When the Limit Breaks

At $x = 1$, $f_n(1) = 1^n = 1$ for every $n$, so the limit is $1$.

At $x = 0$ the limit is trivially $0$. For $0 < x < 1$, write $t = \frac1x > 1$ as $t = 1 + h$ with $h > 0$; Bernoulli's inequality gives $t^n \geq 1 + nh$.

Hence $0 \leq x^n = \frac{1}{t^n} \leq \frac{1}{1+nh} \to 0$, so $x^n \to 0$ by the squeeze theorem.

The pointwise limit is therefore $0$ on $[0,1)$ and $1$ at $x = 1$ — discontinuous at $x = 1$. A first warning that pointwise limits of continuous functions need not be continuous.

On $[0,1)$, $|f_n - f| = x^n$, and it is $0$ at $x = 1$, so $\sup_{[0,1]} |f_n - f| = \sup_{[0,1)} x^n$.

Since $x^n < 1$ for $x < 1$, the supremum is at most $1$. But $x^n \to 1$ as $x \to 1^{-}$ (continuity of the power function), so no number below $1$ is an upper bound.

The supremum is exactly $1$.

Uniform convergence would require $\sup |f_n - f| \to 0$, but here it equals $1$ for every $n$. Hence $(x^n)$ does not converge uniformly.

Let each $f_n$ be continuous and $f_n \to f$ uniformly. Fix $x_0$ and $\varepsilon > 0$.

By uniform convergence there is $N$ with $\sup |f_N - f| < \frac{\varepsilon}{3}$. Fix this $N$.

Since $f_N$ is continuous at $x_0$, there is $\delta > 0$ with $|f_N(x) - f_N(x_0)| < \frac{\varepsilon}{3}$ whenever $|x - x_0| < \delta$.

For $|x - x_0| < \delta$, the triangle inequality gives $|f(x) - f(x_0)| \leq |f(x) - f_N(x)| + |f_N(x) - f_N(x_0)| + |f_N(x_0) - f(x_0)| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$. So $f$ is continuous at $x_0$, and $x_0$ was arbitrary.