Given a set $V$ of non-negative integers, let: $$V\left(t\right)\overset{\textrm{def}}{=}\left\{ v\in V:0\leq v\leq t\right\} ,\textrm{ }\forall t\in\mathbb{R}$$ and let $\left|V\left(t\right)\right|$ denote the number of elements of $V\left(t\right)$. Recall that the (natural/asymptotic) density of $V$ is the quantity $d\left(V\right)$ defined by the limit (if it exists): $$d\left(V\right)\overset{\textrm{def}}{=}\lim_{t\rightarrow\infty}\frac{\left|V\left(t\right)\right|}{t}$$
My question is: if $V$ has positive density, does the inequality: $$\sup_{t\geq0}\left|d\left(V\right)t-\left|V\left(t\right)\right|\right|<\infty$$ necessarily hold? If not, then, for what kinds of $V$ does the inequality hold or fail?
Let $V = \{2^k | k \in \mathbb{N}\}$. Then $d(V) = 0$, but $\sup\limits_{t \geqslant 0}|V(t)| = \infty$. If you want non-zero (and non-one) density - take, for example, set of all even numbers that are not powers of two.
It has nothing specific to do with density. Existence of $\lim\limits_{t \to \infty} \frac{f(t)}{t} = c$ means $f(t) = tc + o(t)$ while $|ct - V(t)|$ been bounded means $f(t) = tc + o(1)$, which is stronger.