# problem about simulating recurrence relation

We have the recurrence relation:

$5x_{n+1}-x_n=\frac{1}{3}$

$x_0=\frac{1}{12}$

solution: $y_h=(\frac{1}{5})^nC$ $y_p: 5A-A=\frac{1}{3}$

$A=\frac{1}{12}$

$y=(\frac{1}{5})^nC+\frac{1}{12}$

$y(0)=C=0$

Will the answer go to $\frac{1}{12}$ or somewhere very close to $\frac{1}{12}$?

The answer sheet says very close to. I thought it would be to exactly $\frac{1}{12}$ because for large n the computer would read the fraction as 0 even though it is not? And how is this not right? Please also explain why the fact that for large power n on fractions like here the fact that the number would be read as 0 would not alter this as that was my reasoning.

$5(x_{n+1}+c)=x_n+c \Rightarrow 5x_{n+1}-x_n=-4c$

Namely,$c=-1/12$

Let $b_{n+1}=x_{n+1}-1/12$, then $b_{n+1}=(1/5) b_n$

So we can achieve the following formula

$b_n=b_0 (\frac 1 5)^{n}$

where, $b_0=x_0-\frac 1 {12}=0 \Rightarrow b_n=0$,

Lastly,

$0=b_n=x_{n}-1/12$,

so we achieve $x_n=1/12$