@Al86 

It is easy to compute by hand if you know elementary Calculus. E.g. the limit of the numerator is:
exp(-t) - exp(-infinity)  = exp(-t).

@Al86 You can try (e.g. for t=2)

Student[Calculus1][LimitTutor]();

but AFAIK it works for (very) simple limits.

@digerdiga 

In the future it would be nice to mention the method(s) you know about for your problem.
Otherwise, the answers seem to be useless because you knew them.

@Carl Love 

Nice and professional (as always :-)).

@digerdiga 

asympt cannot switch branches. You can obtain the same result without puiseux.

restart;
sol:=RootOf((8*n-8)*_Z^6+(n^4+36*n^2-68*n+56)*_Z^5+(n^5+10*n^4+80*n^3-200*n^2+224*n-152)*_Z^4+(n^6+28*n^5+69*n^4-268*n^3+468*n^2-356*n+200)*_Z^3+(3*n^7+32*n^6+7*n^5-204*n^4+380*n^3-544*n^2+272*n-128)*_Z^2+(3*n^8+14*n^7-20*n^6-32*n^5+252*n^4-240*n^3+304*n^2-80*n+32)*_Z-n^9-12*n^8-44*n^7-40*n^6-4*n^5-128*n^4+48*n^3-64*n^2):
f:=eval(op(sol), [_Z=z,n=1/x]):
g:=numer(eval(f, z=a/x+b+c*x)):
S:=solve({seq(coeff(g,x,k),k=0..2)}):
subs(S[],x=1/n, a/x+b+c*x);

@Kitonum 

I know. SymFun just calls solve/identity which could (and should) be improved a lot (you are free to send a SCR for it, and if successful, SymFun will work for your examples). Actually, if we are interested only in an odd/even result, this is easy to obtain.
BTW, the "hack" part was included just to answer (c). SymFun can be of course improved, but I prefer a capable solve/identity.

@Carl Love 

To find the initial point of a p-cycle simply do:

f:=y-> 4*y*(1-y):
p:=2;
solve((f@@p)(x)=x); # or fsolve

@tomleslie 

No need for DirectSearch. The min of expr^2 is of course 0 (attained for all variables 0).

@mamehrashi 

Write your problem in the text area, not in the title.
Maybe a smaller example will help.
If   P = 2*x*y + 3*x*z + 4*y*t + 5*t^2   is the polynomial in the variables x,y,z,t,
what result do you expect?

You just posted a polynomial in 16 variables, homogeneous, total degree 4,  with float coefficients.
What do you want to do with it?

@acer 

You have used the parametric version

plot3d([1,u,v],u=0..2*Pi-2*v,v=0..Pi, coords=spherical)

instead of the explicite

plot3d(1, u=0..2*Pi-2*v,v=0..Pi, coords=spherical)

It seems that you knew that the latter does not work correctly. Is it documented, or just a bug?

@Carl Love 

If we want separated colors, without interpolation, I don't think that an alternative to grid exists.

c:=piecewise(u+v<Pi, 1/6, u+v<2*Pi, 2/3, 1):
plot3d(1,u=0..2*Pi,v=0..Pi, coords=spherical, color=c, grid=[300,300], style=surface);

Compare with the default grid:

@acer Consider for simplicity  plot3d(f(x,y), x=0..1, y=0..1,  color = [c, colorscheme = [c1,c2,c3,c4]])

c = c(x,y) being the color and c1,c2,... fixed colors.

I can imagine a (useful) meaning of this for c = x (i.e. a kind of xgradient), even for c = c(x), but not for c = c(x,y).
What would be your definition? To me, it's nonsense for the moment.

plot3d( 1,u=0..2*Pi,v=0..Pi, coords=spherical, 
colorscheme= ["zgradient",[black, red, yellow, white],  colorspace="HSV"] , 
grid=[300,300]);

rr:=(x,y,z) -> piecewise(x<-1/3,0, x<1/3,1, x<=1,1):
gg:=(x,y,z) -> piecewise(x<-1/3,0, x<1/3,1, x<=1,0):
bb:=(x,y,z) -> piecewise(x<-1/3,1, x<1/3,0, x<=1,0):
plot3d(1,u=0..2*Pi,v=0..Pi, coords=spherical,
colorscheme=["xyzcoloring",[rr,gg,bb]], grid=[300,300]);

@dharr 

Yes, noticing that a "1" must exist in each row and column allows to reduce the number of the candidates.
This number can be halved (so, also the timing). And it is actually necessary to do so because otherwise each A appears twice (there are only 40320).
It is enough to change
for m to n^2 do
to

for m from k+1 to n^2 do

And preferably also k to n^2 - 1.

Edit. It's possible to reduce further the candidates by identifying two matrices A,B with A=B^+. (Maybe other symmetries too.)
BTW, for the second polynomial f:=x^8+x^4+x^3+x^2+1, A does not exist.