ThmDex – An index of mathematical definitions, results, and conjectures.
Result R3316 on D2943: Continuous function
Pointwise limit of continuous functions need not be continuous
Formulation 0
Let $f_1, f_2, f_3, \dots : [0, 1] \to \mathbb{R}$ each be a D4364: Real function such that
(i) $f_n(x) = x^n$ for all $x \in [0,1]$ and $n \in \mathbb{N} + 1$
Then
(1) $f_1, f_2, f_3, \dots$ are each a D2943: Continuous function
(2) $\lim_{n \to \infty} f_n(x) \neq \emptyset$ is not a D2943: Continuous function
Proofs
Proof 0
Let $f_1, f_2, f_3, \dots : [0, 1] \to \mathbb{R}$ each be a D4364: Real function such that
(i) $f_n(x) = x^n$ for all $x \in [0,1]$ and $n \in \mathbb{N} + 1$
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
(i) R2923: Basic real polynomial function is continuous
(ii) R394: Domain-restriction of continuous map is continuous

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 $1$. $\square$