HW 3 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson

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1 HW 3 Solutios Math 5, Witer 009, Prof. Yitzhak Katzelso 7.4 a) Let x =. The lim x = 0 (as you may easily check - for ǫ > 0, just let N be ǫ ), which is ratioal, eve though all the x are irratioal. b) Cosider the umber ad its decimal expasio.44..., ad th3 let x =.4, x =.4, x 3 =.44, et cetera. The error x is less tha 0 at every step, ad so if ǫ > 0, pick N = log 0 ǫ; the if > N, x < 0 log 0 ǫ = ǫ. Thus the x coverge to the irratioal umber eve though the x themselves are all ratioal. 7.5 a) Fid lims where s = + (o formal proof). To do this, u-ratioalize the deomiator by multiplyig by ; we see that s = + = ++. Now the deomiator is greater ++ tha + =, so we kow that 0 < s <. But lim / = 0, ad lim 0 = 0, so this meas that lims = 0. b) Fid lim s where s = + (o formal proof). As with a), u-ratioalize the deomiator by multiplyig ad dividig by the cojugate + +. We see that s =. Divide ++ both the umerator ad deomiator by to get that s = + +. Now as goes to ifiity, / goes to zero, so the deomiator gets closer ad closer to + =, so the lims itself is. c) Fid lim s where s = 4 + (o formal proof). First u-ratioalize the deomiator, ad the as i part b), divide both the umerator ad the deomiator by. You should get that s = As gets bigger, the deomiator goes to 4 + = 4, so we see that lim s = Let s be a sequece of oegative real umbers, ad suppose that lim s = 0. Prove that lim s = 0. To do this, verify the defiitio of the limit. Pick ǫ > 0. The ǫ > 0; sice lim s = 0, we may fid N such that > N implies that s < ǫ. s is positive, so s = s < ǫ. Take the square root of both sides to fid that s < ǫ, ad thus s < ǫ. So we have foud N such that if > N, the s < ǫ; therefore lim s = 0.

2 8.4. Let (t ) be a bouded sequece; i.e. there exists M such that t M for all, ad let (s ) be a sequece with lim s = 0. Prove that lim(s t ) = 0. Pick ǫ > 0. The as log as M > 0, ǫ/m > 0 as well (if M is zero, the t are all zero, so the theorem is obvious. M caot be egative because the o such t exist). Sice lim s = 0, there exists a N such that > N implies that s < ǫ/m. But the if > N, s t = s t s M < ǫ M M = ǫ. So if > N, the s t < ǫ; this meas exactly that lim(s t ) = a) Cosider three sequeces (a ), (b ), ad (s ) such that a s b for all ad lim a = lim b = s. Prove that lims = s. Pick ǫ > 0. Sice lim a = s, there exists N such that > N implies that a s < ǫ; i particular, > N implies that a > s ǫ. Similarly, sice lim b = s, there exists N such that > N implies that b s < ǫ; i particular > N implies that b < s + ǫ. Now let N = max{n,n }. The if > N, we kow that a > s ǫ ad also b < s + ǫ. But s a > s ǫ ad s b < s + ǫ, so s ǫ < s < s + ǫ. This meas precisely that s s < ǫ. So > N implies that s s < ǫ. Thus lim s = s. b) Suppose that (s ) ad (t ) are sequeces such that s t for all ad lim t = 0. Prove that lims = 0. First we will show that lim( t ) = 0. To show this, pick ǫ > 0; sice lim t = 0, there is N such that > N implies t < ǫ. But the if > N, t = t < ǫ as well. So lim( t ) = 0. Now we otice that s t meas exactly that t s t (ote that the coditio s t meas that t caot be egative, so this is true). But the sice lim( t ) = 0 ad limt = 0, part a) tells us that lim s = a) Prove that lim[ + ] = 0. First give the terms a ame; call ( + ) s. The as i Exercise 7.5, s =. Notice that s ++ > 0; ad also, sice + >, + > =, so the deomiator is greater tha, so 0 < s. Now we claim lim = 0. Ideed, if we pick ǫ > 0, the let N = ; ǫ > N implies < = ǫ. So the claim is true. N Ad ow we may apply Exercise 8.5a); sice lim 0 = 0 (why? otice ay N works for ay ǫ!), ad lim = 0, we kow that lims = 0. b) Prove that lim[ + ] =.

3 First call the th term s. The as i 7.5b), rewrite s = + +. If we could use the limit laws, this would be easy to compute to be ; but we ca t do that with this sectio. There are several ways to proceed from here. I prefer to take advatage of Exercise 8.5a). To do this, we eed sequeces a ad b with a s b ad a ad b both covergig to. Oe of these is easy; we otice that the deomiator of s is at least, sice + > =. So s, ad we ca just let b =. The other directio is trickier; we eed a lower boud for s that coverges to. A lower boud for s is like a upper boud for the deomiator. So otice that sice + >, we have + < +, so + + < +. Thus s. Let a + =. If we ca show + that lim a =, we will be doe by Exercise 8.5a). So: pick ǫ > 0. The we wat to fid N such that > N implies that a < ǫ, which is the same thig as showig that ( + < ǫ, ) i.e. that < ǫ, i.e. (sice every term is positive) that < ǫ, i.e. that 4+, i.e. that > ǫ 4 ( ). So pick N = ǫ mi{0, 4 ( )}; ǫ the > N meas > 4 ( ), which is the same thig as a ǫ < ǫ (the oly reaso to iclude the zero is that egative values of N do t really make sese. It s actually ot strictly logically ecessary, though). So lim a =, ad this is what we wated to show. c) Prove that lim[ 4 + ] = 4. This is essetially the same problem as b), sice s = 3 The + +. exact same method will work, ad so I will ot write out the details. There s a extra or here ad there, that s all Let (s ) be a coverget sequece, ad suppose that lims > a. Prove that there exists a umber N such that > N implies s > a. Sice (s ) is coverget, let s call the limit s. The s > a, so s a > 0. Let ǫ = s a; sice lim s = s ad ǫ > 0, there exists N such that > N implies s s < ǫ. I particular, > N implies that s > s ǫ. But ǫ = s a, so > N implies that s > s (s a) = a. 9. a) Prove that lim + =.

4 4 Notice that lim + = lim( + ). By limit law 9.3, this equals lim + lim = + 0 =. b) Prove that lim 3+7 =. 6 5 Divide both umerator ad deomiator by : lim 3+7 = lim The deomiator is ever 0 as is ever 5, so we may apply Theorem to get that this is equal to lim(3+ 7 ) lim(6 5 ). By limit law 9.3, this is equal to lim3+lim 7. Now by theorem 9., lim 7 = 7 lim = 7 0 = 0, ad lim6 lim 5 similarly lim 5 = 0. So our origial limit is equal to 3 =.. 6 c) Prove that lim = Sketch of proof: divide both umerator ad deomiator by 5. The the deomiator is always positive, so we ca apply Theorem 9.6 to break up the limit of a fractio ito the fractio of the limits. Cotiue as above, applyig limit laws 9.3 ad 9., util we see that the oly ozero terms o the top ad the bottom are the oes that origially had 5 ; this shows that the limit will be 7. [You are of 3 course expected to provide the details] Suppose that lima = a, lim b = b, ad that s = a3 +4a b +. Prove that lims = a3 +4a carefully, usig the limit theorems. b + OK. The first thig we otice is that o matter what b is, b + is ever 0. So we may apply theorem 9.6 to show that: lim s = lim(a3 + 4a ) lim(b + ). The apply theorem 9.3 to see that lim s = lim a3 + lim 4a lim b + lim. Now evaluate each piece: lim a 3 = lim(a (a a )). By Theorem 9.4, this is equal to lima lim(a a ), ad by the same theorem agai, this is equal to lima (lim a lim a ) = a (a a) = a 3. Similarly, lim b = b. By Theorem 9., lim 4a = 4 lima = 4a. Ad of course, lim =. So puttig it all together: lim s = a3 + 4a b +.

5 9.4. Let s = ad for, let s + = s +. a) List the first four terms of (s ). s =, s =, s 3 = +, s4 = + +. b) It turs out that (s ) coverges. Assume this fact ad prove that the limit is ( + 5). This is seaky. Lookig at the first four terms should covice you that ay attempt at a direct proof with ǫs, Ns, et cetera will likely fail. It s way too ugly, ad the recursive defiitio poses serious issues. Fortuately, there is a trick. We re actually told that (s ) coverges, so call the limit s. What is s? Cosider the expressio lim s +. This is the limit of the sequece s, s 3, s 4,..., which is just the same thig as the limit of the origial sequece - i.e. it s just s. If you do t believe me, you ca prove it easily; all that happes is the Ns shift by. (I d recommed doig this, i fact. It s ot hard). But we kow that lim s + = lim( s + ). By Example 5 i Sectio 8, is is equal to lim(s + ), which by limit law 9.3 is equal to lim s + = s +. So we ve ow show that lims + is equal both to s ad to s +. So those two umbers are equal to each other(!) ad s = s +. Squarig both sides ad rearragig the terms gives that s s = 0. By the quadratic formula, the solutios are ( ± 5). Now ( 5) is ot a solutio of the origial equatio s = s + (squarig both sides ofte adds additioal extraeous solutios like this), ad so we may throw it out. Thus s = ( + 5). This trick is quite geeral ad works for a lot of recursively defied sequeces. 5