Prove that if \(f'\) exists and is bounded on \((a,b]\), then \(\lim_{x\to a^+}f(x)\) exists.
Solution
1. \(f\) is continuous on \((a, b]\) and differentiable on \((a, b)\).
differentiability on \((a,b]\) implies continuity at every point of \((a,b]\), including \(b\).
2. \(f\) is Lipschitz on \((a, b]\) with constant \(M = \sup_{(a,b]}|f'| < \infty\): for \(a < x < y \le b\), \(|f(x) - f(y)| \le M|x - y|\).
the Mean Value Theorem applied to \(f\) on \([x, y] \subset (a, b]\) gives \(f(x) - f(y) = f'(\xi)(x - y)\) for some \(\xi \in (x, y)\), and \(|f'(\xi)| \le M\).
3. \(f\) satisfies the Cauchy criterion at \(a^+\): for every \(\eps > 0\) there is \(\delta > 0\) such that \(a < x, y < a + \delta\) implies \(|f(x) - f(y)| < \eps\).
by 2, \(|f(x) - f(y)| \le M|x - y| < M\delta\); choose \(\delta = \eps/M\).
4. \(\lim_{x \to a^+} f(x)\) exists.
the Cauchy criterion 3 is equivalent to the existence of the limit (pick a sequence \(x_n \to a^+\); 3 makes \((f(x_n))\) Cauchy in the complete space \(\RR\), so it converges, and the limit is independent of the sequence by 3 again).