Suppose \(f\) and \(g\) are functions mapping \(\mathbb{R}\) into itself with \(\lim_{x\to 0} f(x) = 0\).
Prove from the \(\varepsilon\)–\(\delta\) definition that if \(g\) is bounded, then \(\lim_{x\to 0} f(x)g(x) = 0\) as well.
Give an example to show that the boundedness hypothesis cannot be omitted from part (a).
Solution
Suppose \(\exists M_g \suchthat \forall x,~ g(x) < M\). Then let \(\varepsilon > 0\) be arbitrarily chosen; we want to show that there exists a \(\delta\) such that \(\abs{x} \leq \delta \implies \abs{f(x)g(x)} \leq \varepsilon\). Since \(\lim_{x\to 0} f(x) = 0\), choose a \(\delta_f\) such that \(\abs{x} \leq \delta_f \implies \abs{f(x)} \leq \frac{\varepsilon}{M_g}\). So letting \(\delta = \delta_f\), we have \[ \abs{x} \leq \delta \implies \abs{f(x)g(x)} = \abs{f(x)} \abs{g(x)} \leq {\frac{\varepsilon}{M_g}}\abs{g(x)} \leq \frac{\varepsilon}{M_g}M_g = \varepsilon. \qed \] Let \(f(x) = x\) and \(g(x) = \frac{1}{x}\). Note that \(g(x)\) is unbounded in any neighborhood of 0, and \(f(x)g(x) = 1 \not\to 0\).