x^n converges pointwise but not uniformly on [0, 1]

Counterexample

Let $f_n(x) = x^n$ on $[0, 1]$. Let $f(x) = 0$ for $0 \le x < 1$ and $f(1) = 1$. For $0 \le x < 1$, $x^n \to 0$ by the geometric sequence limit. So $f_n \to f$ pointwise on $[0, 1]$. But $\sup_{0 \le x < 1} |x^n - 0| = 1$ for every $n$. So the convergence is not uniform.

Remarks

Each $f_n$ is continuous, but $f$ is not. This also shows the failure, by the uniform limit theorem. On $[0, r]$ with $0 < r < 1$, the convergence is uniform, since $\sup_{[0, r]} x^n = r^n \to 0$.

Explanations

L1 human

Near $x = 1$, the values $x^n$ stay close to $1$ for a long time before they drop to $0$. So no single $n$ makes all the values small at once.