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Limit of abc-formula for a->0

  1. Jul 20, 2004 #1
    I was wondering someting that is so simple that it baffled me...

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

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

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

    So what's wrong here? Why does this limit not exist?
     
  2. jcsd
  3. Jul 20, 2004 #2

    Galileo

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    The limit does exist.

    [tex]\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}}=[/tex]
    [tex]\frac{b^2-(b^2-4ac)}{2a(-b-\sqrt{b^2-4ac})}=\frac{2c}{-b-\sqrt{b^2-4ac}}[/tex]

    which goes to [tex]\frac{-c}{b}[/tex] when a goes to zero. The same goes for the other square root.
     
    Last edited: Jul 20, 2004
  4. Jul 20, 2004 #3
    But for the other root, I get:

    [tex]\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}}=[/tex]

    [tex]\frac{b^2-(b^2-4ac)}{2a(-b+\sqrt{b^2-4ac})}=\frac{2c}{-b+\sqrt{b^2-4ac}}[/tex]

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

    [tex]\frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}[/tex]

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

    So I don't think you can do this:

    [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \rightarrow \frac{2c}{-b\mp \sqrt{b^2-4ac}}[/tex]

    Right?
     
    Last edited: Jul 21, 2004
  5. Jul 21, 2004 #4

    mathman

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    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.
     
  6. Jul 22, 2004 #5
    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 [itex]\sqrt{b^2}=b[/itex]) will spoil this. Oh well... too bad (for what I wanted).
     
  7. Jul 22, 2004 #6

    Galileo

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    Well, since [tex]\frac{-b+\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}[/tex]
    can only be zero when [itex]ac=0[/itex] this step is perfectly valid. We may assume [itex]a\neq 0[/itex] when calculating the limit. So that leaves [itex]c\neq 0[/itex] 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 [itex]\frac{-c}{b}[/itex], if b is negative the expression with the positive sign converges to [itex]\frac{-c}{b}[/itex]
     
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