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.
Russell Campbell
Sun Oct 26 1997