1. If we understand this problem as literally written "to find the supremum of the sequence of the function", then the task is trivial, because the limit function is 0 if x<>0 and 1 if x=0. Unfortunately, Maple incorrectly calculates the limit at the point x=0 :
limit((x^2+4)/(2*x^2+(n*x+2)^2), n = infinity);
Output is 0
Should be piecewise(x = 0, 1, 0) .
So the right answer is 1 .
Below, the less trivial problem is solved: to find the maximum of the function f(x) in the range [1,infinity) as a function of the parameter n (at first, looking at OP's code, I understood the problem in this very wording.).
2. Note that the correct command to find the maximum (symbolically) will be maximize not maximum. It is obvious that Maple does not directly cope with this problem. Therefore, we present a manual solution in which all the calculations are done using Maple. The case n = 0 is obvious, because we get an even function that decreases on the range [1,infinity] . Therefore, the maximum value of the function in this interval is equal to f(1). We show that in all other cases (n<0 or n>0) also the maximum value of the function on the range [1,infinity] is equal to f(1) . If n<>0 , then the function always has 2 critical points, and if n<0 , then both critical points are less than 1, and if n>0 , then only one critical point is greater than 1. In the second case, the second derivative at this point is positive , so the function has a minimum at this point. Therefore, in all cases, it is enough for us to compare 2 values f(1) and f(infinity) , and this is easily verified that f(1)>f(infinity). So the final answer is f(1) = 5/(n^2+4*n+6) .