On the continuity of a function

This function will be continuous at 0 and 1 and no other points. To see this intuitively, you can just note that x² and x³ are unequal for x≠0,1, so your function will be somehow "jumping around" near these points.

To prove continuity at 0, you can simply note that as x-->0, both x² and x³ go to zero. Therefore f(x) also goes to 0 as x-->0. Also f(0)=0 so we have continuity at 0 (if you want an epsilon-delta proof, just pick delta<min{sqrt(epsilon),1} so |delta²| and |delta³| are both <epsilon). Continuity at 1 is basically the same.

To prove discontinuity away from 0,1, let a≠0,1 be our fixed point of consideration. Probably the easiest way forward is to find sequences (q_n) and (r_n) of rational and irrational numbers respectively, which both converge to a (existence of such sequences follows as Q and R\Q are dense in R). Then, if our function f were continuous at a, we'd have f(q_n) and f(r_n) both converging to f(a). But these limits are a³ and a² respectively, which are unequal. Contradiction so f is not continuous at a
The graph is not a proof. The definition of continuity of f(x) at a is that the limit of f(x) as x->a exists and is equal to f(a).

So you need to do a formal epsilon-delta proof about the limit of f(x) at those two points. And show why such a proof fails for any other point.

Have you ever seen / done a limit proof involving epsilon-delta? The thinking process can be a little difficult for beginners to get used to.

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