Re: Same Limit: OK in 5.2, fails in 6.0; Packages gone in 6.0 ???

*To*: mathgroup at smc.vnet.net*Subject*: [mg77982] Re: [mg77959] Same Limit: OK in 5.2, fails in 6.0; Packages gone in 6.0 ???*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 21 Jun 2007 05:31:34 -0400 (EDT)*References*: <200706200941.FAA10343@smc.vnet.net>

On 20 Jun 2007, at 18:41, jrc wrote: > Why? > > I have, > > $Assumptions = {a > 0, k1 > 0, k2 >0, x e(in) Reals} > (repeats ok) > > i1b = (1/a)* Integral from -a/2 to +a/2 of integrand: > > exp(- i k2 x) exp(i k1 x) dx > > with the expected result, > > i1b = fraction with numerator = 2 Sin[(1/2)a(k1-k2)] > and denominator = a(k1-k2) > > Now I want the limit of this result, as the parameter a goes to > infinity: > > Limit[i1b, a -> Infinity] > > Mathematica 5.2 gives correct result, zero; > (note this is just lim(sin(x)/x, x -> inf), which is obviously > zero) > > Mathematica 6.0 is unable to evaluate the limit. > > Ruskeepaa's "Navigator", 2nd ed, (written for v5.2) claims there > is a package, "Calculus`Limit`" that makes 'Limit' work better > (p. 395 and p. 430). However, no such package seems to exist in > v6.0. > > How many packages no longer exist in 6.0 ???? Is there a list ???? > > Can anyone give me a reason for this obvious failure? > > jrc > Whose failure? Integrate[(Exp[(-I)*k2*x]*Exp[I*k1*x])/a, {x, -a/2, a/2}] (2*Sin[(1/2)*a*(k1 - k2)])/(a*k1 - a*k2) Assuming[Element[k1 | k2, Reals], Limit[%, a -> Infinity]] 0 The result is, of course, not true without the assumption (e.g. put k1=2I, k2 = I). Andrzej Kozlowski

**References**:**Same Limit: OK in 5.2, fails in 6.0; Packages gone in 6.0 ???***From:*jrc <jrchaff@mcn.net>