(January 2008 #6b) Suppose that \(\lim_{n\to\infty}s_n=s\) and \(\lim_{n\to\infty}t_n=t\) with \(s\ne t\) and \(s_n\ne t_n\) for all \(n\). Use an \(\epsilon\)-\(\delta\) proof to show that \[ \lim_{n\to\infty}\frac{s_n+t_n}{s_n-t_n}=\frac{s+t}{s-t}. \]
Solution
1. Reduce to a direct \(\epsilon\)-\(N\) estimate on the difference.Proof: put the two quotients over a common denominator: \[\frac{s_n + t_n}{s_n - t_n} - \frac{s + t}{s - t} = \frac{(s_n + t_n)(s - t) - (s + t)(s_n - t_n)}{(s_n - t_n)(s - t)} = \frac{2(s t_n - s_n t)}{(s_n - t_n)(s - t)}.\] Adding and subtracting \(st\) in the numerator, \(s t_n - s_n t = s(t_n - t) - t(s_n - s)\), so \[\frac{s_n + t_n}{s_n - t_n} - \frac{s+t}{s-t} = \frac{2\big(s(t_n - t) - t(s_n - s)\big)}{(s_n - t_n)(s - t)}.\]
2. Bound the numerator.Proof: since \(s \ne t\), \(\delta := |s - t|/2 > 0\); and \(s_n \to s\), \(t_n \to t\), so for large \(n\), \(|s_n - s| < \delta/2\) and \(|t_n - t| < \delta/2\), giving \(|s_n - t_n| \ge |s - t| - |s_n - s| - |t_n - t| \ge \delta > 0\); the denominator is bounded below in modulus by \(\delta|s-t| = 2\delta^2\)… precisely \(|(s_n - t_n)(s-t)| \ge \delta \cdot |s-t|\).
3. \(\epsilon\)-\(N\) conclusion.Proof: let \(\epsilon > 0\).Choose \(N\) so that for \(n \ge N\): \(|s_n - s| < \frac{\epsilon\,\delta\,|s-t|}{4|t| + 4|s| + 1}\) (and similarly for \(t_n - t\)) and \(|s_n - t_n| \ge \delta\).Then \[\left|\frac{s_n + t_n}{s_n - t_n} - \frac{s+t}{s-t}\right| \le \frac{2\big(|s||t_n - t| + |t||s_n - s|\big)}{\delta\,|s - t|} < \epsilon.\] Hence the quotient converges to \(\frac{s+t}{s-t}\).
4. Q.E.D.