Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

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.
>
> 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 <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[1]:= 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[1]= (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[2]:= Reduce[ForAll[x, x >= 0, 1/x > a]]

Out[2]= False

In[3]:= Reduce[ForAll[x, x >= 0, 1/x > 0]]

Out[3]= True

Maxim Rytin
m.r at inbox.ru


  • Prev by Date: Re: Piecewise with ODE and constant
  • Next by Date: Re: Cumulative probability that random walk variable exceeds given value
  • Previous by thread: Re: Wrong limit?
  • Next by thread: Re: Re: Wrong limit?