Give an \(\epsilon\)-\(\delta\) proof that \(f(x)=x^{1/3}\) is continuous on \([0,1]\).
Solution
1. Use the factorization \(u^3 - v^3 = (u-v)(u^2 + uv + v^2)\).Proof: for \(x, a \in [0,1]\) with \(u = x^{1/3}\), \(v = a^{1/3}\): \[|x^{1/3} - a^{1/3}| = \frac{|x - a|}{x^{2/3} + x^{1/3}a^{1/3} + a^{2/3}}.\]
2. Case \(a = 0\): continuous at \(0\).Proof: the denominator collapses: \(|x^{1/3} - 0| = x^{1/3}\).Given \(\epsilon > 0\), choose \(\delta = \epsilon^3\); then \(|x - 0| < \delta\) gives \(x^{1/3} < \epsilon\).
3. Case \(a > 0\): continuous at \(a\).Proof: the denominator satisfies \(x^{2/3} + x^{1/3}a^{1/3} + a^{2/3} \ge a^{2/3} > 0\), so \[|x^{1/3} - a^{1/3}| \le \frac{|x - a|}{a^{2/3}}.\] Given \(\epsilon > 0\), choose \(\delta = \epsilon\,a^{2/3}\); then \(|x - a| < \delta\) gives \(|x^{1/3} - a^{1/3}| < \epsilon\).
4. Q.E.D. Proof: \(f\) is continuous at every \(a \in [0,1]\), so continuous on \([0,1]\).(The same factorization shows \(x \mapsto \sqrt{x}\) is continuous on \([0,\infty)\): for \(a > 0\) the denominator \(\sqrt{x} + \sqrt{a} \ge \sqrt{a}\); at \(0\) take \(\delta = \epsilon^2\).)