145 - 2 If f(xn) does not converge to f(c), then for some epsilon there will be an f(xn) outside the epsilon neighborhood of f(c) with n greater than any N specified (i.e., there will be an infinite number of such n's). Because those xn's (with f(xn) outside the epsilon neighborhood) converge to c, there will be one (actually infinitely many) in any delta neighborhood of c, hence f is not continuous. If f is discontinous, choose an epsilon for which there does not exist any delta that will keep f(x) within epsilon of f(c). Then choose xn's which are closer than 1/n to c, but with f(xn) further than epsilon from f(c). 5 Yes, the function is equal to x+3 for all x not equal to 2, so define f(2)=5 and the function will be identical to x+3,which is continuous. 13 The only point(s) of continuity will be where 2x = x+3, since both the rationals and irrationals are dense (have we defined that word?). Hence the only point of continuity is x=3.

Sun Oct 26 1997