[4.2.4] Algebraic Limit Theorem for Functional Limits:
Let \(f\) and \(g\) be functions defined on a domain \(A \subseteq \mathbb{R}\), and assume \(\lim_{x \rightarrow c} f(x) = L\) and \(\lim_{x \rightarrow c} g(x) = M\) for some limit point \(c\) of \(A\). Then
Let \(f\) and \(g\) be functions defined on a domain \(A \subseteq \mathbb{R}\), and assume \(\lim_{x \rightarrow c} f(x) = L\) and \(\lim_{x \rightarrow c} g(x) = M\) for some limit point \(c\) of \(A\). Then
- \(\lim\limits_{x \to c} kf(x) = kL\) for all \(k \in \mathbb{R}\),
- \(\lim\limits_{x \to c} [f(x) + g(x)] = L + M\),
- \(\lim\limits_{x \to c} [f(x)g(x)] = LM\), and
- \(\lim\limits_{x \to c} f(x)/g(x) = L/M\), provided \(M \neq 0\).
Definition of functional limits
Proof
(a) Let \(\epsilon > 0\). We want to show that there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\), then
$$
\begin{align*}
|f(x) + g(x) - (L+M) | &< \epsilon
\end{align*}
$$
But since \(\lim\limits_{x \rightarrow c} f(x) = L\). This means that there exists a \(\delta_1 > 0\) such that whenever \(0 < |x - c| < \delta_1\), then
$$
\begin{align*}
|f(x) - L | &< \frac{\epsilon}{2}
\end{align*}
$$
Similarly since \(\lim\limits_{x \rightarrow c} g(x) = M\), there exists a \(\delta_2 > 0\) such that whenever \(0 < |x - c| < \delta_2\), then
$$
\begin{align*}
|g(x) - M | &< \frac{\epsilon}{2}
\end{align*}
$$
Now, set \(\delta = \min(\delta_1, \delta_2)\). Then for \(0 < |x - c| < \delta\), we must have
$$
\begin{align*}
|f(x) + g(x) - (L+M) | &= |(f(x) - L) + (g(x) - M) | \\
&\leq |(f(x) - L)| + |(g(x) - M) | \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \\
\end{align*}
$$
This implies that \(\lim\limits_{x \rightarrow c} f(x)+g(x) = L+M\) as we wanted to show. \(\blacksquare\)
References