a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.


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1 MAS221(21617) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by zero) but not below. (ii) Suppose tht for some ɛ > such n cnnot be found, then for ll A, sup(a) ɛ. But then sup(a) ɛ is smller upper bound for A thn sup(a), nd we hve contrdiction. (2) (iii) The sequence ( n ) is monotonic incresing if n+1 n for ll n N. (iv) Since ( n ) is bounded bove, α exists by the completeness property. By (ii) given ny ɛ > there exists N N such tht N > α ɛ. But ( n ) is monotonic incresing, so for ll n > N we hve n N+1 N > α ɛ. Since n α, nd the result follows. (2) (v) () It s clerly true for n = 1. of n. Then α n = α n < ɛ, for ll n N, n = 1 58 n+1 1 = 57 n + 1 n Assume true for some vlue >, = 56( n 1) n (1 1) =, 57 so n+1 1, nd the required result is true, by induction. (b) For ll n N, n+1 n = 57 n + 1 n + 57 n = 57 n n 57 n n + 57 = 1 2 n n + 57 (by ()), 1
2 nd so n+1 n for ll n N, i.e. ( n ) is monotonic incresing. (c) By () the sequence is bounded bove (by 1), nd by (b) it is monotonic incresing. So it converges to limit by (iv). Let α = lim n = lim n+1. Now tke limits on both sides of the generl formul, nd use lgebr of limits to get α = 57α + 1 α From this, we get α 2 = 1, i.e. α = ±1. but 1 = nd the sequence is monotonic incresing. So we must hve lim n = 1. (2) 2. (i) () f is bounded if there exists K so tht f(x) K for ll x [, b], OR f is both bounded bove nd below. (2) (b) The set A := {f(x), x [, b]} is nonempty, nd bounded both bove nd below. So both the sup nd the inf exist by the completeness xiom. (c) There exist c, d [, b] with f(c) = sup x [,b] f(x) nd f(d) = inf x [,b] f(x). (2) (d) f must be continuous. (e) No, e.g. f(x) = 1/x on (, 1). (2) (ii) () The limit does not exist.. To see this consider e.g the sequences (x n ) nd (y n ) where x n = 1 nd y 1 2nπ n = for π/2+2nπ ll n N. Both converge to zero. But lim g(x n ) = 1 while lim g(y n ) =. (2) (b) x sin(1/x) is continuous s it is the composition of two continuous functions. Then f is continuous by lgebr of limits s it is the product of two continuous functions. (c) Since 1 sin(1/x) 1, we hve x 2 x 2 sin(1/x) x 2 for ll x. (2) But lim x x 2 =, so by the sndwich rule, lim x f(x) =. (2) Then the required continuous extension is given by { f(x) if x f 1 (x) = if x =. (d) For x, by product nd chin rules of differentition f 1(x) = 2x sin(1/x) cos(1/x) (2) 2
3 3. (i) () (b) When x =, f 1() = f 1 (h) f 1 () lim h h = lim h sin(1/h) =, h by similr rgument to the result of (c). (2) But f 1 is not continuous t x = s lim x f 1(x) does not exist, by (). L U { } f(t) dt = sup s(t) dt s is step function, s(t) f(t) for ll t. { f(t) dt = inf f : [, b] R is Riemnn integrble if L } s(t) dt s is step function, s(t) f(t) for ll t. f(t) dt = U f(t) dt. r(t) dt = 1 + e + e 8 + e 27, (2) s(t) dt = e + e 8 + e 27 + e 64. (2) (c) The function f is continuous, so it is Riemnn integrble on bounded intervl. (d) The function r is step function with r(t) e t3 for ll t [, 4], since the two functions hve common vlues t t =, 1, 2, 3 nd e t3 is incresing. Hence 1 + e + e 8 + e 27 = r(t) dt L e t3 dt = Similrly, s is step function with s(t) e t3 [, 4] nd e t3 dt = U e t3 dt 3 e t3 dt. for ll t s(t) dt = e + e 8 + e 27 + e 64.
4 (ii) () The sequence (f n ) converges pointwise to function f : [, b] R if for ech t [, b], we hve lim f n(t) = f(t). [This form is fine. Or they might write out the ɛ, N version of the limit.] (b) The sequence (f n ) converges uniformly to function f : [, b] R if for ll ε >, we hve N N such tht f n (t) f(t) < ε for ll n N nd ll t [, b]. (iii) () The grphs of f 1, f 2, f 3 re below. The slope chnges t (1/4, 1/4) for f 2 nd t (1/8, 1/8) for f 3. (3) (b) The sequence (f n ) converges pointwise to the function f : [, 1] R with { 1, if t =, f(t) =, if t. To see this, firstly, f n () = 1 for ll n, so f n () 1. Now suppose t >. Then there is some N such tht 1 < t. 2 N So for n > N, f n (t) = 1 t s n. (2) 2 n 1 (c) No. If the sequence converged uniformly this would be to the sme limit function f. For ech n, the function f n is continuous. This is cler on [, 1 ) nd on ( 1, 1], since f 2 n 2 n n is liner function there. So it s sufficient to check continuity t t = 1. For this, 2 n lim f n (t) = 1 (2 n 1) 1 t 1 2 = 1 n 2 = f( 1 ). n 2n 2 n A result from the course sys tht the uniform limit of sequence of continuous functions is continuous. But f is not continuous t, since for ny sequence (x n ) converging to with ech x n >, we hve lim f(x n ) = lim = f() = 1. 4
5 4. (i) () (b) lim n+1 n = lim (2n)!((n + 1)!) 2 (n!) 2 (2(n + 1))! = lim = lim (1 + 1/n)(1 + 1/n) (2 + 1/n)(2 + 2/n) = 1 4 So by the rtio test, the series converges. R = lim n n+1 = 4. (2) (n + 1)(n + 1) (2n + 1)(2n + 2) < 1. (2) (ii) Let f n : [, 2π] R be given by f(t) = cos(nt), continuous function. n 2 Note tht f n (t) = cos(nt) n 1 for ll t [, 2π] 2 n 2 nd 1 n=1 converges. n 2 Hence, by the Weierstrss Mtest, the series n=1 f n(t) converges uniformly nd we cn define function by f(t) = n=1 cos(nt) n 2 The uniform limit of sequence of continuous functions is continuous, so this function is continuous. (iii) We hve n = x 2 n + y 2 n C, so x n C nd y n C for ll n. Thus (x n ) is bounded sequence of rel numbers nd so it hs convergent subsequence, sy (x nk ). Consider the subsequence (y nk ) of (y n ). This is gin bounded sequence of rel numbers so it hs convergent subsequence (y nkl ). The sequence (x nkl ) is subsequence of the convergent sequence (x nk ), so it converges. Thus the subsequence ( nkl ) = ((x nkl, y nkl )) converges in ech component, so it converges. (iv) () The open bll with rdius r round x is the set B(x, r) = {y R k x y < r}. 5
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