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 $1$.

You've done most of the work already.  Since $a_n$ is bounded and monotone increasing, then a limit $L$ exists.  Further, since $a_{n+1}$ is a subsequence of $a_n$, then they have the same limit.  That is, $\lim(a_n) = \lim(a_{n+1}) = L$.  Taking a limit of both sides of your recursion, we get:
$$\lim(3a_{n+1}) = \lim(2+a_n^3)$$
And applying limit arithmetic to this, we arrive at:
$$3L = 2 + L^3$$
Now it's just a matter of solving for $L$.  
