<s>[INST] how to prove that a solution of the equation is 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 $x = 5$. 
 [/INST] $ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} -  \cdots 4x-4=0  $ 
$ x^n$ - 4{ $x^{n-1} +  x^{n-2} +  x^{n-3} + \cdots+x+1$}=0
Using the property
$1+x+x^2......x^n-1 $= $\frac{1-x^n}{1-x}$
$ x^n$ - 4{$\frac{1-x^n}{1-x}$} =0 
$x^{n+1} - 5x^n + 4 $= 0
$x - 5 + \frac{4}{x^n} $= 0 
now if n tends to infinity x will tend to five.
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