Question

1-> X- Let f :S → R and g:S → R be functions and c be a cluster point. Assume lim f (x), lim g(x) exists. Using the definition of the limit prove the following lim(af (x) + Bg(x)) =a lim f(x) + Blim g(x) for any a,ßeR xc XC X-> b. lim( f(x))} = (lim f(x)) f(x) lim f (x) c. If (Vxe S)g(x) # () and lim g(x)() then prove lim X-C XC 10 g(x) lim g(x) X-C

Answer 1

(a). Since at sen) and Lt x) exists so let it slu) = l and St =m. Nc Hc Nc ( where a to) 2141 Now, f&f(x) + f g(x)} - {<l+ßm} =1 & 10) -24)+($300) -- Bm ) = 12 (frans -- ) + (24) - m) < 12 () - 1)] + 1 $(30) -m) 1211-11 + 161 1962) -- ml. (1). Let Eyo be a real number. Since tt fex] =d so I a positive si such that I f(x) - el. < for all x E N'(c, s, ons Again, it glx) = m so I a positive & co such that I gen) - ml < pipi for all REN'( e, f, ons (where & 7o), on Let da min {di, &; S;} Then I f(a)-11 < En and I gen-ek e for all ne N'(e, sons. Then from (1) we get 1 (240) +fJea)-(024 pm) for all X EN(CS) ns.- eur switing Xc 2101 211 E 12121 +/Bl. € 2181 -€

Nc it (& flex) + ß 6.30) did + ß.m et ( 2 fex) + f Jens) = 2. It film) + ß. postane gex] If=p=0, then prove is clear. Hence, t (2fm+pjes) It fn) + ß Xe Lt Jx); X, BEIR. Ne (b). Let at fears = d.. xc Now, | (tm) – (1)? l Hay+lll frag-el ---(0) Since St fem) exists so there exists a positive number B and a posittue s, such that I fromgall< B+111, for all XE N(e, signs. cet éto be a real no. since at fm)=l so I a positive Sy such that I trasiek for all Ball XEN (C, s, ns. Let sa min {21, S₂}. Then I deras td l< B+111 and I fea) -1 Kotle for all me N'(es) ns. NAC

From og we 어 xe ie Ne Ne Xc m get, I fees j² (121= 1 fraseel I seasonal < (B+11). B +11 for all x E NCC fons. et (tens) ² = 1 = (at feug) 2 LA (tend ) = (at the tems)? (e). Let It f(x) = 1 and at gen) = m. first we prove that it gent KC glad 1 gen)-ml Hay 190)|m1 svias det p= l ml. Then p7o. Since xt gex] = m, there exists a positive s, such that 18x) - ml <p for all XE N' (C, Signs. fince | Igens 1-1 mils I geas_m/, it follows that 119(x)) - 1m11<p for all x E N'(c, s.)ns. :. Imi-p<lgens 1 <lm ltp for all NE N' (C, Sons. > 19(x) > Ilmi for all KEN'Cesions. NC

هما a real no. Me det tyo since it gex) = m, there exists a positive Sg such that Igen) - ml < ₃ Impre for all re det s= min {fi, da} . to not Then I fees - m Ike for all ne N'ce, sins. for all & EN'CC, f₂) ns. at ging Noc [we know that if fisa IR and gistik and if Lt. fla) = land at Halam xe then at (sa) gens) = lm nac we consider two function flag and ging then using the result with in the third bracket we get Lt gas LA Ne at tex) et (Hans) = 1. (Ym) = 1/m nnc dose LA dens e/m Nc glx) Hence xt day &t forget nae noc geng hoe geus