Let $A$ be a commutative ring with unit, $J = (x)$ an ideal of $A[x]$.
Thus we can consider the inverse system defined as $$\psi_{n,m}: A[x]/J^m \to A[x]/J^n$$ $$g(x) + J^m \to g(x) + J^n$$ $$\forall \ m\geq n$$ Then I have to prove that $$A[[x]] \cong \underleftarrow{\lim}A[x]/J^n$$
$A[[x]]$ is a cone over this inverse system via the projections $\pi_n : A[[x]] \to A[[x]]/ J^n \ \ \forall \ n$.
But how to prove that it is an universal cone ? i.e. for every other cone $(M, \delta_n)_{n \in \mathbb{N}}$ over the inverse system there is an unique map $\lambda : M \to A[[x]]$ such that $$\pi_n \circ \lambda = \delta_n$$
I would proceed as follows, identify $A[[x]]$ with the sub-ring of the infinite product $\prod_{n}A[X]/J^n$ of all elements $(p_n)_{n}$ such that $\psi_{n,m}(p_m)=p_n$ for all $n$ and $m$.
Then, you can define a map $\lambda:M\to \prod_{n}A[x]/J^n$ just sending $m\in M$ to $(\delta_n(m))_n$. Just verify that the image of this morphism is contained in $A[[x]]$. Uniqueness also follows easily because any morphism $M\to \prod_{n}A[x]/J^n$ is determined by the compositions $M\to \prod_{n}A[x]/J^n\to A[x]/J^n$.