(June 2013 #1b) Prove that \[\lim_{x\to-\infty}\frac{x-1}{x}=1.\]
Solution
1. Simplify the difference.Proof: \(\frac{x-1}{x} - 1 = \frac{x-1-x}{x} = -\frac{1}{x}\), so \(\left|\frac{x-1}{x} - 1\right| = \frac{1}{|x|}\) for \(x \ne 0\).
2. \(\epsilon\)-\(M\) argument for \(x \to -\infty\).Proof: let \(\epsilon > 0\) and choose \(M = 1/\epsilon\) (any \(M > 1/\epsilon\) works).For \(x < -M\): \[\left|\frac{x-1}{x} - 1\right| = \frac{1}{|x|} = \frac{1}{-x} < \frac{1}{M} \le \epsilon.\] Hence \(\lim_{x\to-\infty}\frac{x-1}{x} = 1\).
3. Q.E.D.