Recurrence relation limit outside range of values

Question

I Have a recurrence relation defined by:

• $a_1=0$
• $a_{n+1}=\frac{3a_n+1}{a_n+3}$
• $0\leq a_n \lt 1$

When trying to find the limit of this by n$\rightarrow\infty$ and using algebra of limits i get:

$l=\frac{3l+1}{l+3}$

$l^2+3l=3l+1$

$l=\pm1$

Now since $0\leq a_n \lt 1$ the limit can not be -1, but it also appears to not be 1.

Can anyone help me determine if the limit is 1 or something else?

Answer 1

Just because a sequence is bounded strictly, does not prevent its limit from being equal to that bound.

For example $x_n=\frac{1}{n} > 0$ for all $n$ yet $x_n \to 0$

In general if $a<x_n<b$ and $x_n \to x$ then all we can deduce is $a \le x \le b$.