The behavior of quadratic formula in the limit $a\to 0$

Question

$$\lim\limits_{a\to0}\frac{-b+\sqrt{b^2-4ac}}{2a}$$

$b$ and $c$ are constants. As $a$ approaches $0$, what does the formula approach?

Example: $$\lim\limits_{a\to0}\frac{-5+\sqrt{5^2-4·0.001·3}}{2·0.001}\approx-0.6$$

Answer 1

$$\begin{align*} \lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a} &= \lim_{a\to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}\right)\\ &= \lim_{a\to 0}\frac{-b^2+(b^2-4ac)}{2a(b+\sqrt{b^2-4ac})}\\ &= \lim_{a\to 0}\frac{-2c}{b+\sqrt{b^2-4ac}}\\ &= \frac{-2c}{b+|b|} \end{align*}$$ If $b>0$, then you get $-\dfrac cb$, the root of the linear equation hinted in another answer; but

if $b<0$, then the limit does not exist, because

• the larger one of the two quadratic roots becomes $+\infty$ when $a\to 0^+$;
• the smaller one of the two quadratic roots becomes $-\infty$ when $a\to 0^-$;

Answer 2

HINT: What happens to the equation $ax^2+bx+c=0$ as $a\to0$?

Answer 3

Let $f(x)=\sqrt{b^2-2cx}$, then your expression is the definition of $f'(0)$.

$$f'(0)=\lim_{h\to0} \frac{f(h)-f(0)}{h} = \lim_{a\to 0}\frac{f(2a)-f(0)}{2a}$$

We know that $f'(x)=(-2c)\cdot\frac{1}{2\sqrt{b^2-2cx}}=\frac{-c}{\sqrt{b^2-2cx}}$

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Edit: Reading the accepted answer causes me to notice that $f(0)=|b|,$ not $f(0)=b.$ So this proof only works when $b>0.$

When $b<0,$ then $\lim_{a\to 0}\frac{f(2a)-|b|}{2a}$ still converges to $f'(0),$ but $$\frac{f(2a)-b}{2a}=\frac{f(2a)-f(0)}{2a}+\frac{b-|b|}{2a}=\frac{b}{a}$$ and $\lim_{a\to 0}\frac ba$ does not exist, so your limit does not exist in that case.

Answer 4

Let $b \neq 0$; let $c \in \mathbb{R}$; let $N(0)$ be a neighborhood of $0$; let $f: a \mapsto -b + \sqrt{b^{2} - 4ac}$ on $N(0)$; and let $g: a \mapsto 2a$ on $N(0)$. Then $g \neq 0$ on $N(0)$. Since $$ \frac{f'(a)}{g'(a)} = (-4)\frac{(b^{2}-4ac)^{-1/2}/2}{2} = -(b^{2}-4ac)^{-1/2} \to \frac{-1}{|b|} $$ as $a \to 0$, we have $f(a)/g(a) \to -1/|b|$ as $a \to 0$.