# Limit of abc-formula for a->0

1. Jul 20, 2004

### suyver

I was wondering someting that is so simple that it baffled me...

When I have the equation
$$a x^2+b x+c=0$$
this obviously has the solutions
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

And when I have the equation
$$b x+c=0$$
this has the solution
$$x=\frac{-c}{b}$$

My problem now is the limiting case $a\rightarrow 0$ in the upper situation:
$$\lim_{a\rightarrow 0}\frac{-b\pm\sqrt{b^2-4ac}}{2a}\rightarrow -\infty\neq\frac{-c}{b}$$

So what's wrong here? Why does this limit not exist?

2. Jul 20, 2004

### Galileo

The limit does exist.

$$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot \frac{-b-\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=$$
$$\frac{b^2-(b^2-4ac)}{2a(-b-\sqrt{b^2-4ac})}=\frac{2c}{-b-\sqrt{b^2-4ac}}$$

which goes to $$\frac{-c}{b}$$ when a goes to zero. The same goes for the other square root.

Last edited: Jul 20, 2004
3. Jul 20, 2004

### suyver

But for the other root, I get:

$$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdot \frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}=$$

$$\frac{b^2-(b^2-4ac)}{2a(-b+\sqrt{b^2-4ac})}=\frac{2c}{-b+\sqrt{b^2-4ac}}$$

which seems ok... but I don't think it is! After all, I multiply by

$$\frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}$$

which is $0/0$ in the case when $a\rightarrow 0$, and that's not allowed

So I don't think you can do this:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \rightarrow \frac{2c}{-b\mp \sqrt{b^2-4ac}}$$

Right?

Last edited: Jul 21, 2004
4. Jul 21, 2004

### mathman

One root goes to -c/b. The other root becomes infinite. Which is which depends on the sign of b, since the square root goes to |b| as a goes to 0.

5. Jul 22, 2004

### suyver

OK, that makes sense. I had originally expected that both roots would converge to -c/b, but now that I think about it, it seems clear that the fact that b can be negative (so that I cannot just say $\sqrt{b^2}=b$) will spoil this. Oh well... too bad (for what I wanted).

6. Jul 22, 2004

### Galileo

Well, since $$\frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}$$
can only be zero when $ac=0$ this step is perfectly valid. We may assume $a\neq 0$ when calculating the limit. So that leaves $c\neq 0$ as a restriction. Although it would still be allowed when b is negative.

This expression is correct, but you have to take cases. If b is positive, then you'll find the expression with the negative sign to converge to $\frac{-c}{b}$, if b is negative the expression with the positive sign converges to $\frac{-c}{b}$