For writer #1170 - Essay Example

for this 'e' then it obviously holds for larger numbers also. (We have deliberately added 1 on the rhs, taking into account the possibility of both F, G being = 0. so we need to make rhs some strictly positive quantity). See that (4) e. (2(|F| +|G|) + 1) and this we need it to be < , that is we need e < / (2(|F| +|G|) +1). So the procedure is to start with e = minimum {/ (2(|F| +|G|) +1), (|F| +|G| +1)} which is clearly > 0 and so this has its 1, 2 satisfying our (1) and choose to be the minimum of these two.

The steps in order now are:- | f.g -F.G| e.(e + |F| +|G| ) by (4). and e / (2(|F| +|G|) +1), so [/ (2(|F| +|G|) +1)]. (e + |F| +|G|) and again e is also (|F| +|G| +1), so [/ (2(|F| +|G|) +1)]. (2(|F| +|G|) +1) = , everything holding for our as chosen. Thus for any given > 0 we are able to find out a satisfying the limit condition, f.g