I'm trying to prove the following result:
Let $n$ be a fixed natural number and $a_1, a_2, \cdots, a_n$ and $b_1, b_2, \cdots, b_n$ fiexed real values.
Then given $\varepsilon>0$, there exits a $\delta>0$ such that if $\left| r-s \right| < \delta$ then $$ \left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right| < \varepsilon$$
where
$$\left( a_1, b_2|a_2, \cdots, b_n|a_n \right)=a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cfrac{b_4}{\ddots +\cfrac{b_n}{a_n}}}}$$
$$\left( a_1, b_2|a_2, \cdots, b_n|a_n, \xi \right)=a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cfrac{b_4}{\ddots +\cfrac{b_n}{a_n+\xi}}}}$$
I also want to prove that, given $\varepsilon>0$ there exits a $\delta>0$ such that if $$ \left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right| < \delta$$ then $\left| r-s \right| < \varepsilon$
I've proven that if
$$(a_1,b_2|a_2, \cdots, b_n|a_n)=\dfrac{p_n}{q_n}$$
then if we denote $p_0=1$, $p_1=a_1$, $q_1=1$ and $q_2=a_2$ for $n \geq 2$
$$p_n=a_np_{n-1}+b_np_{n-2}$$ $$q_n=a_nq_{n-1}+b_nq_{n-2}$$
Also, I've got that for $n \geq 2$
$$\left( a_1, b_2|a_2, \cdots, b_n|a_n, \xi \right)=\dfrac{p_n+\xi p_{n-1}}{q_n+\xi q_{n-1}}$$
And, from the previous result, I've expanded the difference
\begin{align} {\left| \left( a_1, b_2|a_2, \cdots, b_n|a_n,r \right) - \left( a_1, b_2|a_2, \cdots, b_n|a_n,s \right) \right|=\left| \dfrac{p_n+r p_{n-1}}{q_n+r q_{n-1}}-\dfrac{p_n+s p_{n-1}}{q_n+s q_{n-1}} \right|=\\ =|r-s| \left| \dfrac{p_nq_{n-1} - p_{n-1}q_n}{(q_n+rq_{n-1})(q_n+sq_{n-1})} \right|\\} \end{align}
Now, I don't know how to bound the last fraction, I guess it should be less than $1$, but I don't know how to continue. Any ideas?