Let $A$ be a unital commutative complex Banach algebra. Suppose that $x \in A$ such that the Banach algebra generated by $\{1_A,x\}$ equals $A$. Show that the natural map
$$Sp(A) \to Sp_A(x): \chi \mapsto \chi(x)$$
is a homeomorphism.
I managed to show that this map is well-defined, continuous and injective. Homeomorphism will follow from the compact domain and the Hausdorff codomain once I have shown this is surjective.
In order to show this is surjective, I'm trying to show that if $\lambda \in Sp(A)$ and $J:= \overline{(x-\lambda)} \unlhd A$, then
$$A/J \cong \mathbb{C}$$
I try to apply isomorphism theorems for this. In order to do this, I need to define a surjective map $A \to \mathbb{C}$ with kernel equal to $J$.
So let $a \in A$. Then there are polynomials $P_n \in \mathbb{C}[X]$ such that $a = \lim_n P_n(x)$. I guess the obvious image of a map with codomain $\mathbb{C}$ would be $\lim_n P_n(\lambda)$, but then I have to show two things:
(1) This does not depend on the chosen sequence.
(2) The limit actually exists.
I don't manage to prove both these things, and I'm unsure it is even true.