       Re: Wrong limit?

• To: mathgroup at smc.vnet.net
• Subject: [mg104846] Re: Wrong limit?
• From: Maxim <m.r at inbox.ru>
• Date: Thu, 12 Nov 2009 05:59:01 -0500 (EST)
• References: <200911061014.FAA07962@smc.vnet.net> <hdbha5\$jg8\$1@smc.vnet.net>

```On Nov 10, 5:03 am, DrMajorBob <btre... at austin.rr.com> wrote:
> 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 <gtu2... 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
>
> --
> DrMajor... at yahoo.com

Interestingly, the original problem can actually be solved using
Reduce:

In:= Reduce[
ForAll[eps, eps > 0, Exists[del, del > 0, ForAll[x,
Implies[0 < Abs[x - a] < del,
Abs[(x^2 - a^2)/(5 x^2 - 4 a x - a^2) - M] < eps]]]], Reals]

Out= (a < 0 && M == 1/3) || (a == 0 && M == 1/5) || (a > 0=
&& M ==
1/3)

One just has to be careful to make sure the expressions do not become
indeterminate, otherwise the meaning of the quantifiers is not really
well defined. E.g.,

In:= Reduce[ForAll[x, x >= 0, 1/x > a]]

Out= False

In:= Reduce[ForAll[x, x >= 0, 1/x > 0]]

Out= True

Maxim Rytin
m.r at inbox.ru

```

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