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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