Real Aalysis Fall 004 Take Home Test SOLUTIONS. Use the defiitio of a limit to show that (a) lim si = 0 (b) Proof. Let ε > 0 be give. Defie N >, where N is a positive iteger. The for ε > N, si 0 < si < ε. Hece lim N = 0. lim 3 + x 3 = Proof. Let A > 0 be give. Defie δ =. The for x 3 < δ, f(x) = A x 3 > δ = A. Hece lim x 3 + x 3 =.. (a) Let {x } be a bouded sequece of real umbers ad {x k } be a mootoe subsequece. Prove that {x k } coverges to a limit. Proof. By Bolzao-Weierstrass Theorem, the bouded subsequece {y } = {x k } has a coverget subsequece {y l } which coverges to x 0. Now either {x k } is odecreasig or oicreasig. Without loss of geerality, assume that {y } is mootoically odecreasig. Give ε > 0, there exists N such that > N implies x 0 ε < y l < x 0 + ε. I particular, x 0 ε < y ln+ < x 0 + ε. For > l N+ we have x 0 ε < y ln+ < y < x 0 + ε. Thus lim y = lim x k = x 0. (b) If lim x = L =, prove that lim = x L Proof. Let ε = L. The there exists N > 0 such that for > N, x L < L. Cosequetly, for > N, x = x L + L L x L > L L = L.
For ay give ε > 0 there exists N > 0 such that for > N, x L < ε L. Let N = max{n, N }. The for > N, Thus lim x = L. x L = x L x L x L L / < ε/l L / = ε. 3. Let x + =, x > 0. Prove that the sequece {x } coverges ad the compute 3 + x the limit of the sequece. < 3 + x 3 = 3 0 for all. Hece 3 0 < x < 3 First Proof (My Preferece). Sice x > 0, x + = x + = 3 + x < 3 + 3 (b) If lim x = L exists, the L = lim x + = lim = 3 + x Thus L = 3 + 3. (a) If x > L, the x = 3 + L for all. Moreover, for all >. lim (3 + x ) = 3 + L = L. () L + 3L = 0. () < 3 + x 3 + L = L from (). More geerally, if x > L, the x = < 3 + x 3 + L = L. O the other had, if x < L, the x < L ad x > L. Without loss of geerality, assume x > L. Defie the sequece {s } as s = x L. The claim that s + < s for all. If ot, the s +. s Assume = k. The s k+ s k = x k+ L L x k = 3+x k L L x k 3L Lx k 3L 3x k + Lx k x k (x k Lx k + L ) L 6L + 3x k + 0 (x k L) 3L + 3x k + 0 (usig ()) (x k L)[x k L + 3] 0.
Sice x k L < 0, x k L + 3 0 or x k L 3 < 0, a cotradictio. A similar coclusio is obtaied for = k. Sice s is bouded ad mootoically decreasig, lim s = α. Thus lim s = α = lim s +. For the case x > L, x + > L ad x < L ad cosequetly (i) s = L x α x L α as, (ii) s + = x + L α x + L + α as. Now lim x + = lim 3 + L α L + α = 3 + x L + 3L + 3α + αl α αl = 0 α(3 α) = 0 α = 0 or α = 3. If α = 3, L α < 0 ad L + α > 3, a cotradictio. lim x = L. Hece α = 0 ad thus Secod Proof. Let f(x) = (3 + x) 3 for all x > 0. Now 3 + x. The f (x) = (3 + x). Sice x > 0, f (x) = x L x L = f(x ) L x L = f (ξ ) 3 by the Mea Value Theorem, where ξ is betwee x ad L. x L 3 x L. Similarly, x 3 L x L = f(x ) L = f (ξ ) x L 3, ξ betwee x ad L. x 3 L ( ) 3 x L x L. By iductio, x L 3 ( ) x L 0 as. Thus lim 3 x = L. 4. Prove that the fuctio f(x) = is cotiuous for all real umbers x 0. x L is defied as the fixed poit of f, i.e., f(l) = L = 3 + L L + 3L = 0 L = 3 + 3
Method. Let ε > 0 be give. Let x 0 R\{0}. Put δ = x 0. The if x x 0 < δ = x 0, the x 0 < x < 3 x 0. Let δ = ε 0 x 0 x 0. For δ = mi(δ, δ ), if x x 0 < δ, the x x = x 0 x 0 x 0x = x x 0 x + x 0 x 0x < x x 0 5 x 0 x x 0 0 4 x x 0 0 x 0 x 0 = ε Method. Cosider the fuctio g(x) = x. Give ε > 0, let δ = x 0. The x 0 < x < 3 x 0 for all x x 0 < δ. Let δ = ε 5 x 0. The for δ = mi(δ, δ ), ad x (x 0 δ, x 0 +δ), we have x x 0 = x x 0 x + x 0 < x x 0 5 x 0 ε. The use a theorem about the reciprocal of a cotiuous fuctio is cotiuous. 5. Suppose that f is cotiuous o [a, b], oe-to-oe (if x x, the f(x ) f(x )), ad f(a) < f(b). Prove that f is mootoically icreasig (strictly) o [a, b]. Proof. Suppose that f is ot mootoically icreasig o [a, b]. b a a b
The there exists x, x [a, b] with x < x but f(x ) > f(x ). We have two cases to cosider: Case (i): If f(a) < f(x ), the f(a) lies betwee f(x ) ad f(x ). Hece by the Itermediate Value Theorem, there exists c betwee x ad x with f(c) = f(a), a cotradictio to the assumptio that f is oe-to-oe. Case (ii): If f(a) > f(x ), the sice f(a) < f(b), the value f(a) lies betwee f(x ) ad f(b). Agai, by the Itermediate Value Theorem, there exists d betwee x ad b with f(d) = f(a), a cotradictio. Hece f is icreasig. 6. Suppose that S, S,..., S are sets i R ad that S = S S S, S. Let B i = sup S i, b i = if S i, i. Fid a formula relatig sup S ad if S i terms of the {b i } ad {B i }. Proof. S = i= S i. Claim that sup S = mi{b i, i }. Let B r = mi{b i : i }. If x S, the x S r, ad hece x B r. Thus sup S B r. O the other had, sup S x S r ad by the defiitio of B r, sup S B r. Hece sup S = B r. Oe may show i a similar fashio that if S = max{b i i }. 7. Suppose that lim f(x) = ad lim g(x) = a. Suppose that for some positive umber x a x M, we have g(x) a for x < M. Prove that lim f(g(x)) =. x Proof. Sice lim f(x) =, for ay A > 0, there exists δ > 0 such that x a < δ x a f(x) > A. For this δ there exists M > 0 such that x < M g(x) a < δ. Let K = max{ M, M}. The for x < K, g(x) a < δ ad thus f(g(x)) > A. Hece lim x f(g(x)) =. 8. If f(x) is cotiuous o [a, b], if a < c < d < b, ad M = f(c) + f(d), prove there exists a umber ξ betwee a ad b such that M = f(ξ). Proof. Let M = f(c)+f(d). If f(c) f(d), the M ad M f(c) + f(d) f(c) + f(c) = = f(c) + f(d) f(d) + f(d) = f(d), = f(c). By the Itermediate Value Theorem (sice f(c) M f(d)), there exists ξ betwee c ad d (hece betwee a ad b) such that f(ξ) = M. Thus M = f(ξ).
9. Suppose that f(x) ad g(x) are fuctios defied for x > 0, lim g(x) exists ad is fiite, x 0 + ad f(b) f(a) g(b) g(a) for all positive real umber a ad b. Prove that lim f(x) x 0 + exists ad is fiite. Proof. Suppose that lim g(x) = L ad f(b) f(a) g(b) g(a). If lim f(x) does x 0 + x 0 + ot exist, there exists ε > 0 ad two sequeces x > 0, z > 0 such that lim x = 0, lim z = 0 ad for all f(x ) f(z ) ε. For this ε > 0 there exists δ > 0 such that 0 < x < δ g(x) L < ε. Now for 0 < x, x < δ, g( x ) g( x ) = g( x ) L g( x ) + L g( x ) L + g( x ) L < ε + ε = ε. For this δ there exists N such that > N, 0 < x, z < δ. Hece for > N f(x ) f(z ) g(x ) g(z ) < ε, a cotradictio. Thus lim f(x) exists. x 0+ + + 3 + + 0. Evaluate lim. (You eed ot give a proof but you should show some work or justificatio. Quote a theorem or what have you. Calculator results or graphical aalysis are ot acceptable.) Proof. Let s = + + +. The < < s < =. Notice lim =. Thus s is bouded. Moreover, we claim that s + < s. Suppose the cotrary, that is s + s. The Hece or ( + + + + + ) + ( + + + ). ( + + + ) + + ( + + + ) + + + + which is a cotradictio. + + + +, > Hece {s } is a mootoe bouded sequece ad thus is coverges by the Bolzao- Weierstrass Theorem. Moveover, lim s =.