@Carl Love The limit is for sqrt(n/3) * f(n,x)  as requested, not for f(n,x) (which is obviouosly 0).

limit( sqrt(n/3)*asympt('rsolve'({f(0) = a, f(n) = sin(f(n - 1))}, f(n)), n), n=infinity );

        1

@Alfred_F  Maple is only able to guess the limit:

Download Jacobi-vv.mw

@Carl Love 

arctan(2/n^2) = arctan(n+1) - arctan(n-1)

Telescoping sum ==>
sum( 2/n^2), n=1..N )  = -Pi/4 + arctan(N) + arctan(N+1) --> -Pi/4 + 2*Pi/2 = 3*Pi/4.
 

@Alfred_F The point (1,1) does not matter, the monotone convergence condition is a.e. (almost everywhere).

@Alfred_F Here are higher precision computations; you may play with Digits and epsilon.

restart;
N:=10^12:
Digits:=50:
exact:=evalf(ln(2026/2025)):
J1:=evalf(Int(N*x^N/(x^N+x+2024), x = 0 .. 1, method=_Dexp, epsilon=1e-20)); 
J2:=evalf(Int(t^(1/N)/(t + t^(1/N) + 2024), t = 0 .. 1, method=_Dexp, epsilon=1e-20));
err1=J1-exact;
err2=J2-exact;

      J1 := 0.00049370526798868160514093629790805923985216893049776
      J2 := 0.00049370526798868160514093619869170079706955297924341
      err1 = -4.9352240248370621056973318812232002987*10^(-16) 
      err2 = -4.9352240248380542692817597073827128422*10^(-16)   

If you are using the good old Derive, be careful with numeric computations because it has a very special implementation with fractions instead of the standard floating-point numbers.

@mmcdara The problem with such transforms is that expr and S(expr) could be totally different; just plot expr - S(expr) for  expr:=sin(6-2*sin(x)) + cos(6*sin(-3+x));

You already have a function z(t) in the system. Do you mean another function, say Z(t) = exp(int(y(u),u=0..t)) ?  Maybe u=t..0 ?

@mmcdara The correct mathematical formulation  AREA  ∈  {4, 8}  was given. The and/or expression corresponds to a  common language and it is permitted. I don't see any reason to be too pedantic.

@mmcdara The standard formula for the area of a triangle in coordinate geometry uses abs:
Area of a triangle - Wikipedia

@mmcdara  abs is necessary because we want an unoriented area (Area(ABC) = Area(ACB)).
See the examples:

A,B,C:=[0, 3*t-3], [1, t+1], [2, t+1]; # oriented area = 8 (at t=10)

A,B,C:=[0, 3*t*(1/5)-3], [1, t+1], [2, t+1]; # oriented area = -4 (at t=10)

@acer Thank you. I think that you are also waiting for smarter simplify and evala commands.

@Alfred_F  Let A,B,C be consecutive vertices and M,N the midpoints of AB, BC.
angle(BAC)=Pi/n implies AC = 2*AB*cos(Pi/n) ==> MN = cos(Pi/n) --> 1.

@Alfred_F  Most of the Kamke's problems are easily soved by Maple. See

comparison_maple_mathmatica_des_kamke.pdf

kamke_differential_equations.pdf

@Rouben Rostamian  Technically this is the Gateaux derivative (differential). Unlike the Frechet differential, it is not always linear.

@Kitonum A nice counterexample is the almost regular "tetrahedron" with sides [4,4,4,4,4,7].