This question is related to this.
Introduce the asymptotic notation Big Oh in probability. Suppose that the real sequence $a_n\to 0$ as $n\to\infty$. If $C_1,C_2$ are constants, independent of $n$, then $$\frac{C_1+O_p(a_n)}{C_2+O_p(a_n)}=\frac{C_1}{C_2}+O_p(a_n).$$ Why is it true?
My proof based on Henry's comments
Let $X_n,Y_n$ be $O_p(a_n)$ just for distinction of asymptotic terms in the numerator and in the denominator. Then \begin{align} \frac{C_1+X_n}{C_2+Y_n}&=\frac{C_1}{C_2+Y_n}+\frac{X_n}{C_2+Y_n}\\ &=\frac{C_1}{C_2}\frac{C_2}{C_2+Y_n}+\frac{X_n}{C_2+Y_n}\\ &=\frac{C_1}{C_2}\bigg(1-\frac{Y_n}{C_2+Y_n}\bigg)+\frac{X_n}{C_2+Y_n}. \end{align}
We claim that $Y_n=O_p(a_n)=o_p(1)$ given that $a_n=o(1)$.
Let $\epsilon,\delta>0$. By hypothesis, $P(\lvert X_n\rvert\geq M_1\lvert a_n\rvert)\leq \epsilon$ for some $M_1>0$ and all sufficiently large $n$. Further, as $a_n\to0$ eventually $\delta\geq M\lvert a_n\rvert$, and then $$P(\lvert Y_n\rvert\geq \delta\rvert)\leq P(\lvert X_n\rvert\geq M_1\lvert a_n\rvert)\leq \epsilon.$$ for all sufficiently large $n$. This is exactly the definition of $Y_n$ being $o_p(1)$.
Then \begin{align} \frac{Y_n}{C_2+Y_n}&=Y_n\frac{\frac{1}{C_2}}{\frac{1}{C_2}}\frac{1}{C_2+Y_n}= O_p(a_n)\frac{\frac{1}{C_2}}{1}\frac{1}{1+O_p(a_n)}\\ &=O_p(a_n)\frac{1}{C_2}\frac{1}{1+O_p(a_n)}=O_p(a_n)\frac{1}{C_2}\frac{1}{1+o_p(1)}\\ &=O_p(a_n)\frac{1}{C_2}O_p(1)=O_p(a_n)O_p(1)=O_p(a_n), \end{align} where the result $\frac{1}{1+o_p(1)}=O_p(1)$ is found in page 13 of Van der Vaart's Asymptotic Statistics.
It follows that $$\frac{C_1+O_p(a_n)}{C_2+O_p(a_n)}=\frac{C_1}{C_2}+O_p(a_n)+O_p(a_n)=\frac{C_1}{C_2}+O_p(a_n),$$ as desired.