uniform limit theorem
Here is the single most important reward of uniform convergence: it cannot break continuity. If every function in your sequence is continuous and they approach a limit uniformly, the limit is continuous too. Pointwise convergence offers no such guarantee — that gap is the whole reason uniform convergence earns its keep.
Statement: let f_n be continuous on a set E (or at a point a) and suppose f_n -> f uniformly on E. Then f is continuous on E. The proof is the famous epsilon/3 argument: write |f(x) - f(a)| <= |f(x) - f_n(x)| + |f_n(x) - f_n(a)| + |f_n(a) - f(a)|. Uniform convergence makes the first and third terms small for large n UNIFORMLY in x, then continuity of that fixed f_n makes the middle term small for x near a.
Read contrapositively, the theorem is a powerful disqualifier: if continuous functions converge POINTWISE to a discontinuous limit, the convergence cannot have been uniform. This is the quickest way to certify that a given pointwise convergence fails to be uniform, without computing any sup norm.
Each f_n(x) = x^n is continuous on [0,1], but their pointwise limit jumps from 0 to 1 at x = 1, so it is discontinuous. By the uniform limit theorem (in contrapositive form), the convergence x^n -> f cannot be uniform on [0,1] — confirming the sup-norm computation that sup |x^n - f| stays at 1.
A discontinuous pointwise limit rules out uniform convergence.