The sequence of functions $f_n: E \rightarrow \mathbb{R}$ where $f(x) = x^n$ converges uniformly to $ g(x) = 0 $ for $E = [0,1-\varepsilon]$, $\forall \varepsilon > 0$. Yet, it doesn't converge uniformly for $[0,1)$, the supremum of all $E$.
Is it possible to use infinitesmials to express the interval of uniform convergence in a closed form? If not, why not? If it is possible to do this, isn't this a great advantage of the hyper reals?
Here's how I interpret your question:
Can we find a hyperreal number $h > 0$ such that the sequence $f_n^\star$ converges uniformly to the zero function on the hyperreal interval $[0,1-h)$ but does not converge uniformly to the zero function on $[0,1-h')$ for any $h' < h$?
No such hyperreal exists.
It's clear that there is no real number r > 0 such that the sequence f_n converges uniformly to the zero function on the real interval $[0,1-r)$ but does not converge uniformly to zero on $[0,1-r')$ for any $r' < r$. Applying the Transfer principle to the sentence asserting the non-existence of real numbers satisfying this property yields a sentence asserting that there cannot be a hyperreal number with this property either.