Limit n->Infinity of recursive sequence

Alternatively, and perhaps more directly, use

RSolve[{a[n] == Sqrt[3] + 1/2 a[n - 1], a[0] == 1}, a[n], n]
(* {{a[n] -> 2^-n (1 - 2 Sqrt[3] + 2^(1 + n) Sqrt[3])}} *)
Limit[a[n] /. %[[1]], n -> Infinity]
(* 2 Sqrt[3] *)

[Possibly this is too cheap and should just be a comment, I'm not sure.]

One way to go about such problems is to realize that "in the limit", a[n]==a[n-1]. So just solve an algebraic equation. If there are multiple solutions then you need to figure out which is correct based on initial values and some other reasoning.

Solve[x == Sqrt[3] + x/2, x]

(* Out[130]= {{x -> 2 Sqrt[3]}} *)

You can attempt to convert your sequence in terms of a function which is not recursive, then take the Limit of that function

a[0] = 1;
a[n_] := a[n] = Sqrt[3] + 1/2 a[n - 1]

seq30 = Table[a[ic], {ic, 0, 30}];
func = FindSequenceFunction[seq30, n]

Limit[func, n -> Infinity]

2^(1 - n) (1 - 2 Sqrt[3] + 2^n Sqrt[3])

2 Sqrt[3]