Asymptotic normality of MLE under classical condition: why can we divide by the remainder? I have been reading the proof of the following theorem.


I do not understand the beginning of the proof.

How do we get from (7.3.5) to (7.3.6)? Technically speaking, all the terms involved are random variables, presumably defined on some sample space $\Omega$. Apparently (7.3.5) holds for all $\omega\in \Omega$. However, when we divide by 
$$R: = -(1/n) l_n''(\theta_0) - (1/2n) \left(\hat{\theta}_n - \theta_0\right) l'''_n (\theta^*_n)$$
what if $R(\omega) = 0$ for some $\omega$?
 A: Since on one effectively answered my question, I will have to answer my own. The point is to notice the following. Suppose $X_n ,Y_n, Z_n$ are three sequences of r.v. defined on $\Omega$ and 
$$X_n (\omega) = Y_n (\omega) Z_n(\omega) \text{ for all } \omega \in \Omega$$
Now, if we want to "divide both sides by $Y_n$", obviously we can only do so for $\omega$ on which $Y_n$ is not zero. That is, at best we have
$$\text{For each }n, \frac{X_n(\omega)}{Y_n(\omega)} = Z_n(\omega) \text{ for all } \omega \text{ such that } Y_n(\omega)\ne 0$$
Suppose that $\frac{X_n}{Y_n}$ converges to another random variable $W$ in distribution. We claim that if for each $n$, $\frac{X_n}{Y_n}=Z_n$ on some set $B_n$ such that $\lim_n \mathbb{P}\left(B_n\right) = 1$, we can still conclude that $Z_n$ converges to $W$ in distribution, too. We formalize it as the following lemma.
Lemma 2: Let $X_n, Y_n, Z_n$ be random variables defined on $\Omega$. Let $S_n \subset \Omega$ be such that $\lim_n \mathbb{P}\left(S_n\right) = 1$. Suppose that for each $n$, $X_n(\omega) = Y_n(\omega)/Z_n(\omega)$ for all $\omega \in S_n$. Suppose that $Y_n/Z_n \overset{d}\to W$. Then $X_n\overset{d}\to W$.
Proof: Let $\epsilon > 0$. For any $t$ and for any $n$, we have
\begin{align*}
\left|\mathbb{P}\left( X_n \le t\right) - \mathbb{P}\left( W \le t\right) \right| &= \left| \mathbb{P}\left(X_n\le t\right)  - \mathbb{P}\left(Y_n/Z_n \le t\right) + \mathbb{P}\left( Y_n/Z_n \le t\right) -\mathbb{P}\left(W \le t\right)\right| \\
&\le \left| \mathbb{P}\left(X_n\le t\right)  - \mathbb{P}\left(Y_n/Z_n \le t\right)\right| + \left|\mathbb{P}\left( Y_n/Z_n \le t\right) - \mathbb{P}\left(W \le t\right)\right|
\end{align*}
Since $Y_n/Z_n \overset{d}\to W$, we have $\lim_n \mathbb{P}\left( Y_n/Z_n \le t\right) = \mathbb{P}\left(W\le t\right)$ for any $t$. This implies that there exists some $N_1$ such that $\left| \mathbb{P}\left( Y_n/Z_n \le t\right)  - \mathbb{P}\left(W\le t\right) \right| < \frac{\epsilon}{2}$ for all $n\ge N_1$.
For the first terms, we have
\begin{align*}
\left| \mathbb{P}\left( X_n \le t\right) - \mathbb{P}\left(Y_n/Z_n \le t \right) \right| &= \left| \mathbb{P}\left(\omega\in S_n\mid X_n(\omega) \le t\right) + \mathbb{P}\left(\omega \notin S_n \mid X_n(\omega) \le t\right) - \mathbb{P}\left( \omega \in S_n \mid Y_n/Z_n \le t \right) - \mathbb{P}\left( \omega \notin S_n \mid Y_n/Z_n \le t \right) \right| \\
&\le \left| \mathbb{P}\left( \omega \in S_n \mid X_n \le t\right) - \mathbb{P}\left( \omega \in S_n \mid Y_n/Z_n \le t \right) \right| + \left| \mathbb{P}\left( \omega \notin S_n \mid X_n \le t \right) - \mathbb{P}\left( \omega\notin S_n \mid Y_n/Z_n \le t\right)\right|
\end{align*}
If $\omega$ is in $S_n$ and satisfies that $X_n(\omega) \le t$, then $Y_n(\omega)/Z_n(\omega) \le t$, and vice versa. As a result, $\mathbb{P}\left( \omega \in S_n \mid X_n \le t\right) = \mathbb{P}\left(\omega\in S_n \mid Y_n/Z_n \le t \right)$. On the other hand, we have
\begin{align*}
\left| \mathbb{P}\left( \omega \notin S_n \mid X_n \le t \right) - \mathbb{P}\left( \omega\notin S_n \mid Y_n/Z_n \le t\right)\right| &\le \mathbb{P}\left( \omega \notin S_n \mid X_n\le t \right) + \mathbb{P} \left(\omega \notin S_n \mid Y_n/Z_n \le t\right) \\
&\le 2\mathbb{P}\left(S_n^C\right)
\end{align*}
Since $\lim_n \mathbb{P}\left( S_n\right) = 1$, $\lim_n \mathbb{P}\left(S_n^C\right) = 0$. Thus there exists $N_2$ such that $\left| \mathbb{P}\left(S_n^C\right) \right| < \frac{\epsilon}{2}$ for all $n\ge N_2$.
Finally, we see that when $n\ge \max\left(N_1, N_2\right)$, we have
\begin{align*}
\left|\mathbb{P}\left( X_n \le t\right) - \mathbb{P}\left( W \le t\right) \right| &\le \left| \mathbb{P}\left(X_n\le t\right)  - \mathbb{P}\left(Y_n/Z_n \le t\right)\right| + \left|\mathbb{P}\left( Y_n/Z_n \le t\right) - \mathbb{P}\left(W \le t\right)\right| \\
&\le \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
&= \epsilon
\end{align*}
This shows that $X_n \overset{d}\to W$. $\blacksquare$
A: First, you need to know $l_n'(\hat{\theta_n})=0$. This is because  $\hat{\theta}$ is a solution of the maximum of log likelihood funtion $l(\theta_0)$
Then you just rearrage terms of $(7.3.5)$ you will get $(7.3.6)$
$l_n'(\hat{\theta_n})=l_n'(\theta_0)+(\hat{\theta_n}-\theta_0)l_n''(\theta_0)+\frac{1}{2}(\hat{\theta_n}-\theta_0)^2l_n'''(\theta_n^{*}) \tag{7.3.5}$ 
i.e 
$$0=l_n'(\theta_0)+(\hat{\theta_n}-\theta_0)l_n''(\theta_0)+\frac{1}{2}(\hat{\theta_n}-\theta_0)^2l_n'''(\theta_n^{*}) \tag{7.3.5'}$$
Rearrange $(7.3.5')$ 
$$-l_n''(\theta_0)(\hat{\theta_n}-\theta_0)-\frac{1}{2}(\hat{\theta_n}-\theta_0)^2l_n'''(\theta_n^{*})=l_n'(\theta_0)$$
i.e.
$$(\hat{\theta_n}-\theta_0)\left[ -l_n''(\theta_0)-\frac{l_n'''(\theta_n^{*})}{2}(\hat{\theta_n}-\theta_n)\right]=l_n'(\theta_0)$$
Then
$$(\hat{\theta_n}-\theta_0)=\frac{l_n'(\theta_0)}{-l_n''(\theta_0)-\frac{l_n'''(\theta_n^{*})}{2}(\hat{\theta_n}-\theta_n)}$$
Next, multiply $\sqrt{n}$ on both sides you will get $(7.3.6)$ by some aggrangements of numerator and denominator of rigt hand side.
