Lemma (bounded)
If \(\lim\limits_{x \to a} f(x) = L\), then there exists an \(M > 0\) and \(\delta > 0\) such that
$$
\begin{align*}
|f(x)| \leq M \quad \text{ for all } | x - a| < \delta
\end{align*}
$$
Proof
Since \(\lim\limits_{x \to a} f(x) = L\), then when \(\epsilon = 1\), there exists a \(\delta > 0\) such that \(|x - a| < \delta\) implies
$$
\begin{align*}
|f(x) - L| &< 1 \\
|f(x)| - L \leq |f(x) - L| &< 1 \quad \text{(Reverse Triangle Inequality)} \\
|f(x)| &\leq L + 1.
\end{align*}
$$
Hence \(|f(x)| \leq 1 + L\) whenever \(|x - a| < \delta\). \(\blacksquare\)
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai