I just looked at the article but wonder if there is a general derivation for the limit of the ratio.

There does seem to be a pattern

In[15]:= f3 = 
  RSolveValue[{a[n] == a[n - 1] + a[n - 2] + a[n - 3], 
    a[1] == a[2] == a[3] == 1}, a, n];
Limit[f3[n + 1]/f3[n], n -> \[Infinity]] // RootReduce

Out[16]= Root[-1 - #1 - #1^2 + #1^3 &, 1]

In[17]:= f4 = 
  RSolveValue[{a[n] == a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4], 
    a[1] == a[2] == a[3] == a[4] == 1}, a, n];
Limit[f4[n + 1]/f4[n], n -> \[Infinity]] // RootReduce

Out[18]= Root[-1 - #1 - #1^2 - #1^3 + #1^4 &, 2]

RSolve works also, although it gives a more complex expression

In[20]:= sln = 
  RSolve[{a[n] == a[n - 1] + a[n - 2] + a[n - 3], 
    a[1] == a[2] == a[3] == 1}, a[n], n];

In[21]:= f[n_] = a[n] /. sln[[1]] // ToRadicals;

In[22]:= lim = Limit[f[n + 1]/f[n], n -> \[Infinity]]

Out[22]= (-22 2^(5/6) + 4 Sqrt[11] (283 + 21 Sqrt[33])^(1/6) + 
   2 2^(2/3) Sqrt[11] (259 + 45 Sqrt[33])^(1/6) - 
   11 (329 + 57 Sqrt[33])^(1/6) + 
   2 Sqrt[22 (7 - Sqrt[33])] (329 + 57 Sqrt[33])^(1/6) + 
   2 Sqrt[22 (7 + Sqrt[33])] (329 + 57 Sqrt[33])^(1/6) - 
   11 2^(1/6) (329 + 57 Sqrt[33])^(1/3) + 
   2^(1/3) Sqrt[11] (26803 + 4611 Sqrt[33])^(1/6) + 
   2^(2/3) Sqrt[11] (8745739 + 1522395 Sqrt[33])^(
    1/6))/(3 (329 + 57 Sqrt[33])^(
   1/6) (-11 + Sqrt[11] (566 - 42 Sqrt[33])^(1/6) + 
     Sqrt[11] (566 + 42 Sqrt[33])^(1/6)))

In[23]:= N[lim, 30]

Out[23]= 1.83928675521416113255185256465

The MathWorld entries for the tribonacci number, tribonacci constant, and Fibonacci $n$-step number are quite informative on the subject of linear recurrences that generalize the usual Fibonacci sequence. I will only add in addition to Frank's answer that you can use DifferenceRoot[] directly:

tribonacci = DifferenceRoot[Function[{a, n}, {a[n] == a[n - 1] + a[n - 2] + a[n - 3],
                                              a[1] == 1, a[2] == 1, a[3] == 1}]];

and then use it in Limit[] (with some help from FunctionExpand[] and ToRadicals[]):

Limit[tribonacci[n + 1]/tribonacci[n] // FunctionExpand, n -> ?] // ToRadicals
   (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3))/3

N[%, 30]
   1.83928675521416113255185256465