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