(January 2007 #1) Let \[f(x)=\int_1^x\frac1t\,dt\] for \(x>0\). (a) Use an \(\epsilon\)-\(\delta\) proof to show that \(f\) is continuous on \((0,\infty)\). (b) Use an \(\epsilon\)-\(\delta\) proof to show that \(f\) is differentiable on \((0,\infty)\).
Solution
1. (a) \(f\) is continuous on \((0,\infty)\).Proof: fix \(x_0 > 0\) and \(\epsilon > 0\).For \(x\) in the neighborhood \(|x - x_0| < x_0/2\) (so \(x > x_0/2\)), \(1/t \le 2/x_0\) for \(t\) between \(x\) and \(x_0\), and \[|f(x) - f(x_0)| = \left|\int_{x_0}^x \frac{dt}{t}\right| \le \frac{2}{x_0}|x - x_0|.\] Choose \(\delta = \min(x_0/2,\ \epsilon x_0/2)\); then \(|x - x_0| < \delta\) gives \(|f(x) - f(x_0)| < \epsilon\).
2. (b) \(f\) is differentiable on \((0,\infty)\) with \(f'(x) = 1/x\).Proof: fix \(x > 0\) and consider the difference quotient; for \(h \ne 0\) with \(|h| < x/2\), \[\frac{f(x+h) - f(x)}{h} = \frac{1}{h}\int_x^{x+h}\frac{dt}{t}.\] Since \(1/t\) is continuous at \(x\), for every \(\epsilon > 0\) there is \(\delta < x/2\) with \(|1/t - 1/x| < \epsilon\) for \(|t - x| < \delta\).For \(0 < |h| < \delta\), \[\left|\frac{1}{h}\int_x^{x+h}\frac{dt}{t} - \frac{1}{x}\right| = \left|\frac{1}{h}\int_x^{x+h}\left(\frac{1}{t} - \frac{1}{x}\right)dt\right| \le \frac{1}{|h|}\int_x^{x+h}\left|\frac{1}{t} - \frac{1}{x}\right|dt \le \epsilon.\] Hence the quotient tends to \(1/x\), so \(f'(x) = 1/x\).
3. Q.E.D.