(June 2007 #4b part i) Evaluate \[\lim_{n\to\infty}\int_{\pi/2}^{\pi}\frac{n\sin(x/n)}{x}\,dx\] and justify your reasoning.
Solution
Goal: Evaluate \(\lim_{n\to\infty}\int_{\pi/2}^{\pi}\frac{n\sin(x/n)}{x}\,dx\).
1. The answer is \(\pi/2\).
1.1. Pointwise limit: for each fixed \(x \in [\pi/2, \pi]\), \(\frac{n\sin(x/n)}{x} \to 1\).Proof: \(\frac{n\sin(x/n)}{x} = \frac{\sin(x/n)}{x/n} \to \frac{\sin 0}{0}\text{-form} = 1\) since \(\lim_{t \to 0} \sin t / t = 1\) and \(x/n \to 0\).
1.2. Domination: \(\left|\frac{n\sin(x/n)}{x}\right| \le 1\) on \([\pi/2, \pi]\) for all \(n \ge 1\).Proof: \(\frac{n\sin(x/n)}{x} = \frac{\sin(x/n)}{x/n}\) with \(0 \le x/n \le \pi\) (as \(x \le \pi\), \(n \ge 1\)). For \(0 \le t \le \pi\), \(\sin t / t \le 1\) (standard: \(\sin t \le t\) on \([0,\infty)\)), and \(\sin(x/n) \ge 0\) since \(x/n \in [0, \pi]\).So \(0 \le \frac{\sin(x/n)}{x/n} \le 1\).
1.3. Q.E.D. Proof: Dominated Convergence: by 1 and 2, \(\lim_n \int_{\pi/2}^\pi \frac{n\sin(x/n)}{x}\,dx = \int_{\pi/2}^\pi 1\,dx = \pi - \pi/2 = \pi/2\).