What is the Limit of the abc-Formula as a approaches 0?

[suyver]

I was wondering something 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?

 

[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.

 

[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?

 

[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.

 

[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).

 

[Galileo]

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

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.

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

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}