Limit of $a_n$, where $a_1=-\frac32$ and $3a_{n+1}=2+a_n^3$. Let $a_1=-\frac32$, and $3a_{n+1}=2+a_n^3$. I need to show that $\displaystyle \lim_{n\to \infty} a_n = 1$.
I can show that the sequence is monotonically increasing and bounded as follows:
By the critical value method, $\frac13\left(2+x^3\right)>x$ for $x\in (-2,1)$. So, $a_n<a_{n+1}$ for all $n$. Also, if $f(x)=\frac13\left(2+x^3\right)$, then $f$ is increasing and $f(1)=1$. So, $a_n<1$ implies $a_{n+1}<1$. Hence the sequence is convergent, but I can't seem to find a good way to show that the $a_n$ are getting arbitrarily close to $1$.

Now that you know that your iteration is convergent, you have that any limit must be a fixed point of the iteration function. That is, if $L$ is the limit, then
$$3L = 2+L^3$$
or
$$L^3-3L+2=0.$$
This comes from taking limits on both sides. With some guess and check, you can factor this as
$$(L-1)^2(L+2)=0.$$
So the limit is either $1$ or $-2$. But the sequence is increasing from $-3/2$, so the limit is $1$.
