       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>

```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.

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]
>
> 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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