Let \(f:\mathbb{R}\to\mathbb{R}\) be a differentiable function with \(f'\in C(\mathbb{R})\). Assume that there are \(a,b\in\mathbb{R}\) with \(\lim_{x\to\infty}f(x)=a\) and \(\lim_{x\to\infty}f'(x)=b\). Prove that \(b=0\). Then, find a counterexample to show that the assumption \(\lim_{x\to\infty}f'(x)\) exists is necessary.
Solution
1. \(b = 0\).
by the mean value theorem applied to \(f\) on \([x, x+1]\), for each \(x\) there is \(\xi_x \in (x, x+1)\) with \[ f(x+1) - f(x) = f'(\xi_x) . \] As \(x \to \infty\), the left side tends to \(a - a = 0\) (both \(f(x+1)\) and \(f(x)\) tend to \(a\)). Also \(\xi_x \to \infty\), so \(f'(\xi_x) \to b\) (since \(\lim_{x\to\infty} f'(x) = b\)). Hence \(b = 0\).
2. The assumption that \(\lim_{x\to\infty} f'(x)\) exists is necessary.
\(f(x) = \sin(x^2)/x\) tends to \(0\) as \(x \to \infty\), but \[ f'(x) = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}, \] whose limit as \(x \to \infty\) does not exist (the term \(2\cos(x^2)\) oscillates between \(-2\) and \(2\) while the second term tends to \(0\)). So \(f\) is differentiable, \(\lim f = 0\), yet \(\lim f'\) fails to exist.
3. Q.E.D.