# Answer to Question #8069 in Real Analysis for Mike

Question #8069

Show that K(x-a) = O(k-a) [this is the letter O no the number 0] for any K.

Expert's answer

K(x-a) = O(k-a) x->x0 <=> exists C>0 and exists eps>0: |K(x-a)|<|C(x-a)|, xε(x0-eps,x0+eps). So, let's assume C:=K+1 and eps:=1 for any K.

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