(June 2009 #1) Give an \(\epsilon\)-\(\delta\) proof that \[f(x)=\frac{x^2}{1-x^2}\] is continuous on \((0,1)\). Is \(f\) uniformly continuous on \((0,1)\)? Prove your answer.
Solution
1. \(f(x) = \frac{x^2}{1 - x^2}\) is continuous on \((0,1)\).Proof: fix \(a \in (0,1)\) and \(\epsilon > 0\).The function \(x \mapsto 1 - x^2\) is continuous and \(1 - a^2 > 0\), so pick \(\delta_1 > 0\) with \(|x - a| < \delta_1\) implying \(1 - x^2 > \frac{1 - a^2}{2}\) (hence the denominator stays bounded away from \(0\)). On the neighborhood \(|x-a| < \delta_1\), the numerator \(x^2\) is bounded by \((a + \delta_1)^2\), and the difference \[\left|\frac{x^2}{1-x^2} - \frac{a^2}{1-a^2}\right| = \frac{|x^2 - a^2|}{(1-x^2)(1-a^2)}\] satisfies \(|x^2 - a^2| \le |x-a|\cdot 2(a + \delta_1)\), so choosing \(\delta = \min\big(\delta_1, \frac{\epsilon(1-a^2)^2}{4(a+\delta_1)}\big)\) gives \(|f(x) - f(a)| < \epsilon\) whenever \(|x - a| < \delta\).Hence \(f\) is continuous at every \(a \in (0,1)\).
2. \(f\) is NOT uniformly continuous on \((0,1)\).Proof: take \(x_n = 1 - \frac{1}{n}\) and \(y_n = 1 - \frac{1}{2n}\) (so \(y_n < x_n < 1\)). Then \(|x_n - y_n| = \frac{1}{2n} \to 0\), but \[f(x_n) = \frac{(1 - 1/n)^2}{1 - (1 - 1/n)^2} \sim \frac{1}{2/n} = \frac{n}{2} \to \infty, \quad \text{and similarly } f(y_n) \to \infty,\] so \(|f(x_n) - f(y_n)| \to \infty \not\to 0\).No \(\delta\) can control pairs \((x_n, y_n)\) with \(|x_n - y_n| \to 0\), so \(f\) is not uniformly continuous.(Equivalently, a uniformly continuous function on a bounded interval is bounded — here \(f\) is unbounded near \(1\).)
3. Q.E.D.