Oh, of course not! Perhaps even $N(x) < 3x + x^{\epsilon}$ for large $x$. But this should be really far from what can be proven with current methods :)

Using $x = k! + 1$, we can even show that $\limsup_{x \rightarrow \infty} N(x) - 3x = \infty$. In fact, using an estimate of Rankin on large prime gaps, we should get $N(x) - 3x > c\log x \log \log x \log \log \log \log x (\log \log \log x)^{-2}$ infinitely often.

I very much do believe that it is possible to prove that the limit is finite. But how to do that is probably beyond me. Conceivably Vinogradov's method or Schnirelmann's approach works.

yeah, I meant to say 'binary Goldbach for all large enough values'. I edited my answer accordingly.

No, it also implies binary Goldbach. If every number larger than $(3 + \epsilon)n$ can be written as a sum of primes that are all larger than $n$, then every even number between $(3 + \epsilon)n$ and $4n$ must be a sum of two primes.

We definitely have $d(n) = o(n^{\epsilon})$. So $min[d(n), f(x)] = d(n)$ if $f(x) \ge cx^{\alpha}$. Am I missing something?

Maybe they mean the 1973 article of Kløve, titled "Sums of distinct primes"?

Hmm, maybe I understand the question wrong. Please read page 56 of the article of Erdos and Graham (page 52 of this pdf: math.ucsd.edu/~ronspubs/80_11_number_theory.pdf)

@Scott Could you please elaborate? It sounds like a good joke. I just don't get it :p

@Mark Yeah, sorry for this anticlimactic ending. Thanks for accepting my answer though. My first one! Yay!

I see what you're saying, but, to be honest, that's not a good start. Because unless you prove that the difference between $p^2(N)$ and $p^D(N)$ is bounded by a finite amount (which doesn't sound very reasonable), you either show no asymptotics at all, or you prove that their difference can be arbitrarily large, which is impossible without resolving the twin prime conjecture.