finding limit equation of a convergent sequence (vn) in math exercise

Question

$(v_n)$ is a sequence, defined by :

$v_0 = 1$

$v_{n+1} = 9/6 - v_n$

After doing the first parts of my exercise, I know that :

$0 \lt v_n \lt 3$

$v_{n+1} - v_n = (3-v_n)^2/6-v_n$

$(v_n)$ is monotonically increasing

$(v_n)$ is convergent

I must then find the equation of $(v_n)$'s limit $1$, which is $l^2 - 6l + 9 = 0$

$v_n$ and $v_{n+1}$ have the same limit, so $l = 9/6 - l$, but I don't know how to get the equation of the limit I am asked to find.

Answer 1

But you have already found the equation. You have

$v_{n+1}-v_{n}=\frac{(3-v_{n})^{2}}{6-v_{n}}$ and both $v_{n+1}$ and $v_n$ have the value $l$ as $n\rightarrow \infty$, hence just put $l$ in above equation in appropriate places.