consider the equation
$ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots-4x-4=0$
for $n = 1$ :: solution is : $x = 4$
for $n = 2$, ($x^2 - 4 x - 4 = 0$) :: solution is : $x = 4.8$
for $n = 3$, ($x^3 - 4x^2 - 4 x - 4 =0 $) :: solution is : x = $4.96$
how to prove that as $n \to\infty$ (what ever it means) the solution is $x = 5$.