Some Proofs on Limits and Continuity

Here are a couple of proofs that I think are nice. By nice, I mean, I like how I proved the claim because the proof is concise, clear, and rigorous.

1. Let f:R->R be a continuous function. For all r in Q, let f(r)=0.
   Claim: f(x)=0 for all x in R.
   Pf: Because f is continuous, for all c in R, for all V=N(f(c);ß), there exists U=N(c;∂) such that f(U) is a subset of V. So, V=(f(c)-ß, f(c)+ß); U=(c-∂, c+∂); and f(U) is contained in V. By the density of the rationals, there is at least one rational number in U, so f(U)=0&f(U\Q). Now, if c is rational, then V=(-ß, ß), which obviously contains 0, but must also contain f(U\Q) for all ß>0. That is, f(U\Q) is between 0 and ß, for all ß, so f(U\Q)=0. If c is irrational, then V=(f(c)-ß, f(c)+ß), which must contain 0. But if f(c)≠0, then, if ß≤f(c), 0 is not in V. So, f(c)=0. qed.
2. This one is pretty facile, but I still liked how I proved it.
   Let f:D->R, with c in D', and lim(x->c)f(x)=L>0.
   Claim: There is V=N*(c, ß): f(x)>0, for all x in R.
   Pf: Pick ß such that ß0. Then, ß>|f(x)-L| implies that f(x)-L>-ß, so f(x)>L-ß. But, by choice, L>ß>0, so L-ß>0, so f(x)>0. That is, let V be the deleted neighborhood of c, N*(c;ß), were L>ß>0. Thus, for all x in V&D, ß>|f(x)-L| implies that f(x)>L-ß<, so f(x)>0. qed.