) We hve ( n A)b n + A(b n B) = n b n Ab n + Ab n AB = n b n AB. (c) Let ɛ = ɛ. Since lim K n n = A, we cn find positive integer N such tht n A < ɛ = ɛ for ll n N K. (d) Let ɛ = ɛ. Since lim K n b n = B, we cn find positive integer N such

22 tht b n B < ɛ = ɛ K for ll n N. (e) Let N = N + N. Then n N implies n N nd n N. In this cse, n b n AB = ( n A)b n + A(b n B) n A b n + A b n B < ɛ K M + ɛ K A < ɛ + ɛ = ɛ where we hve used the fct tht M K = M M+ A < nd A K = A M+ A < Exercise 4.3 Give n exmple of two divergent sequences { n } n= nd {b n } n= such tht { n b n } nd { n + b n } re convergent. Let n = ( ) n nd b n = ( ) n+. Both sequences re divergent. Moreover, n b n = ( ) n+ = for ll n. Hence, lim n n b n =. Finlly, n + b n = ( ) n ( ) n = 0 for ll n. Therefore, lim n ( n + b n ) = 0 Exercise 4.4 Let k 0 be n rbitrry constnt nd lim n n = A. Show tht lim n k n = ka. Let ɛ > 0 be rbitrry. There is positive integer N such tht n A < ɛ k for ll n N. Moreover, for n N we hve k n ka = k( n A) = k n A < k This estblishes the desired result ɛ k = ɛ. Exercise 4.5 Suppose tht lim n n = 0 nd {b n } n= is bounded. Show tht lim n n b n = 0. Since {b n } n= is bounded, we cn find positive constnt M such tht b n M for ll n. Let ɛ > 0 be rbitrry. Since lim n n = 0, there is

23 positive integer N such tht n 0 = n < ɛ M n N, we hve n b n 0 = n b n = n b n < ɛ M M = ɛ. This shows tht lim n n b n = 0 Exercise 4.6 sin n () Use the previous exercise to show tht lim n = 0. n sin n (b) Show tht lim n = 0 using the squeeze rule. n for ll n N. Thus, for () Let n = nd b n n = sin n. Then lim n n = lim n = 0 nd n b n = sin n. Now the result follows from the previous exercise. (b) Since sin n, we obtin sin n. But lim n n n n = n lim n = 0 so tht by the squeeze rule n sin n lim n n = 0 Exercise 4.7 Suppose tht lim n n = A, with A 0. Show tht there is positive integer N such tht n > A for ll n N. Hint: Use Exercise.8. Let ɛ = A > 0. Then there is positive integer N such tht n A < A for ll n N. By Exercise.8, we hve A n < A or n > A for n N Exercise 4.8 Let { n } n= be sequence with the following conditions: () n 0 for ll n. () lim n n = A, with A 0. () Show tht there is positive integer N such tht for ll n N we hve n A < A n A. (b) Let ɛ > 0 be rbitrry. Show tht there is positive integer N such tht for ll n N we hve n A < A ɛ. 3

24 (c) Using () nd (b), show tht lim n n = A. () From the previous exercise, we cn find positive integer N such tht n > A for n N. Thus, for n N we obtin n A = n A n A < A n A. (b) Let ɛ > 0 be rbitrry. Since lim n n = A, we cn find positive integer N such tht for ll n N we hve n A < A ɛ. (c) Let ɛ > 0 be rbitrry. Let N = N + N. Then n N implies tht n N nd n N. Moreover, n A = n A < n A A n A < A A ɛ = ɛ. This estblishes the required result Exercise 4.9 Let 0 < <. Show tht lim n n Since 0 < <, we hve =. Hint: Use Exercise 3. (b). >. By Exercise 3.(b), we find lim n ( ) n By the previous exercise, we cn write = lim n n =. lim n n = lim n n = = 4

25 Exercise 4.0 Show tht if lim n n = A nd lim n b n = B with b n 0 for ll n nd B 0, then n lim = A n b n B. Using Exercise?? nd Exercise?? we cn write n lim = lim n = A n b n n b n B = A B Exercise 4. Given tht lim n n = A nd lim n b n = B with n b n for ll n. () Suppose tht B < A. Let ɛ = A B > 0. Show tht there exist positive integers N nd N such tht A ɛ < n < A + ɛ for n N nd B ɛ < b n < B + ɛ for n N. (b) Let N = N + N. Show tht for n N we obtin the contrdiction b n < n. Thus, we must hve A B. () Since lim n n = A, there exists positive integer N such tht n A < ɛ for ll n N. By Exercise.4, this is equivlent to A ɛ < n < A+ɛ for n N. Similrly, since lim n b n = B, there exists positive integer N such tht b n B < ɛ for ll n N. By Exercise.4, this is equivlent to B ɛ < b n < B + ɛ for n N. (b) If n N then n N nd n N. Thus, b n <B + ɛ = B + A B =A A B = A ɛ < n = A + B This contrdicts the fct tht n b n for ll n. Hence, we conclude tht A B Exercise 4. Suppose tht lim n n bn = L nd lim n b n = 0 where b n 0 for ll n N. Find lim n n. 5

26 We hve lim n n = lim b n = n n b n ( lim n ) ( ) n lim b b n = L 0 = 0 n n Exercise 4.3 The Fiboncci numbers re defined recursively s follows: = = nd n+ = n+ + n for ll n N. Suppose tht lim n n+ n = L. Find the vlue of L. We hve n+ = n+ + n Divide through by n+ to obtin n+ = + n. n+ n+ Tke the limit s n to obtin L = + L or L L = 0. Solving this qudrtic eqution for L > 0 we find L = + 5 Exercise 4.4 Show tht the sequence defined by n = n n + + n + 3 ( )n n + 7 hve two limits by finding lim n n nd lim n n+. We hve nd lim n n = lim n n n + + 4n + 3 4n + 7 = lim n + n+ = lim n n n + (n + ) + 3 (n + ) + 7 = 0 6

27 Exercise 4.5 Use the properties of this section to find n + 5n lim. n n + 4 We hve lim n n + 5n n + 4 = lim n = lim + 5 n + 4 n n lim n = lim n = Exercise 4.6 Find the limit of the sequence defined by n = n ln n. + 5 n lim n ( + 4 ) n lim n ( + 5 ) n + lim n 4 n We hve lim n ln n = lim n ln n ln n =. Thus, lim n n = e Exercise 4.7 Consider the sequence defined by n = n. () Show tht n n for ll n N. (b) Show tht the sequence { n } n= is divergent. Hint: Exercise 4.. () By induction on n. If n = we hve = =. Suppose tht n n. Then n+ n+ n+ = n n++ n+ n n+ n+ = n+ n+ = n +. (b) Since n n for ll n N nd lim n n =, we conclude tht lim n n = 7

28 Exercise 4.8 Find the limit of the sequence defined by n = ln (n + n) ln n. We hve lim n [ln (n + ( ) ( ) n) ln n] = lim n ln n+ n = lim n n ln + n = ln Exercise 4.9 Consider the sequence defined by n = n 3 n +. () Show tht 3 < n < 3 n for ll n. (b) Find the limit of n s n. () We hve 3 = n 3 n < n 3 n + < n 3 n + 3 n < 3 n. (b) Since lim n n = we conclude by the Squeeze rule tht lim n n = 3 Exercise 4.0 Let { n } n= be convergent sequence of nonnegtive terms with limit L. Suppose tht the terms of sequence stisfy the recursive reltion n n+ = n + for ll N N. Find L. We hve lim n n n+ = lim n ( n + ) lim n n lim n n+ = lim n n + lim n L L = L + L L = 0. Solving this eqution we find L = nd L =. But n 0 for ll n N so tht L 0. Thus, L = Exercise 4. Find the limit of the sequence defined by n = cos n + sin n n. We hve lim sin n n = lim cos n + lim n n n n = + 0 = 8

29 Exercise 4. Suppose tht n+ = n+ n. Show tht the sequence { n } n= is divergent. Suppose the contrry nd let L = lim n n. Then L = L + or L = L + L which leds to the contrdiction 0 =. Hence, the given sequence must be divergent 9

30 Solutions to Section 5 Exercise 5. Show tht the sequence { n } n= is decresing. We hve n + n n+ n n n+ for ll n. This shows tht the sequence is decresing Exercise 5. Show tht the sequence { +e n } n= is incresing. We hve n n + e n e n+ e (n+) e n + e (n+) + e n +e n +e (n+) n n+ for ll n. This shows tht the sequence is incresing Exercise 5.3 Show tht the sequence { n } n= is bounded from below. bound? Are there more thn one lower bound? Wht is lower Since 0 for ll n, the given sequence is bounded from below with n lower bound 0. Any negtive number is lower bound Exercise 5.4 Show tht the sequence { } +e n n= is bounded from bove. Wht is n upper bound? Are there more thn one upper bound? We hve + e n. Thus, the given sequence is bounded +e n from bove with n upper bound equls to. Any number lrger thn is n upper bound Exercise 5.5 Let { n } n= be n incresing sequence tht is bounded from bove. () Show tht there is finite number M such tht M = sup{ n : n }. (b) Let ɛ > 0 be rbitrry. Show tht M ɛ cnnot be n upper bound of the sequence. 30

31 (c) Show tht there is positive integer N such tht M ɛ < N. (d) Show tht M ɛ < n for ll n N. (e) Show tht M ɛ < n < M + ɛ for ll n N. (f) Show tht lim n n = M. Tht is, the given sequence is convergent. () Since the sequence { n } n= is bounded from bove, by the Completeness Axiom there is finite number M such tht M = sup{ n : n }. We will show tht { n } n= converges to M. (b) Let ɛ > 0 be rbitrry. Consider the number M ɛ. This number is either n upper bound of the sequence or not. If it is n upper bound then we must hve M M ɛ which is n impossible inequlity. (c) If such n N does not exist, then we will hve n M ɛ for ll n. This mens tht M ɛ is n upper bound which is impossible by (b). Thus, there is positive integer N such tht M ɛ < N M. (d) Since the sequence is incresing, for ll n N we cn write N n. Hence, for ll n N we hve M ɛ < n. (e) For ll n N we cn write M ɛ < n M < M + ɛ or equivlently n M < ɛ for ll n N. (f) This follows from the definition of convergence nd (e) Exercise 5.6 Consider the sequence { n } n= defined recursively by = 3 nd n+ = n + for n. () Show by induction on n, tht n+ = n +. n+ (b) Show tht this sequence is incresing. (c) Show tht { n } n= is bounded from bove. Wht is n upper bound? (d) Show tht { n } n= is convergent. Wht is its limit? Hint: In finding the limit, use the rithmetic opertions of sequences. () We hve = + = 3 + = 7 = 3 + = Suppose tht n = n +. Then n n+ = n+ = n + + = ( n+ n + ) + = n+ 3

32 n +. n+ (b) Since n+ n = > 0, the given sequence is incresing. n+ (c) By (b), we hve n+ = n + > n. Solving this equlity for n we find n < so tht the sequence { n } n= is bounded from bove with n upper bound equls to. (d) By the previous exercise, the sequence is convergent sy to A. Using the rithmetic opertions of sequences, we cn write ( ) lim n+ = lim n n n + = lim n n + lim n. Thus, A = A +. Solving this eqution for A we find A = Exercise 5.7 Let { n } n= be decresing sequence such tht M n for ll n. Show tht { n } n= is convergent. Hint: Let b n = n nd use Exercise 5.5 nd Exercise 4.4. Since { n } n= is decresing, the sequence {b n } n= is incresing. Since the sequence { n } n= is bounded from below, the sequence {b n } n= is bounded from bove. By Exercise 5.5, the sequence {b n } n= is convergent. By Exercise 4.4, the sequence { n } n= is lso convergent Exercise 5.8 Show tht monotone sequence is convergent if nd only if it is bounded. Suppose first tht sequence { n } n= is monotone convergent sequence. By Exercise 3.9, the sequence { n } n= is bounded. Conversely, suppose tht { n } n= is bounded sequence. Then there is positive constnt M such tht n M for ll n. By Exercise.4, we hve M n M for ll n. This shows tht the sequence is bounded from below s well from bove. If the sequence is either incresing or decresing, then it is convergent by Exercise 5.5 nd Exercise 5.7 3

33 Exercise 5.9 Let n be defined by = nd n+ = + n for n N. () Show tht n for ll n N. Tht is, { n } n= is bounded from bove. (b) Show tht n+ n for ll n N. Tht is, { n } n= is incresing. (c) Conclude tht { n } n= is convergent. Find its limit. () By induction on n. For n =, we hve =. Suppose tht n. Then n+ = + n + =. Thus, n for ll n N. (b) By induction on n. For n = we hve = + 0. Suppose tht n n 0. Then n+ n = + n + n = n n + n + + n 0. (c) Since the sequence is incresing nd bounded from bove, it is convergent, sy with limit. Thus, we hve = +. Solving this eqution we obtin = or =. Since = nd the sequence is incresing we conclude tht = Exercise 5.0 Let n = n k=. k () Show tht n < for ll n N. Hint: Recll tht n k= (b) Show tht { n } n= is incresing. (c) Conclude tht { n } n= is convergent. () We hve n = n k= k (b) Since n+ = n + (n+) = + n k= + <. k n > n, the given sequence is incresing. =. (n+)n n+ (c) This follows from the fct tht n incresing sequence tht is bounded from bove is convergent Exercise 5. Consider the sequence { n } n= defined recursively s follows = nd 7 n+ = n + 3 for ll n N. () show tht < n < 3 for ll n N. (b) Show tht n+ n for ll n N. (c) Deduce tht { n } n= is convergent nd find its limit. 33

34 () We prove this by induction on n. If n = then < = < 3. Suppose tht < n < 3. Then 7 < 7 n+ < < n+ < 3. (b) We hve n+ n = 7 ( n 7 n + 3) = ( 7 n )( n 3) < 0. Thus, n+ n for ll n N. (c) We deduce tht the given sequence is convergent, sy with limit L. Thus, 7L = L + 3. Solving this eqution for L we find L = nd L = 3. Since the sequence is decresing nd < n < 3 we must hve L = Exercise 5. Let { n } n= be n incresing sequence. Define b n = n. Show tht n the sequence {b n } n= is incresing. note tht s s s n s n+, hence s +s + +s n ns n+. Adding n(s + + s n ) to both sides, we obtin n(s + + s n ) + (s + + s n ) n(s + + s n ) + ns n+, or, in other words, (n + )(s + + s n ) n(s + + s n + s n+ ), Dividing both sides by n(n + ), we obtin s + + s n n s + + s n + s n+, n + or, in other words, b n b n+ for ll n. This proves our clim Exercise 5.3 Give n exmple of monotone sequence tht is divergent. One exmple is the sequence defined by n = n Exercise 5.4 Consider the sequence defined recursively by = nd n+ = 3 + n ll n N. () Show tht n 6 for ll n N. (b) Show tht { n } n= is incresing. (c) Conclude tht the sequence is convergent. Find its limit. for 34

35 () If n = we hve = < 6. Suppose tht n 6. Then n+ = 3 + n = 6. (b) We hve n+ = 3 + n = 6+n n+n = n (c) Since the sequence is incresing nd bounded from bove it is convergent with limit sy equls to L. Thus, L = 3+ L nd solving for L we find L = 6 Exercise 5.5 Give n exmple of two monotone sequences whose sum is not monotone. Let { n } n= = {,,,, 3, 3, } nd {b n } n= = {,,, 3, 3, }. The first sequence is incresing nd the second sequence is decresing. However the sum is the sequence {0,, 0,, 0,, } which is not monotone 35

36 Solutions to Section 6 Exercise 6. Let { nk } k= be subsequence of sequence { n} n=. Use induction on k to show tht n k k for ll k N. For k = we hve n N so tht n. Suppose tht n k k. Then n k+ > n k k. Thus, n k+ k + Exercise 6. Let { n } n= be sequence of rel numbers tht converges to number L. Let { nk } k= be ny subsequence of { n} n=. () Let ɛ > 0 be given. Show tht there is positive integer N such tht if n N then n L < ɛ. (b) Let N be the first positive integer such tht n N N. Show tht if k N then nk L < ɛ. Tht is, the subsequence { nk } k= converges to L. Hence, every subsequence of convergent sequence is convergent to the sme limit of the originl sequence. () This follows from the fct tht the sequence { n } n= converges to L. (b) Let N be such tht n N N. Then if k N we hve n k > n N N. Hence, nk L < ɛ. This shows tht the subsequence { nk } k= converges to L Exercise 6.3 Let { n } n= be sequence of rel numbers. Let S = {n N : n > m for ll m > n}. () Suppose tht S is infinite. Then there is sequence n < n < n 3 < such n k S. Show tht nk+ < nk. Thus, the subsequence { nk } k= is decresing. (b) Suppose tht S is finite. Let n be the first positive integer such tht n S. Show tht the subsequence { nk } k= is incresing. () By the definition of S we hve nk+ < nk since n k+ > n k. (b) Let n be the first positive integer such tht n S. This mens tht there is positive integer n > n such tht n < n. But n S so tht there is positive integer n 3 > n such tht n < n3. Continuing this process we find n incresing subsequence { nk } k= 36

37 Exercise 6.4 (Bolzno-Weierstrss) Every bounded sequence hs convergent subsequence. Hint: Exercise 5.8 By the previous exercise, the sequence hs bounded monotone subsequence. By Exercise 5.8 this subsequence is convergent Exercise 6.5 Show tht the sequence {e sin n } n= hs convergent subsequence. Since sin n we must hve e esin n e for ll n N. Thus, the given sequence is bounded so tht by the Bolzno-Weierstrss theorem it hs convergent subsequence Exercise 6.6 Prove tht the sequence { n } n= where n = cos nπ is divergent. Consider the two subsequences { n+ } n= tht converges to 0 nd { 4 } n= tht converges to. Thus, the originl sequence must be divergent Exercise 6.7 Prove tht the sequence { n } n= where n = (n + 0n + 35) sin n 3 n + n + hs convergent subsequence. Hint: Show tht { n } n= is bounded. We hve n = (n + 0n + 35) sin n 3 n + n + = (n + 0n + 35) sin n 3 n + n + n + 0n + 35 n + n + n + 0n + 35 = + 0 n n + 35 n 56 Hence, by the Bolzno Weierstr theorem the given sequence hs convergent subsequence 37

38 Exercise 6.8 Show tht the sequence defined by n subsequence. = cos n sin n hs convergent We hve cos n nd sin n. Adding we obtin 3 n 3 so tht the given sequence is bounded. By the Bolzno-Weierstrss theorem, the given sequence hs convergent subsequence Exercise 6.9 True or flse: There is sequence tht converges to 6 but contins subsequence converging to 0. Justify your nswer. By Exercise 6., this cnnot hppen Exercise 6.0 Give n exmple of n unbounded sequence with bounded subsequence. Let { 0 if n is odd n = n if n is even Then { n } n= is unbounded. However, the subsequence { n+ } n= = {0, 0, 0, } is bounded Exercise 6. Show tht the sequence {( ) n } n= is divergent by using subsequences. This sequence hs two subsequences { n } n= tht converges to nd { n+ } n= tht converges to. Thus, the originl sequence cnnot be convergent 38

39 Solutions to Section 7 Exercise 7. Consider the sequence whose n th term is given by n =. Let ɛ > 0 be n rbitrry nd choose N >. Show tht for m, n N we hve ɛ m n < ɛ. Tht is, the bove sequence is Cuchy sequence. Hint: Exercise.7. Let n, m N. Then by Exercise.7 we hve n m n + m < ɛ + ɛ == ɛ Exercise 7. Show tht ny Cuchy sequence is bounded. Hint: Let ɛ = nd use Exercise.8. Let { n } n= be Cuchy sequence. Let ɛ =. There is positive integer N such tht whenever n, m N we hve n m <. In prticulr, letting m = N we cn write n N < for ll n N. By Exercise.8 we cn write n N < for ll n N or n < + N for ll n N. Let M = mx{,,, N, N + }. Then n M for ll n. Tht is, { n } n= is bounded Exercise 7.3 Show tht if lim n n = A then { n } n= is Cuchy sequence. Thus, every convergent sequence is Cuchy sequence. Let ɛ > 0 be rbitrry. Then there is positive integer N such tht n A < for ll n N. Thus, for ll n, m N we hve ɛ n m = ( n A) + (A m ) n A + m A < ɛ + ɛ = ɛ. This shows tht { n } n= is Cuchy sequence 39

40 Exercise 7.4 () Using Exercise 7., show tht for ech n, the sequence { n, n+, } is bounded. (b) Show tht for ech n the infimum of { n, n+, } exists. Cll it b n. () By Exercise 7., there is positive constnt M such tht n M for ll n. In prticulr, n M, n+ M, n+ M,. Tht is, { n, n+, } is bounded. (b) Since { n, n+, } is bounded, it is bounded from below. By the Completeness Axiom, b n = inf{ m : m n} exists Exercise 7.5 () Show tht the sequence {b n } n= is bounded from bove. (b) Show tht the sequence {b n } n= is incresing. Hint: Show tht b n is lower bound of the sequence { n+, n+, }. () By the definition of infimum, we hve b n n for ll n. Thus, b n n n M. This, shows tht {b n } n= is bounded from bove. (b) We know tht b n m for ll m n. In prticulr b n m for ll m n +. This shows tht b n is lower bound of { n+, n+, }. But b n+ is the gretest lower bound of { n+, n+, }. Hence, b n b n+ nd the sequence {b n } n= is incresing Exercise 7.6 Show tht the sequence {b n } n= is convergent. Cll the limit B. This follows from Exercise 5.5 Exercise 7.7 () Let ɛ > 0 be rbitrry. Using the definition of Cuchy sequences nd Exercise??, show tht there is positive integer N such tht N ɛ < n < N + ɛ for ll n N. (b) Using (), show tht N ɛ is lower bound of the sequence { N, N+, }. Thus, N ɛ b n for ll n N. (c) Agin, using () show tht b n N + ɛ for ll n N. Thus, combining 40

41 (b) nd (c), we obtin N ɛ b n < N + ɛ. (d) Using Exercise 4., show tht N ɛ B N + ɛ. (e) Using (), (d), nd Exercise.7, show tht lim n n = B. Thus, every Cuchy sequence is convergent. () Let ɛ > 0 be rbitrry. Since { n } n= is Cuchy, there is positive integer N such tht n m < ɛ for ll n, m N. In prticulr, for ll n N we hve n N < ɛ which, by Exercise.4, we hve N ɛ < n < N + ɛ for ll n N. (b) From prt (), we hve N ɛ < n for ll n N. This shows, tht N ɛ is lower bound of the sequence { n, n+, } n=n for n N. But b n is the gretest lower bound of the sequence { n, n+, } n=n for n N. We conclude tht N ɛ b n for ll n N. (c) For n N we hve b n n < N + ɛ. (d) Tking the limit s n we obtin N ɛ B N + ɛ. (e) Using (), (d) nd Exercise.7, we hve tht for n N n B = ( n N ) + ( N B) n N + N B < ɛ + ɛ = ɛ Exercise 7.8 () Show tht if { n } n= is Cuchy then { n} n= is lso Cuchy. (b) Give n exmple of Cuchy sequence { n} n= such tht { n } n= is not Cuchy. () Since { n } n= is Cuchy then it is convergent. Since the product of two convergent sequences is convergentm the sequence { n} n= is convergent nd therefore is Cuchy. () Let n = ( ) n for ll n N. The sequence { n } n= is not Cuchy since it is divergent. However, the sequence { n} n= = {,, } converges to so it is Cuchy Exercise 7.9 Let { n } n= be Cuchy sequence such tht n is n integer for ll n N. Show tht there is positive integer N such tht n = C for ll n N, where C is constnt. 4

42 Let ɛ =. Since { n} n= is Cuchy, there is positive integer N such tht if m, n N we hve m n <. But m n is n integer so we must hve n = N for ll n N Exercise 7.0 Let { n } n= be sequence tht stisfies n+ n+ < c n+ n for ll n N where 0 < c <. () Show tht n+ n < c n for ll n. (b) Show tht { n } n= is Cuchy sequence. () The proof is by induction on n. For n = we hve 3 < c. Suppose tht n+ n+ < c n+. Then n+3 n+ < c n+ n+ < c n+. (b) Let ɛ > 0 be given. Since lim n c n = 0 we cn find positive integer N such tht if n N then c n < ( c)ɛ. Thus, for n > m N we hve n m m+ m + m+ m+ + + n n <c m + c m+ + + c n <c m ( + c + c + ) = cm c < ɛ It follows tht { n } n= is Cuchy sequence Exercise 7. Wht does it men for sequence { n } n= to not be Cuchy? A sequence { n } n= is not Cuchy sequence if there is rel number ɛ > 0 such tht for ll positive integer N there exist n, m N such tht n, m N nd n m ɛ Exercise 7. Let { n } n= nd {b n } n= be two Cuchy sequences. Define c n = n b n. Show tht {c n } n= is Cuchy sequence. 4

43 Let ɛ > 0 be given. There exist positive integers N nd N such tht if n, m N nd n, m N we hve n m < ɛ nd b n b m < ɛ. Let N = N + N. If n, m N then c n c m = n b n m b m ( n b n ) + ( m b m ) n m + b n b m < ɛ. Hence, {c n } n= is Cuchy sequence Exercise 7.3 Suppose { n } n= is Cuchy sequence. Suppose n 0 for infinitely mny n nd n 0 for infinitely mny n. Prove tht lim n n = 0. Let ɛ > 0 be given. Since { n } n= is Cuchy, there is positive integer N such tht if n, m N then n m < ɛ. Let n N. Then the element n is positive, negtive or zero. Cse : Suppose n 0. Since m 0 for infinitely mny m, there is m N such tht m 0 (else, there would be less thn N nonpositive terms in the sequence, which contrdicts the ssumption). Since n, m N, n n m = n m < ɛ. Cse : Suppose n < 0. Since m geq0 for infinitely mny m, there is m N such tht m 0 (else, there would be less thn N non-negtive terms in the sequence, which contrdicts the ssumption). But then, since n, m N, n < n n + m n m < ɛ. So, in ny cse, for ny given ɛ we cn find N N such tht if n N then n < ɛ, tht is, lim n n = 0 Exercise 7.4 Explin why the sequence defined by n = ( ) n is not Cuchy sequence. We know tht every Cuchy sequence is convergent. We lso know tht the given sequence is divergent. Thus, it cn not be Cuchy 43

44 Solutions to Section 8 Exercise 8. Show tht lim x x x =. Let ɛ > 0 be rbitrry. We note first tht x x = (x )(x + ) x = x + = x. Thus, choose δ ɛ. If 0 < x < δ (which is ɛ) we obtin x x = x < ɛ which is the required result Exercise 8. Let f(x) = x. Suppose tht lim x x 0 f(x) = L. () Show tht there is positive number δ such tht if 0 < x < δ then x L x <. 4 (b) Let x = δ nd x 4 = δ. Compute the vlue of f(x 4 ) f(x ). (c) Use () to show tht f(x ) f(x ) <. x (d) Conclude tht L does not exist. Tht is, lim x 0 does not exist. x () Since lim x 0 f(x) = L, for ɛ = we cn find δ > 0 such tht x L 4 x < whenever 0 < x 0 < δ. 4 (b) We hve f(x ) f(x ) = ( ) =. (c) Since both x nd x stisfy 0 < x < δ nd 0 < x < δ, we hve f(x ) L < nd f(x 4 ) L <. Thus, 4 f(x ) f(x ) = (f(x ) L) (f(x ) L) () f(x ) L + f(x ) L < = But by (b), we hve tht f(x ) f(x ) =. We conclude tht < which in impossible. x (d) The contrdiction obtined in (c) shows tht lim x 0 does not exist x 44

45 Exercise 8.3 Let f(x) = sin ( x). Suppose tht limx 0 f(x) = L. () Show ( tht there is positive number δ such tht if 0 < x < δ then sin x) L < 4. (b) Let n be positive integer such tht x = < δ nd x (n+)π = δ. Compute the vlue of f(x ) f(x ). (c) Use () to show tht f(x ) f(x ) <. (d) Conclude tht L does not exist. Tht is, lim x 0 sin ( x) does not exist. (n+)π < () Since lim x 0 f(x) = L, for ɛ = we cn find δ > 0 such tht 4 sin ( x) L < whenever 0 < x 0 < δ. 4 (b) We hve f(x ) f(x ) = 0 =. (c) Since both x nd x stisfy 0 < x < δ nd 0 < x < δ, we hve f(x ) L < nd f(x 4 ) L <. Thus, 4 f(x ) f(x ) = (f(x ) L) (f(x ) L) () f(x ) L + f(x ) L < = But by (b), we hve tht f(x ) f(x ) =. We conclude tht < which in impossible. (d) The contrdiction obtined in (c) shows tht lim x 0 sin ( x) does not exist Exercise 8.4 Suppose tht lim x f(x) exists. Also, suppose tht lim x f(x) = L nd lim x f(x) = L. So either L = L or L L. () Suppose tht L L. Show tht there exist positive constnts δ nd δ such tht if 0 < x < δ then f(x) L < L L nd if 0 < x < δ then f(x) L < L L. (b) Let δ = min{δ, δ } so tht δ < δ nd δ < δ. Show tht if 0 < x < δ then L L < L L which is impossible. (c) Conclude tht L = L. Tht is, whenever function hs limit, tht limit is unique. () Since L L, we hve ɛ = L L > 0. Since lim x f(x) = L, there 45

46 is δ > 0 such tht if 0 < x < δ then f(x) L < L L. Similrly, since lim x f(x) = L there is δ > 0 such tht if 0 < x < δ then f(x) L < L L. (b) Suppose 0 < x < δ. Then L L = f(x) L + L f(x) f(x) L + f(x) L < L L + L L (c) The contrdiction obtined in (b) implies L = L Exercise 8.5 Using the ɛδ definition of limit show tht lim x (x + x + ) =. = L L Let ɛ > 0 be given. Note first tht x + x + = (x )(x + ) = x x + ( x + ) x +. If 0 < x + < then x < nd this implies x <. So choose δ = min{, ɛ }. Clerly, δ nd δ ɛ. So if < x+ < δ we hve x +x+ ( x +) x+ < 5 x+ 5 ɛ = ɛ 5 Exercise 8.6 x Prove directly from the definition tht lim x =. x+3 4 First note tht x x = 3 x 4 x + 3. If x < then 0 < x <. Thus, x + 3 = x + 3 > 3. Let ɛ > 0 be given. Choose δ = min{, 4ɛ}. Then if 0 < x < δ we hve x = 3 x 3 x = x < 4ɛ = ɛ x+3 4 < 4 x+3 Exercise 8.7 In this exercise we discuss the concept of sided limits. () We sy tht L is the left side limit of f s x pproches from the left if nd only if 46

47 ɛ > 0, δ > 0 such tht 0 < x < δ f(x) L < ɛ nd we write lim x f(x) = L. Show tht lim x 0 x x =. (b) We sy tht L is the right side limit of f s x pproches from the right if nd only if ɛ > 0, δ > 0 such tht 0 < x < δ f(x) L < ɛ nd we write lim x + f(x) = L. Show tht lim x 0 + x x =. Exercise 8.8 Prove tht L = lim x f(x) if nd only if lim x f(x) = lim x + f(x) = L. Suppose first tht lim x f(x) = L. Let ɛ > 0 be given. Then there is δ > 0 such tht 0 < x < δ f(x) L < ɛ. Suppose tht 0 < x < δ. Then 0 < x = x < δ f(x) L < ɛ. This shows tht L = lim x f(x). Likewise, one cn show tht L = lim x + f(x). Conversely, suppose lim x f(x) = lim x + f(x) = L. Let ɛ > 0 be given. Then there exist δ > 0 nd δ > 0 such tht 0 < x < δ f(x) L < ɛ nd 0 < x < δ f(x) L < ɛ. Let δ = min{δ, δ }. Suppose 0 < x < δ. Since x = ±(x ) we must hve f(x) L < ɛ. This shows tht lim x f(x) = L Exercise 8.9 Using ɛ nd δ, wht does it men tht lim x f(x) L? If lim x f(x) L then there is n ɛ > 0 such tht for ll δ > 0 there is n x in the domin of f such tht 0 < x < δ but f(x) L ɛ 47

48 Solutions to Section 9 Exercise 9. Suppose tht lim x f(x) = L nd lim x g(x) = L. Show tht lim [f(x) ± g(x)] = L ± L. x Let ɛ > 0 be rbitrry. Since lim x f(x) = L, we cn find δ > 0 such tht 0 < x < δ = f(x) L < ɛ. Similrly, since lim x g(x) = L, we cn find δ > 0 such tht 0 < x < δ = g(x) L < ɛ. Let δ = min{δ, δ }. Notice tht δ δ nd tht δ δ. Thus, if 0 < x < δ then (f(x) + g(x)) (L + L ) = (f(x) L ) + (g(x) L ) f(x) L + g(x) L < ɛ + ɛ = ɛ A similr rgument holds for f(x) g(x) Exercise 9. Suppose tht lim x f(x) = L nd lim x g(x) = L. Show the following: () There is δ > 0 such tht 0 < x < δ = f(x) < + L. Hint: Notice tht f(x) = (f(x) L ) + L. (b) Given ɛ > 0, there is δ > 0 such tht 0 < x < δ = f(x) L < ɛ ( + L ). 48

49 () Let ɛ =. Since lim x f(x) = L, we cn find δ > 0 such tht But then (b) Since 0 < x < δ = f(x) L <. f(x) = (f(x) L ) + L f(x) L + L < + L. ɛ > 0 nd lim (+ L ) x f(x) = L, we cn find δ > 0 such tht 0 < x < δ = f(x) L < ɛ ( + L ) Exercise 9.3 Suppose tht lim x f(x) = L nd lim x g(x) = L. () Show tht f(x)g(x) L L = f(x)(g(x) L ) + L (f(x) L ). (b) Show tht f(x)g(x) L L f(x) g(x) L + L f(x) L. (c) Show tht lim x f(x)g(x) = L L. Hint: Use the previous exercise. () We hve f(x)(g(x) L ) + L (f(x) L ) =f(x)g(x) L f(x) + L f(x) L L =f(x)g(x) L L. (b) By () nd Exercise.7/Exercise.7, we hve f(x)g(x) L L = f(x)(g(x) L ) + L (f(x) L ) f(x)(g(x) L ) + L (f(x) L ) = f(x) (g(x) L ) + L (f(x) L ) (c) Let ɛ > 0 be rbitrry. By Exercise 9.(), there is δ > 0 such tht 0 < x < δ = f(x) < + L. Also, by Exercise 9.(b), there exist δ > 0 nd δ 3 > 0 such tht 0 < x < δ = f(x) L < 49 ɛ ( + L )

50 nd 0 < x < δ 3 = g(x) L < ɛ ( + L ). Let δ = min{δ, δ, δ 3 }. Notice tht δ δ, δ δ, nd δ δ 3. Suppose tht 0 < x < δ. Using (b) nd the bove inequlities we find f(x)g(x) L L f(x) g(x) L + L f(x) L ɛ <( + L ) ( + L ) + L ɛ <( + L ) ( + L ) + ( + L ) = ɛ + ɛ = ɛ This estblishes the required result ɛ ( + L ) ɛ ( + L ) Exercise 9.4 () Suppose tht f(x) M for ll x in its domin nd lim x g(x) = 0. Show tht lim f(x)g(x) = 0. x Hint: Recll Exercise 4.5 (b) Show tht lim x 0 x sin ( x) = 0. () Let ɛ > 0 be rbitrry. Since lim x g(x) = 0, there is positive number δ such tht if 0 < x < δ then g(x) 0 = g(x) < ɛ. Thus, for M 0 < x < δ we hve f(x)g(x) 0 = f(x)g(x) = f(x) g(x) < M ɛ M = ɛ. This shows tht lim x f(x)g(x) = 0. (b) Let f(x) = x nd g(x) = sin ( ( x). Note tht limx 0 f(x) = 0 nd g(x) = sin x). It follows from () tht ( ) lim x sin = lim f(x)g(x) = 0 x 0 x x 0 50

51 Exercise 9.5 Suppose tht lim x f(x) = L with L 0. Show tht there exists δ > 0 such tht 0 < x < δ = f(x) > L > 0. Hint: Recll the solution to Exercise 4.7 Let ɛ = L L > 0. Then there exists δ > 0 such tht f(x) L < whenever 0 < x < δ. But by Exercise.8, we hve L f(x) < L L or f(x) > whenever 0 < x < δ Exercise 9.6 Let g(x) be function with the following conditions: () g(x) 0 for ll x in the domin of g. () lim x g(x) = L, with L 0. () Show tht there is δ > 0 such tht if 0 < x < δ then g(x) L < L g(x) L. (b) Let ɛ > 0 be rbitrry. Show tht there is δ > 0 such tht if 0 < x < δ then g(x) L < L ɛ. (c) Using () nd (b), show tht Hint: Recll Exercise 4.8 lim x g(x) = L. () From Exercise 9.5, we cn find δ > 0 such tht g(x) > L whenever 0 < x < δ. Thus, for 0 < x < δ we obtin g(x) L = g(x) L < g(x) L L g(x) L. 5

52 (b) Let ɛ > 0 be rbitrry. Since lim x g(x) = L, we cn find positive number δ such tht if 0 < x < δ then g(x) L < L ɛ. (c) Let ɛ > 0 be rbitrry. Let δ = min{δ, δ }. Notice tht δ δ nd δ δ. Now, if 0 < x < δ then g(x) L = g(x) L < g(x) L L g(x) L < L L ɛ = ɛ. This estblishes the required result Exercise 9.7 Show tht if lim x f(x) = L nd lim x g(x) = L where g(x) 0 in its domin nd L 0 then f(x) lim x g(x) = L. L Hint: Recll Exercise 4.0. Using Exercise 9.6 nd Exercise 9.3 we cn write [ ] f(x) lim x g(x) = lim f(x) = L = L x g(x) L L Exercise 9.8 Let f(x) nd g(x) be two functions with common domin D nd point in D. Suppose tht f(x) g(x) for ll x in D. Show tht if lim x f(x) = L nd lim x g(x) = L then L L. Hint: Recll Exercise 4. Since lim x f(x) = L, there exists positive number δ such tht f(x) L < ɛ whenever 0 < x < δ. By Exercise.4, this is equivlent to L ɛ < f(x) < L + ɛ whenever 0 < x < δ. Similrly, since lim x g(x) = L, there exists positive number δ such tht g(x) L < ɛ whenever 0 < x < δ. By Exercise.4, this is equivlent to L ɛ < 5