What is the difference between $O(N/ \log_2(N))$ and $N-o(N)$?

Question

On the second page of this paper under the introduction section they say

"We first show that for the set of parameters considered by [16], the function family has $O(N/ \log_2(N))$ simultaneously hardcore bits. Next we introduce a new parameter regime for which we prove that the function family is still trapdoor one-way and has up to $N - o(N)$ simultaneously hardcore bits."

What are the authors trying to say about their new parameter regime? That's it's better or worse? I tried to graph $O(N/ \log_2(N))$ to see how it looks and I can see that the growth rate is slower than a linear function like O(N).

So $N-o(N)$ = something that is linear - something that grows slower than linear = something else that grows slower than linear. So the new parameter regime is better than $O(N/ \log_2(N))$ ?

I'm not sure I am following what they are trying to say.

Answer 1

Every sequence in $N-o(N)$ is in $O(N)$ but not in $O(N/\log N)$.

Note that "something that is linear" minus "something that grows slower than linear" is "something else that grows" LINEAR, not "slower than linear". So the new parameter regime corresponds to sequences growing faster than those in $O(N/\log N)$.