How do the roots change How do the roots of the  quadratic  equation    $ax^2+bx+c=0$ change when $b$ and $c$  retain constant values and $a$ tends to zero?      ($b\neq 0$)
 A: The roots are given (on any field of characteristic $\neq2$) by:
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Taking this as a function of $\;a\;$, we can apply L'Hôpital's rule for one of the roots:
$$\lim_{a\to0}\frac{-b+\sqrt{b^2-4ac}}{2a}=\lim_{a\to0}-\frac{c}{\sqrt{b^2-4ac}}=-\frac c{|b|}$$
assuming, of course, $\;b\neq0\;$.
For the other one the numerator doesn't vanish but the denominator does, so the limit is $\;\pm\infty\;$, depending on the sign of $\;b\;$.
A: If $|a|$ is small enough, the discriminant $b^2-4ac$ will be positive, so the equation can be considered to have two distinct roots, by keeping $a\in(-\delta,\delta)$, $a\ne0$, for some $\delta>0$.
We can assume $b>0$ (just change $x$ to $-x$ otherwise)
The roots are
$$
\frac{-b-\sqrt{b^2-4ac}}{2a},
\qquad
\frac{-b+\sqrt{b^2-4ac}}{2a},
$$
We see that
$$
\lim_{a\to0}\frac{\sqrt{b^2-4ac}-b}{2a}=
\lim_{a\to0}\frac{-4ac}{2a(\sqrt{b^2-4ac}+b)}=-\frac{c}{b}
$$
whereas
$$
\lim_{a\to0^+}\frac{\sqrt{b^2-4ac}+b}{2a}=-\infty
\qquad
\lim_{a\to0^-}\frac{\sqrt{b^2-4ac}+b}{2a}=\infty
$$

In my old high school book, when discussing “parametric degree two equations” it was always said that “when $a=0$ one root becomes infinite”, which has a justification: the parametric equation was essentially a pencil of parabolas, the degenerate ones being a line counted twice (case of discriminant $0$) and the union of a line and of the improper line (case $a=0$). Of course the book didn't mention this.
A: $$x_{1/2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
If you are looking at the limit for $a\to 0$ then the equation becomes a linear equation $bx+c=0$.
A: If $a$ is infinitesimal, then $b^2 - 4ac \to b^2$, so $$x \to \frac {-b \pm b} {2a},$$ which is either $\infty$ or we have a $0/0$ case, in which we are allowed to use L'Hopital's rule. Taking $b$ and $c$ as constants, we let $a$ vary and differentiate $$\frac {-b + \sqrt {b^2 - 4ac}} {2a} \to \frac {-c} {\sqrt {b^2 - 4ac}} \to - \frac {c} {b}.$$
