Question

Find the limit of the sequence as n to or indicate that it does not converge en2 0 (0,0,1) O Does not converge 0 (0, 1, 7) 0 (0,0,0)

real analysis

Answer 1

де Solution: 4 let the given Sequence be: ( en ) and consider these sequence separately. Because in IRN, a sequeme (x ) converges to (x0, x . ., M) (= x + 1 = 1.2......(4) so, consider, (x) = h = ( 2 1 1 1 1 1 1 1 1 1 1 1 2 3 , ...) so, consider (sun) = ( 2 1 14 17 & ..) and (12) = (-+ 1 2 3 ...) Now, heima ( ) = lime Eu) >0 and home made line ( 1) O. And Sime Both the sequence of eventems and odd terms converges to o , So the original sequence converges to o, so (0) o d Now, consider the sequence, (x) = ( en ) = ( en ) go, lim ena = 0 and so the sequence (un) o ano Now, consider the sequence, (x) = (T) =(5, TT, IT, ...) Since the given sequence is a constant sequence and S convera n-

So, dim (7) in. So using o. (i) & (i) we conclude that the sequences (x)= ( 6 ) o (n)= (12) 0 and (x)= (17) - as n o, and so wsing the prop. og fint (N=3). We conclude that (eg, en ) – (0,0.77) so option (1) is correct to proved; (B); Giren statement is that Is it true that any unbounded sequence in 1R can't have a convergent subsequence? (a) is not true, because, convergence for the vectors is defined ees, the sequence of vectors say x (x, ... Xo) GIRA is convergent to a vector ( x, & ... DFK) JERK for agiven ESO, I NEW such that 11(x) - (+311 LE whenever, man and pl. llp is p- nom given in 18k such that for 1sp 200 11(n) =p =P |(62_109) + cartes y yo and for p= 0o, 11.(n) - Colla 2 Sup 1 (2. - (1) I Lick

So the convergence for vectors are well defined. (b) is false, because if we consider the sequence, (Xw) = (0,1 0,2 0,3 0,4, 0.5, --- ) Then the given sequence is unbounded but it has Convergent subsequence say, (Mr) = (0,0,0...) ; (c) is true If (n) is an unbounded Sequence Subsequence is also unbounded Then such that Every it cant have a convergent subsequeme for if we take the sequeme of even and odd numbered say (92) and (2n-1), Then both these sequences are unbounded and so the original sequence is also unbounded - The subsequences of (n) and (x2-1) are also unbounded, and subsequences of further subsequences are also unbounded. So, Ary possible subsequence of xn) must be one of them They are unbounded = they can't be convergent. ce of the sequence has all its unbounded subsequences Then it can't have convergent subsequence 9 There it a

(d) is false, Because if we take the sequence of the form of (n)= (1,2,3,4,5, ...) Then it can't have convergent subsequence El o Consider the sequence (x) = ( 520, 533 ) = ( 20 ) So In order to complete the proof that the sequence (an) = ( 21 I) (0,0,0) in R3 wrt. 11.11: let us fix & So, we need to seek notlist sus no, 1180- (0,ooilly LE;

Son Now, let 116)- (oooolly = 1 ano) It lan y cojl + 1(3)) -(0)1 where, (x ) = ( 12), (x) = (0) and (x) = (1/h) So 11am-2000 11. = 152-01 +10-01 +1th of = h hatua Now, since in Eur - - 2 So 1 2 + 1 - 2 1 1 2 3 4 2 2 2 2 i 11 2n-6000)|l, = the terms in <E So, by choosing the I 2 2 2 2 2 In> 2 2

So By choosing nos f / 2 th, or by Reoranging Alc to the answeri no= (1+1, since i t was rence proved so option (d) is correct it, (0) Find dins (x-y)? (x.p→(0,0) So consider the point (1.b) = (0,0) and fra.y) = (x-y32 Then ca.y= x + y ² ay din flag - dim (x²+42 - 2ay) = 0+0+0=0, (x:7)_) (08) (x.) (0) So The Simultaneous limit for function f. 1R²R is given by dim fex,y)=0 h. J (XV) -(09 fonce option (6) is correct a fome proved