Question

Analysis problem

(b) Let f, q be defined on A to R and let c be a cluster point of A i. Show that if both lim f and lim (f + g) exist, then lim g exists. c I>c ii. If lim f and lim fg exist, does it follow that lim g exists? -c (c) Suppose that f and g have limits in R as x -> o and that f(x) < g(x) for all x € (a, ). Prove that lim f< lim g

Answer 1

Page No-D Solution: - (b) (i) Given that, fig be defined on a tor bet c be a cluster point of A. it lin FCL) and lim (f+9) exists. *>C RDC KSC ie, lim fcr=l ; lim (f+g) (k)=m let {ts} be a sequence in A-{c} Converging to co since lim f(x) = e and lim (+19) CK) :m RSC KSC He have, lim f(xn)= e and lim (fig ) (Rn) = m 270' by sequential criteúon of limits is lim [6f7g) (xn) - fokn]] cm-e as Ling [(fig -t) (xn) Jamak > line g( xn) = m-d no

Page No - since , {xn} is an arbitarry sequence in A-{c} converging to c. It follows from the sequential criterion that, lim g(x) = me (tinite) C RSC Hence proved. (i) It lim te t and is fg exists , but it do not follows lim gas exists. a nec v If we take, czo f(x)=x and go)= ceste then, lim L=0 K30 lim rcos L=0 so but, lim ces ť does not exists > (o suppose, im o fer- im gW) = m x 500 f(x) < ger) # le caro)

Page No-B bet, h(x) = g(c)- fex). :. her) 20 #ke Caro) KO 19 SO let us choose aso since lim fex)= Ja leal number G,>a . such that , I f(x) - el cal t XJG, since iim g&> am then, I a leat number Gy> a such that 1 g(x) -ml 24 FK > GL let G = max { G., Gz} ::|f(x) -e12 412 *x>G 19(K)-ml 24/2 * KSG 19-f)(x)-[m-e) | < 1 g(x)= ml + HCK) -el < GlZ+G tvG = 1 hex)- (m_e) | LG ** SG i lim hee)= m-l yo a lavol and limu and lim hau 20 4 ne (aro) since h(K) = m-e is meso i mze : Ling fa) - gau) Hence proved