Question

XL Xa 12. (a) Suppose that f(x) = g(x) for all x. Prove that lim f(x) < lim g(x), provided that these limits exist. (b) How can the hypotheses be weakened? (c) If f(x) < g(x) for all x, does it necessarily follow that lim f (x) < lim g(x)? Ya X-

Answer 1

1 2 @ Since the theorem provides two possible outcomes and only excludes one outcome we will show that the excluded autcome is not consistent with the given statement. so the exclude autcome 18 aux beqinning assumption let lim frx) = L and lim g(x)=M x-a we will assume that LM and shaw that this produces a contro diction. It will be sufficient to show that there exist one epsilon that produces the controduction wechavre E = L-M YO since lim [ g(x) - f1x)] = M-L ( Difference law for limits) Then there exist od no such that for every x 021x-cl<d => 1 9(*) - fra) - (M-L)) <€ hus -ek gix) - f(x) - MtL <t 9x) -f(x) -MtL SL-M gre) -f(x) <6. > gir) <for) which is con trodiction TLSM or lim telim gia - kogu 1- 0

If we remove I with the the relation a rela a n d Then the hypothesis is weaked с аkе f(x) = - 1x and gra) = x and a=0 since f(x) < 9(x) Vo but um xto fuo - lim x-o 9(x)=0 So, which is contradiction fa&gra) # lem for homegra)