Can we say something about roots of quadatric equation whose coefficient consists of terms that tend to 0?

Question

I am particularly interested in discussing the solution of $\epsilon_1 x^2+(\epsilon_2 +a)x+(\epsilon_3+b)=0$ where $a>0,b>0$ are constants and $\epsilon_1,\epsilon_2,\epsilon_3\to 0$. Looking at the equation , I can think that one solution is $-\frac{b}{a}$ and other solution must tend to $\infty$ (similar things happen for $\epsilon x^2+ax+b=0$). But, how can i prove this mathematically?

Answer 1

Clearly, for any $\epsilon_1\neq 0$, the solutions of your quadratic equation are $$x_{1,2} = \frac{1}{2\epsilon_1} \left(-(\epsilon_2+a) \pm \sqrt{(\epsilon_2+a)^2 - 4\epsilon_1(\epsilon_3+b) }\right).$$

For $\epsilon_1 \rightarrow 0$, we can use $\sqrt{x+a} = \sqrt{x}\sqrt{1+\frac ax} = \sqrt{x} \left(1 + \frac{a}{2x} + \mathcal{O}\{\left(\frac{a}{x}\right)^2\}\right)$ for $a/x \ll 1$ and thus $$x_{1,2} = \frac{1}{2\epsilon_1} \left(-(\epsilon_2+a) \pm (\epsilon_2+a) \left(1 - \frac{ 2\epsilon_1(\epsilon_3+b) }{(\epsilon_2+a)^2} + \mathcal O\left\{\left(\frac{\epsilon_1(\epsilon_3+b) }{(\epsilon_2+a)^2}\right)^2\right\}\right)\right).$$

For $a\neq 0$ the first-order expansion becomes asymptotically exact for $\epsilon_1 \rightarrow 0$ and we thus have

$$x_1 \rightarrow -\frac{1}{2\epsilon_1}\frac{2\epsilon_1(\epsilon_3+b)}{\epsilon_2+a} = -\frac{\epsilon_3+b}{\epsilon_2+a},$$

converging to $-b/a$. For $x_2$ we have

$$x_2 \rightarrow -\frac{1}{2\epsilon_1}\left(-2(\epsilon_2+a)+\frac{2\epsilon_1(\epsilon_3+b)}{\epsilon_2+a}\right) = \frac{\epsilon_2+a}{\epsilon_1} - \frac{\epsilon_3+b}{\epsilon_2+a},$$

which diverges for $\epsilon_1 \rightarrow 0$ provided that $a \neq 0$.