For each integer $n \geq 1$, the function $f_n$ is a monomial restricted to the closed interval $[0, 1]$ and therefore continuous by the results
Next, if $x \in [0, 1]$ and $n \geq 1$ is an integer, then
\begin{equation}
\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} x^n =
\begin{cases}
1, \quad & x = 1 \\
0, \quad & x \in [0, 1)
\end{cases}
\end{equation}
Thus, if $ \varepsilon = 1/2$, then
\begin{equation}
|1 - f(x)| = |1 - 0| = 1 > \frac{1}{2}
\end{equation}
for all $x \in [0, 1]$ for which $0 < |1 - x|$. Thus, result
R3312: Characterisation of continuity in Euclidean space states that $f$ is not continuous at $1$. $\square$
