The problem at hand is to show that for $X_j$ iid holds
$$\lim_{n \to \infty}\frac{P(\bar{X}_n \leq x)}{\frac{1}{ \sqrt{2\pi(\sigma^2/n)}} \int_{-\infty}^x \exp\Big(\frac{-(t-\mu)^2}{2(\sigma^2/n)}\Big)\ \text{d}t}=1$$
Apparently, one needs to notice that $$\{\bar{X}_n \leq x\}=\bigg\{\sqrt{n}\frac{\bar{X}_n - \mu}{\sigma} \leq \color{red}{\sqrt{n}}\frac{x - \mu}{\sigma} \bigg\}$$ and then apply the CLT. Yet I don't understand why can we apply it if the RHS of the inequality depends on $n$. So applying the CLT, we'd get
$$ \lim_{n \to \infty} P\bigg\{\sqrt{n}\frac{\bar{X}_n - \mu}{\sigma} \leq \sqrt{n}\frac{x - \mu}{\sigma} \bigg\} =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\sqrt{n}\frac{x - \mu}{\sigma}} e^{-t^2/2} \ \text{d}t $$ Which doesn't quite make sense.
We assume that $x\gt 0$, $\mu=0$ and $\sigma=1$. After a substitution, we notice that the denominator goes to $1$ as $n$ goes to infinity. So the task is to show that $\mathbb P\{\bar{X_n}\leqslant x\}\to 1$. To see that, fix a positive $R$. For $n$ such that $\sqrt n \cdot x\geqslant R$, we have $$\mathbb P\{\bar{X_n}\leqslant x \}=\mathbb P\{ \sqrt n\bar{X_n}\leqslant \sqrt n x\}\geqslant \mathbb P\{ \sqrt n\bar{X_n}\leqslant R\}.$$ Using the central limit theorem and denoting $N$ a standard normal random variable, we obtain $$\liminf_{n\to\infty}\mathbb P\{\bar{X_n}\leqslant x \}\geqslant \mathbb P\{N\leqslant R\}.$$ We are able to conclude now because $R$ was arbitrary.