An interesting limit from quadratic functions...

Eliot Rosewater · 20-12-2010 12:30AM #1

Given that the rightmost root of [LATEX]\displaystyle ax^2+bx+c[/LATEX] is

[LATEX]\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}[/LATEX]

and that the only root of [LATEX]\displaystyle bx+c[/LATEX] is

[LATEX]\displaystyle \frac{-c}{b}[/LATEX]

Then surely

[LATEX]\displaystyle \lim_{a \to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)=\frac{-c}{b}[/LATEX]

for b>0! What a bizzare limit! Like, it comes out of nowhere! How does the b even go below the line?!

I wrote a short C++ program to test it out for b=2; c=1 and it started converging (slowly).

Wolfram Alpha plot:
gif&s=41&w=410&h=190

equivariant · 20-12-2010 12:43AM

Eliot Rosewater wrote: »

Given that the rightmost root of [LATEX]\displaystyle ax^2+bx+c[/LATEX] is

[LATEX]\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}[/LATEX]

and that the only root of [LATEX]\displaystyle bx+c[/LATEX] is

[LATEX]\displaystyle \frac{-c}{b}[/LATEX]

Then surely

[LATEX]\displaystyle \lim_{a \to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)=\frac{-c}{b}[/LATEX]

for b>0! What a bizzare limit! Like, it comes out of nowhere! How does the b even go below the line?!

I wrote a short C++ program to test it out for b=2; c=1 and it started converging (slowly).

Wolfram Alpha plot:

I guess that multiplying above and below by the conjugate (i.e. [latex]-b - \sqrt{b^2-4ac}[/latex]) will explain it?

FoxT · 20-12-2010 01:32PM

yes, it jumps out immediately. Interesting, counterintuitive limit though!

Eliot Rosewater · 20-12-2010 07:49PM

Ah damn, now it's not a mystery any more.

:pac:

Seriously though, cheers for that - even with the proof, it still looks counter-intuitive on the face of it, as FoxT says.

An interesting limit from quadratic functions...

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