Given $$N_{t+1}=\frac{rN_t}{1+bN_t^2}$$ for $r>0$ and $b>0$
I need to:
$1$. Find the limit of the recursion.
$2$. Prove that: $$\frac{2r^2}{(4+r^2)\sqrt{b}}\le N_t \le \frac{r}{2\sqrt{b}}$$ for large enough $t$.
$3$. Prove that if $b>4$ then for some $t$ we have $N_t<1$.
Thanks a lot.
The limit (if there is one) of any continuous recurrence relation must be a fixed point. That means in your case that if there is a limit $N$, it needs to statisfy $$ N = \frac{rN}{1+bN^2} $$ For $b = 4$ and $r = 2$, both sides of the inequality in point 2 is equal to $1/2$, while the sequence itself will never be exactly equal to that unless it starts out with $N_1 = \frac{1}{2}$, so the inequality will never hold in that case (I'm sure there are other cases as well).