[PDF] letra cancion bandolero paris latino
[PDF] letter and sound assessment form
[PDF] letter coding examples
[PDF] letter coding tricks
[PDF] letter granting permission to use copyrighted music
[PDF] letter identification assessment free
[PDF] letter identification assessment pdf
[PDF] letter of acceptance of appointment as director
[PDF] letter of appointment of additional director in private company
[PDF] letter of consent for child to travel with grandparents
[PDF] letter of consent to travel with one parent
[PDF] letter of permission to travel
[PDF] letter of permission to travel from employer
[PDF] letter pattern programs in java
[PDF] letter stating i pay rent
MA 101 (Mathematics I)
Hints/Solutions for Practice Problem Set - 2Ex.1(a)State TRUE or FALSE giving proper justication: If (xn) is a sequence inRwhich
converges to 0, then the sequence (xnn) must converge to 0. Solution: The given statement is TRUE. Ifxn!0, then there existsn02Nsuch thatjxnj<12 for allnn0and so 0 jxnnj<(12 )nfor allnn0. Since (12 )n!0, by sandwich theorem, it follows thatjxnnj !0 and consequentlyxnn!0. Ex.1(b)State TRUE or FALSE giving proper justication: There exists a non-convergent se- quence (xn) inRsuch that the sequence (xn+1n xn) is convergent. Solution: The given statement is FALSE. If possible, let there exist a non-convergent sequence (xn) such that the sequence (yn) is convergent, whereyn=xn+1n xn= (1 +1n )xnfor alln2N.
Then, sincexn=yn1+
1n for alln2Nand since (1+1n ) converges to 16= 0, it follows that (xn) must be convergent, which is a contradiction. Ex.1(c)State TRUE or FALSE giving proper justication: There exists a non-convergent se- quence (xn) inRsuch that the sequence (x2n+1n xn) is convergent. Solution: The given statement is TRUE, because ifxn= (1)nfor alln2N, then (xn) is not convergent, but (x2n+1n xn) = (1 +(1)nn ) is convergent (with limit 1), since(1)nn !0. Ex.1(d)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the sequence ((1)nxn) converges to`2R, then`must be equal to 0. Solution: The given statement is TRUE. Since (1)nxn!`, the subsequences ((1)2nx2n) = (x2n) and ((1)2n1x2n1) = (x2n1) of ((1)nxn) must also converge to`. Sincex2n>0 for alln2N, `0 and sincex2n1<0 for alln2N,`0. Hence`= 0. Ex.1(e)State TRUE or FALSE giving proper justication: If an increasing sequence (xn) in Rhas a convergent subsequence, then (xn) must be convergent. Solution: The given statement is TRUE. Let (xnk) be a convergent subsequence of (xn). Then (xnk) is bounded above,i.e.there existsM >0 such thatxnkMfor allk2N. For eachk2N, knkand since (xn) is increasing, we getxkxnkM. Thus (xn) is bounded above and consequently (xn) is convergent. Ex.1(f)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that lim n!1(n32 xn) =32 , then the series1P n=1x nmust be convergent. Solution: The given statement is TRUE. Since the sequence (n32 xn) is convergent, it is bounded and so there existsM >0 such that 0n32 xnMfor alln2N. Hence 0xnMn
3=2for all
n2N. Since1P n=1Mn
3=2is convergent, by comparison test,1P
n=1x nis convergent. Ex.1(g)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1n2x2nconverges, then the series1P n=1x nmust converge.
Solution: For eachn2N, we havenP
k=1x k=nP k=11k kxk(nP k=11k 2)12 (nP k=1k2x2k)12 (using Cauchy-
Schwarz inequality). Since both the series
1P n=11n
2and1P
n=1n2x2nare convergent, their sequences of partial sums are bounded. Hence the sequence nP k=1x k 1 n=1of partial sums of the series1P n=1x nis bounded above. Therefore by monotonic criterion for series, the series 1P n=1x nis convergent. Ex.1(h)State TRUE or FALSE giving proper justication: If (xn) is a sequence inRsuch that the series 1P n=1x3nis convergent, then the series1P n=1x4nmust be convergent. Solution: The given statement is FALSE. Ifxn=(1)nn
1=4for alln2N, then1P
n=1x3n=1P n=1(1)nn 3=4is convergent by Leibniz's test (we note that the sequence ( 1n
3=4) is decreasing and converges to 0),
but 1P n=1x4n=1P n=11n is not convergent. Ex.1(i)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1x3nis convergent, then the series1P n=1x4nmust be convergent.
Solution: The given statement is TRUE. If1P
n=1x3nis convergent, thenx3n!0. So there exists n
02Nsuch thatx3n<1 for allnn0. Hencexn<1 for allnn0and therefore 0< x4n< x3n
for allnn0. Since1P n=1x3nis convergent, by comparison test,1P n=1x4nmust be convergent. Ex.1(j)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1x4nis convergent, then the series1P n=1x3nmust be convergent.
Solution: The given statement is FALSE. Ifxn=1n
1=3for alln2N, then1P
n=1x4n=1P n=11n 4=3is convergent, but 1P n=1x3n=1P n=11n is not convergent. Ex.1(k)State TRUE or FALSE giving proper justication: Iff:R!Ris continuous at both 2 and 4, thenfmust be continuous at somec2(2;4). Solution: The given statement is FALSE. Letf(x) =(x2)(x4) ifx2Q;
0 ifx2RnQ:
Let (xn) be any sequence inRsuch thatxn!2. Sincejf(xn)j j(xn2)(xn4)j !0, f(xn)!0 =f(2). This shows thatf:R!Ris continuous at 2. Similarlyfis continuous at 4. Letc2(2;4). Then there exist sequences (rn) inQand (tn) inRnQsuch thatrn!candtn!c. Sincef(rn) = (rn2)(rn4)!(c2)(c4)6= 0 and sincef(tn)!0, it follows thatfcannot be continuous atc. Ex.1(l)State TRUE or FALSE giving proper justication: There exists a continuous function f:R!Rsuch thatf(x)2Qfor allx2RnQandf(x)2RnQfor allx2Q. Solution: The given statement is FALSE. If possible, let there exist a continuous functionf:R! Rsuch thatf(x)2Qfor allx2RnQandf(x)2RnQfor allx2Q. Letg(x) =xf(x) for allx2R. Theng:R!Ris continuous andg(x)2RnQfor allx2R. By the intermediate value theorem, it follows thatgmust be a constant function. Henceg(x) =g(0) for allx2Rand sof(x) =x+f(0) for allx2R. In particular, we getf(f(0)) = 2f(0), which is a contradiction, sincef(0) =g(0)2RnQ. Ex.1(m)State TRUE or FALSE giving proper justication: Iff: [1;2]!Ris a dieren- tiable function, then the derivativef0must be bounded on [1;2]. Solution: The given statement is FALSE. Letf(x) =(x1)2sin1(x1)2if 1< x2;
0 ifx= 1:
Clearlyf: [1;2]!Ris dierentiable on (1;2] withf0(x) = 2(x1)sin1(x1)22x1cos1(x1)2for allx2(1;2]. Also, sincef(x)f(1)x1=jx1jjsin1(x1)2j jx1jfor allx2(1;2], it follows that lim x!1+f(x)f(1)x1= 0 and hencefis dierentiable at 1 (withf0(1) = 0). Ifxn= 1 +1p2nfor all n2N, thenxn2[1;2] for alln2Nandf0(xn) =2p2n! 1, which shows thatf0is not bounded on [1;2]. Ex.1(n)State TRUE or FALSE giving proper justication: Iff: [0;1)!Ris dierentiable such thatf(0) = 0 = limx!1f(x), then there must existc2(0;1) such thatf0(c) = 0. Solution: The given statement is TRUE. If possible, letf0(x)6= 0 for allx2(0;1). Then by the intermediate value property of derivatives, eitherf0(x)>0 for allx2(0;1) orf0(x)<0 for allx2(0;1). We assume thatf0(x)>0 for allx2(0;1). (The other case is almost similar.) Thenfis strictly increasing on [0;1) and sof(x)> f(1)> f(0) = 0 for allx2(1;1). This contradicts the given fact that limx!1f(x) = 0. Hence there existsc2(0;1) such thatf0(c) = 0. Ex.1(o)State TRUE or FALSE giving proper justication: Iff:R!Ris dierentiable, then for eachc2R, there must exista;b2Rwitha < c < bsuch thatf(b)f(a) = (ba)f0(c). Solution: The given statement is FALSE. Letf(x) =x3for allx2R, so thatf:R!Ris dier- entiable. If possible, let there exista;b2Rwitha <0< bsuch thatf(b)f(a) = (ba)f0(0). Thenb3a3= (ba)0 = 0)b3=a3, which is not true, sincea <0 andb >0. Ex.1(p)State TRUE or FALSE giving proper justication: The functionf:R!R, dened by f(x) =x+ sinxfor allx2R, is strictly increasing onR. Solution: The given statement is TRUE. Sincef0(x) = 1+cosx0 for allx2R,fis increasing onR. If possible, let there existx1;x22Rwithx1< x2such thatf(x1) =f(x2). Thenfmust be constant on [x1;x2] and sof0(x) = 0 for allx2[x1;x2]. This implies that cosx=1 for all x2[x1;x2], which is not true. Thereforefis strictly increasing onR. Ex.1(r)State TRUE or FALSE giving proper justication: Iff: [0;1]!Ris a bounded function such that lim n!11n n P k=1f(kn ) exists (inR), thenfmust be Riemann integrable on [0;1]. Solution: The given statement is FALSE. Iff(x) =0 ifx2[0;1]\Q;
1 ifx2[0;1]\(RnQ);
thenf: [0;1]!Ris a bounded function and we know thatfis not Riemann integrable on [0;1].
However, sincef(kn
) = 0 fork= 1;:::;nand for alln2N, limn!11n n P k=1f(kn ) = 0.
Ex.2(a)For alln2N, letan=n+1n
andxn=1n
2(a1++an). Examine whether the
sequence (xn) is convergent. Also, nd the limit if it is convergent.
Solution: For alln2N,xn=1n
2[(1+2++n)+(1+12
++1n )] =12 (1+1n )+1n 1+12 ++1nn . Since 1n !0, by the solution of Ex.4 of Practice Problem Set - 2, we get1n (1+12 ++1n )!0. It follows (by limit rules for algebraic operations) that (xn) is convergent with limit12 (1 + 0) + 0:0 =12
Alternative solution: We can show that limn!11n
2(1 +12
++1nquotesdbs_dbs6.pdfusesText_12
[PDF] Week 3 Solutions Page 1 Exercise (241) Prove that ) is Cauchy