(%i12) limit(sin(sin(sin(x)^5)^3)/x^15,x,0);
(%o12) 1
O.K.
(%i13) limit(sin(sin(sin(x)^5)^4)/x^20,x,0);
(%o13) lim(sin(sin(sin(x)^5)^4)/x^20,x,0)
Very weak
The following is really horrible
(%i3) tlimit(sin(sin(sin(x)^5)^3)/x^15,x,0);
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(x)^5)^3 Assumed to be zero in `taylor'
sin(sin(sin(x)^5)^3) Assumed to be zero in `taylor'
(%o3) 0
A new rule is necessary:
Replace sin(u) with u in products when lim u = 0, if you don't know if
lim u = 0 then try the replacement and if it gives finally (recursing) zero
then you win, all the replacement are correct, that is the last one pay for the
beer.
I think that in about 1/2 - 1 hour I can write some batteries: The rule is to
replace in products and powers f(u) with the first non zero taylor polynomial of
f. But to be smarter follow replacing to see if last one pay the beer.