Fix $\alpha$ sufficiently small such that $0<\alpha\leq \bar{\alpha}$ for some $\bar{\alpha}$. Let $a, b, c, d \in \mathbb{R}$ be given. Suppose the following holds: $$ \frac{-\alpha a}{1-\alpha c} \leq x < \frac{-\alpha (a+b)}{1-\alpha (c+d)}, \quad dx - b>0 $$
Is it possible to find an $M>0$ such that $0\neq|x|< M \alpha$? ($M$ is a constant independent of $\alpha$)
My try
Suppose $a>0$ and $a+b < 0$. When $\alpha$ is very small both denominators are like 1 and we can see that we can find an $M$.
Question
My try is not rigorous at all and I want to use some kind of continuity argument to manage denominators. Also, I do not impose any condition on $a$ and $b$. Additionally, I do not know how to use $dx - b>0$. How can I utilize $dx - b>0$ with no condition on $a$ and $b$ to show $0 \neq |x|< M \alpha$ rigorously?