[The Algebraic Limit Theorem] Let \(\lim a_n = a\) and \(\lim b_n = b\). Then,
- \(\lim (ca_n) = ca\) for all \(c \in \mathbb{R}\);
- \(\lim (a_n + b_n) = a + b\);
- \(\lim (a_nb_n) = ab\);
- \(\lim (a_n/b_n) = a\);
Previous Definitions:
- Absolute value definition and other properties
- Sequence definitions and convergence definitions
- Subsequences
- Show the limit template
Problem Discussion
For (i) we want to show that \(\lim (ca_n) = ca\). So for any \(\epsilon > 0\), we want to find \(N\) such that
$$
\begin{align*}
\big\lvert ca_n - ca \big\rvert < \epsilon
\end{align*}
$$
Solving for \(n\):
$$
\begin{align*}
\big\lvert ca_n - ca \big\rvert &< \epsilon \\
\big\lvert c \big\rvert \big\lvert a_n - a \big\rvert &< \epsilon \\
\big\lvert a_n - a \big\rvert &< \frac{\epsilon}{|c|}.
\end{align*}
$$
So now we are ready to write a formal proof.
Formal Proof
Let \(\epsilon > 0\) be arbitrary. If \(c = 0\), then the result follows trivially since \(ca_n = ca = 0\) for all \(n\). So we may assume \(c \neq 0\). Now, since \(\lim a_n = a\), then by the definition of convergence, there exists a natural number \(N\) such that for all \(n \geq N\), we must have
$$
\begin{align*}
|a_n - a| &< \frac{\epsilon}{|c|}.
\end{align*}
$$
We now verify that this choice has the desired property. Let \(n \geq N\). Then,
$$
\begin{align*}
\big\lvert ca_n - ca \big\rvert = \big\lvert c \big\rvert \big\lvert a_n - a \big\rvert < \big\lvert c \big\rvert \frac{\epsilon}{|c|} = \epsilon. \\
\end{align*}
$$
Thus,
$$
\begin{align*}
\big\lvert ca_n - ca \big\rvert < \epsilon. \\
\end{align*}
$$
as required. \(\blacksquare\)