Can we deduce that there are infinitely many integers $n$ such that $detA_{n}=0$?

Question

Let $δ_{n},θ_{n},ω_{n}$ be three real sequences converging to $δ,θ,ω$ respectively. Define the following matrix

$$A_{n} = \begin{bmatrix} δ_{n-1} & θ_{n-1} & ω_{n-1} \\ δ_{n} & θ_{n} & ω_{n} \\ δ_{n+1} & θ_{n+1} & ω_{n+1} \end{bmatrix}$$

Assume that $detA_{n}=0$ for some integers $n$ but there is no method to see that this determinant is zero infinitely many times. Since the sequences are convergent, we get $$lim_{n→∞}A_{n}=A=\begin{bmatrix} δ & θ & ω \\ δ & θ & ω \\ δ & θ & ω \end{bmatrix}$$

and we get $detA=0$ by some relations between $δ,θ,ω$.

My question is: Based on this last result can we deduce that there are infinitely many integers $n$ such that $detA_{n}=0$

Answer 1

No. Perhaps the first $20$ terms of each sequence is zero.
From then on, say $\delta_n=\exp(-2n)$, $\theta_n=\exp(-n)$ and $\omega_n=2$. The determinant is never zero again.