# [Maxima] Learning that limit is 0 from above

Stavros Macrakis macrakis at alum.mit.edu
Thu Nov 1 10:18:28 CDT 2007

```tlimit (which uses taylor) also works, if you're willing to answer an
unnecessary question:

tlimit(1/e,e,inf)

asks the sign of a^(k/(k-1)+1)*k^2-2*a^(k/(k-1)+1)*k+a^(k/(k-1)+1)
and then gives a reply.  Of course, it would have been nicer if Maxima could
recognize that that expression is positive, which it can if you factor it as
Barton shows below.

-s

On 10/31/07, Barton Willis <willisb at unk.edu> wrote:
>
> Try something like this:
>
> (%i1) assume(k>1/2,k<1,a>0,b>a)\$
> (%i2) e : (1/(b-a))*integrate((1-k)*((2*t-a)/k)^(k/(k-1)),t,a,b)\$
>
> (%i3) factor(taylor(e,b,inf,0));
> (%o3) (b^(k/(k-1))*(k-1)*2^(k/(k-1)))/(k^(k/(k-1))*(2*k-1))
>
> (%i4) sign(%);
> (%o4) neg
>
> BW
>
> -----maxima-bounces at math.utexas.edu wrote: -----
>
> >To: maxima at math.utexas.edu
> >From: "Paul Smith" <phhs80 at gmail.com>
> >Sent by: maxima-bounces at math.utexas.edu
> >Date: 10/31/2007 05:53PM
> >Subject: Re: [Maxima] Learning that limit is 0 from above
> >
> >On 10/31/07, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
> >> > (%i1) f(x) := exp(-x);
> >> > (%o1)                          f(x) := exp(- x)
> >> > (%i2) limit(f(x),x,inf);
> >> > (%o2)                                  0
> >> > (%i3)
> >> >
> >> > In this example, the convergence to zero is from above (0^+). Is it
> >> > possible to have maxima tell us it?
> >>
> >> Yes, compute limit(1/f(x),x,0,plus).  If the result is inf, then the
> >> original converges from above; if minf, from below.
> >>
> >> By the way, Maxima is generally better at working with expressions than
> >> functions, so I would have written the above as
> >>
> >>     f: exp(-x)\$
> >>     limit(f,x,inf);
> >>     limit(1/f,x,inf);
> >>
> >> Hope this helps,
> >
> >However, your technique not always work. See this code:
> >
> >assume(k>1/2,k<1,a>0,b>a);
> >f(b):= (1/(b-a))*integrate((1-k)*((2*t-a)/k)^(k/(k-1)),t,a,b)\$
> >limit(f(b),b,inf);
> >limit(1/f(b),b,inf);
> >
> >The first limit is zero, but the second one cannot be calculated by
> >maxima. Any further ideas?
> >
> >Paul
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