Why doesn't Mathematica evaluate this simple limit?

On a different occasion (Dirichlet coefficients as limits: wrong) I have shown that the sometimes limited capabilities of the function Limit[] can be improved by using an intermediate Series[].

Following this idea we can write for the limit in question

Limit[Expand[
  Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 2}]] /. 
   x -> 10/ n + Sin[n]/n^2], n -> \[Infinity]]

(* 25 *)

In this manner we can even calculate the limit with a symbolic parameter "a"

Limit[Expand[
  Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 2}]] /. x -> a/ n + Sin[n]/n^2],
  n -> \[Infinity]]

(* a^2/4 *)

Also, a general function is permissible (provided Limit[f[n]/n,n -> \[Infinity]] == 0)

Limit[
 Expand[Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 2}]] /. 
   x -> a/ n + f[n]/n], n -> \[Infinity]]

(* Limit[a^2/4 + 1/2 a f[n] + f[n]^2/4, n -> \[Infinity]] *)

Where the final Limit can only be assessed once f[n] is given explicitly.

Modification of the OP.

Taking f[n] = Sin[n] (instead of f[n] = Sin[n]/n as in the OP) we find

Limit[Expand[
  Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 2}]] /. x -> a/ n + Sin[n]/n], 
 n -> \[Infinity]]

(* Limit[a^2/4 + 1/2 a Sin[n] + Sin[n]^2/4, n -> \[Infinity]] *)

Taking the x-expansion beyond x^2 we get for all higher powers

Limit[Expand[
  Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 3}]] /. x -> a/ n + Sin[n]/n], 
 n -> \[Infinity]]

(* 1/4 (a + Interval[{-1, 1}])^2 *)

The problem here is the Sin[n] which has no limit since it is an oscillating function, but it is always bounded by $\pm 1$: if you change you code with the following:

Limit[(n - Sqrt[1 + 10 n + n^2])^2, n -> Infinity]

with 1 in place of Sin (or -1 if you want), you get the result:

(*25*)

Since the functionality of Limit[] has been improved in version 11.2, the limit is now evaluated rather easily:

Limit[(n - Sqrt[Sin[n] + 10 n + n^2])^2, n -> ∞]
   25