Step 1 (Recast the problem in a simpler form). Let $p_n(x) := x^n-4x^{n-1}-4x^{n-2}-\cdots -4x-4$ and observe that neither $x=0$ nor $x=1$ solve the equation $p_n(x)=0$ for any value of $n$, hence you can assume $x\neq 0,1$. Now, rewrite: $p_n(x) = x^n-4\sum_{k=0}^{n-1} x^k\; ;$ from the rule for the sum of a geometric progression you get: $\begin{split} p_n(x) &= x^n -4\frac{1-x^n}{1-x} &= \frac{x^{n+1}-5x^n+4}{x-1} \; , \end{split}$ thus $p_n(\bar{x})=0$ iff $f_n(\bar{x})=0$, where: $f_n(x) := x^{n+1}-5x^n+4\; .$
Step 2 (Existence and properties of the solutions of $f_n(x)=0$). By the IVT, function $f_n$ has some zeros in the oper interval $I:=]4,5[$ because $f_n(4)<0; more precisely, there is only one zero in $I$ and it lies in the subinterval $]\frac{5n}{n+1} , 5[$.
Let $x_n$ be the zero of $f_n$ (and a fortiori of $p_n$) which lies in $I$. Since the sequence $f_n(x)$ is strictly decreasing for any $x\in I$, the sequence $x_n$ is strictly increasing and therefore it has a limit $\bar{x}$; from the upper bound $x_n<5$, you infer the bound $\bar{x} \leq 5$ thus $\bar{x}$ is finite.
Step 3 (Evaluation of $\bar{x}$). Now, you have to evaluate $\bar{x}$. From $f_n(x_n)=0$ and $x\neq 0,1$ you get: $= \frac{1}{x_n^n\ (x_n-1)}\ \left( x_n-5+\frac{4}{x_n^n}\right) =0 \qquad \Leftrightarrow \qquad x_n=5-\frac{4}{x_n^n} \; ;$ since $4, the sequence $x_n^n$ tends to zero, hence finally: $\bar{x} = \lim_{n\to \infty} x_n =\lim_{n\to \infty} 5-\frac{4}{x_n^n} =5$ as you claimed.