Question whether dominated convergence can be used

Let $$g$$ be some function and $$X$$ a random variable. Assume $$g_n (y) \sim h(y) \, n^{-1/2},$$ where $$h(0)=1$$ and $$\sim$$ stands for asymptotic equivalence ($$\lim_n \frac{g_n(y)}{ h(y) n^{-1/2}} = 1$$). Further, there exists a constant $$c> 0$$ such that for all $$y$$ and $$n \in \mathbb{N}$$ $$g_n(y) \leq c\, h(y) \, n^{-1/2}.$$ Let $$k \in \mathbb{N}$$. I am asking myself whether there is a way to use the dominated convergence theorem so that $$\lim_{n \to \infty} E\left( \frac{g_{n-k} (X) } {g_n (0) } \right) = E \left( \lim_{n \to \infty} \frac{g_{n-k} (X) } {g_n (0) } \right) = E (X).$$

I mean, I could use the upper bound for the numerator, but what to do about the denominator? The denominator is also deterministic whereas the numerator is random, but I can not just pull the denominator out of the expectation and use DCT, can I?

Assuming $$g_n$$ is non-negative and $$h(X)$$ has finite first moment, then yes. The trick is to multiply both numerator and denominator by $$n^{1/2}$$, then pull the (deterministic) denominator out. Specifically, we have

$$E\left(\frac{g_{n-k}(X)}{g_n(0)}\right) = \frac1{n^{1/2}g_n(0)}E\Big(n^{1/2}g_{n-k}(X)\Big).$$

The term outside the expectation converges to $$1$$ as $$n\to\infty$$. On the other hand, we have

$$n^{1/2}g_{n-k}(X) \to h(X)$$

and

$$0 \le n^{1/2}g_{n-k}(X) \le c\left(\frac{n}{n-k}\right)^{1/2}h(X) \le c\sqrt{2} h(X)$$

for all $$n \ge 2k$$, and this right hand side is integrable. Thus, applying the dominated convergence theorem we find $$\lim_{n\to\infty}E\Big(n^{1/2}g_{n-k}(X)\Big) = E\big(h(X)\big),$$

and hence

$$\lim_{n\to\infty}E\left(\frac{g_{n-k}(X)}{g_n(0)}\right) = \lim_{n\to\infty}\frac1{n^{1/2}g_n(0)}\cdot\lim_{n\to\infty}E\Big(n^{1/2}g_{n-k}(X)\Big) = E\big(h(X)\big).$$