Prove this set has zero g-density

$\omega=\{0,1,2,...\}$

We define g-density for a set A like this:

$$d(A)= \limsup_{n\to\infty} \dfrac{|A\cap \{1,2,...,n\}|}{g(n)}$$

When $g: \omega \to [0,\infty) $ with $\lim\limits_{n \to \infty} g(n)= \infty$ and $\lim\limits_{n \to \infty} \frac{n}{g(n)}\neq 0 $.

Let $A:= \{n_0<n_1<...\} \subset \omega $ is infinite.

Since $\lim\limits_{n \to \infty} g(n)= \infty$ then we can choose a subsequence$\{n_{i_k}\}_{k \in \omega}$ of $A$ such that $g(n_{i_k})\ge 2^k$ for every $k \in \omega$ .

Let $$B := \{n_{i_k} : k \in \omega \} $$

I need to prove that $d(B)=0$

I tried to prove like this:For any $n\in \omega$ we have:$$ |B\cap \{1,2,...,n\}|= \left\{ \begin{array}{lr} 0 & \mbox{,if } n < n_{i_1} \\ k & \mbox{, if } n_{i_k} \le n< n_{i_{k+1}} \end{array}\right. $$So

$0 \le \lim\limits_{n\to\infty} \dfrac{|B\cap \{1,2,...,n\}|}{g(n)}=\lim\limits_{n\to\infty} \dfrac{|B\cap \{1,2,...,n_{i_k}\}|}{g(n_{i_k})} =\lim\limits_{n\to\infty} \dfrac{k}{g(n_{i_k})} \le \lim\limits_{n\to\infty} \dfrac{k}{2^k} =0 $

$\Rightarrow d(B)=0$

Is this true ? I am not sure if I really can write $g(n_{i_k})$ instead $g(n)$ in limits above. Is that true? If it isn't how can prove that density is zero?