Question

If K is a constant, show that K(x-a) = o(x-a) [yes that is a "o" not a "0"] if and only if K=0.

Answer 1

K(x-a) = o(x-a), x->x0 <=> for any C>0 exists eps>0: |K(x-a)|<|C(x-a)|, xε(x0-eps,x0+eps) <=> for any C>0 exists eps>0: K|(x-a)|<C|(x-a)|, xε(x0-eps,x0+eps) <=> for any C>0: K<C <=> K=0.