fₙ(x) = xⁿ converging to f(x) = 0
Pointwise on (−1, 1): For any fixed x, xⁿ → 0. But as x → 1⁻, convergence gets arbitrarily slow.
The red region shows where fₙ escapes the ε-band. No matter how large n is, there's always an x close enough to 1 where |xⁿ| > ε. So sup|fₙ − f| stays near 1 — not uniform.