Proving formally that this set of functions converges pointwise

Question

For
$$f_n(x)= \left\{ \begin{array}{ll} n & x\geq n \\ 1 & x< n \\ \end{array} \right. $$ Graphically (is this usually how it's first approached), I can see that $f_n$ approaches to the function $f(x) = 1$ pointwise.
This is my proof (formal) for this limit, but I'm stuck on one part.
A set of functions $\{f_{n}\}_{n\in \mathbb{N}}$ converges to a function $f$ pointwise $$\iff \forall x \in I, \forall \epsilon > 0 \quad \exists N : n \geq N \implies |f_n (x) - f(x)| < \epsilon. $$ By inspection of the function $f_n$, we can see that picking $N = x$ implies that if $n \geq x$, then $x \leq n$. Now for the case that $x < n$, then $|f_n (x) - f(x)| = |1 - 1| = 0$ which is less than $\epsilon$ by definition.
However, I can't seem to figure out what if $x = n$? i.e. if $n = N$? This means I have to show that $|n - 1| < \epsilon$, which I'm not sure how to.

Answer 1

Let $x \in \mathbb R$. Then there is $ N \in \mathbb N$ such that $x<N$. For $n \ge N$ we then get

$x<n$ and therefore $f_n(x)=1$.

Result: to each $x$ there is $N$ such that $f_n(x)=1$ for all $ n \ge N$