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Re: Wrong limit?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104812] Re: [mg104654] Wrong limit?
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Tue, 10 Nov 2009 06:03:10 -0500 (EST)
  • References: <200911061014.FAA07962@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

One could hope that Mathematica covered all cases in every symbolic  
calculation, but it's not possible, and if it were, it would frequently be  
inefficient.

Consider the simple quadratic:

Solve[a x^2 + b x + c == 0, x]

{{x -> (-b - Sqrt[b^2 - 4 a c])/(
    2 a)}, {x -> (-b + Sqrt[b^2 - 4 a c])/(2 a)}}

That's wrong when a == 0, just as in your Limit problem.

Reduce is more complete, if that's what we want... but do we really want  
every computation cluttered to the maximum degree?

Reduce[a x^2 + b x + c == 0, x]

(a != 0 && (x == (-b - Sqrt[b^2 - 4 a c])/(2 a) ||
      x == (-b + Sqrt[b^2 - 4 a c])/(2 a))) || (a == 0 && b != 0 &&
    x == -(c/b)) || (c == 0 && b == 0 && a == 0)

Bobby

On Fri, 06 Nov 2009 04:14:48 -0600, wiso <gtu2003 at alice.it> wrote:

> Look at this:
>
> Limit[(x^2 - a^2)/(5 x^2 - 4 a x - a^2), x -> a]
> Mathematica answer = 1/3
>
> this is ok for a !=0, but if a = 0 the value is
>
> Limit[(x^2 - a^2)/(5 x^2 - 4 a x - a^2) /. a -> 0, x -> 0]
> 1/5
>


-- 
DrMajorBob at yahoo.com


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