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
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- Wrong limit?
- From: wiso <gtu2003@alice.it>
- Wrong limit?