Convergence of ratio of Big O-terms in probability

Say we have constants $$v \in \mathbb{R}$$ and $$w >0$$. Assume that $$a_n \to 0$$ and $$b_n \to 0$$ as $$n \to \infty$$. I wish to show that

$$\frac{v+O_p(a_n)}{w+O_p(b_n)}=\frac{v}{w}+O_p(a_n)$$

where we assume that $$w+O_p(b_n)\neq 0$$ for all $$n$$. Now, I am not 100% sure the statement is even correct, but so far I have tried the following:

Letting $$Y_n =O_p(b_n)$$ and $$X_n = O_p(a_n)$$ we show that $$\frac{v+X_n}{w+Y_n}-\frac{v}{w}=\frac{wX_n-vY_n}{w(w+Y_n)} \in O_p(a_n)$$

Since $$Y_n \in O_p(b_n)$$ and $$b_n \to 0$$ we in particular have that $$Y_n \in o_p(1)$$, i.e. $$Y_n \to 0$$ in probability. Thus, picking $$0 < \epsilon < w$$

\begin{align} P\left( \left| \frac{wX_n-vY_n}{w(w+Y_n)} \right| > Ca_n \right) &\leq P\left( \left| \frac{wX_n}{w(w-\epsilon)} \right| +\left| \frac{vY_n}{w(w-\epsilon)} \right| > Ca_n,|Y_n| \leq \epsilon \right)+P(|Y_n| > \epsilon) \\ &\leq P\left( \left| X_n \right| > CK_1 a_n \right) + P\left( \left| Y_n \right| > CK_2 a_n \right)+P(|Y_n| >\epsilon) \end{align} We need the last expression to converge to $$0$$ as first $$n \to \infty$$ and then $$C \to \infty$$. The first and last term both go to zero - but the problem is the term $$P\left( \left| Y_n \right| > CK_2 a_n \right)$$ which I believe will only converge to $$0$$ if $$Y_n \in O_p(a_n)$$. Since $$Y_n \in O_p(b_n)$$, we will have $$Y_n \in O_p(a_n)$$ only when $$b_n = O(a_n)$$. My question is then if it is possible in the case where we do not have $$b_n = O(a_n)$$.

Your calculation seems correct to me! Notice that if in your calculation $$X_n \equiv 0$$, then clearly the rate is going to depend on $$b_n$$. I.e. the statement that the rate only depends on $$a_n$$ will not hold in general. You might see if you can show that the result can be established with rate $$O_p( \max\{a_n,b_n\})$$.