[The Algebraic Limit Theorem] Let $\lim a_n = a$ and $\lim b_n = a$. Then,

$\lim (ca_n) = ca$ for all $c \in \mathbf{R}$;
$\lim (a_n + b_n) = a + b$;
$\lim (a_nb_n) = ab$;
$\lim (a_n/b_n) = a$;

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Problem Discussion

Similar to the previous two examples in here and here, we’re going to use the template after working on finding the right $N$ value. For (i) we want to show that $\lim (ca_n) = ca$. So we want to find $N$ such that for any $\epsilon > 0$,

$$ \begin{align*} \big\lvert ca_n - ca \big\rvert < \epsilon. \end{align*} $$

always holds. 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. Choose a natural number $N$ satisfying

$$ \begin{align*} n &> \frac{\epsilon}{|c|}. \end{align*} $$

We now verify that this choice has the desired property. Let $n \geq N$. Then,

$$ \begin{align*} \big\lvert ca_n - ca \big\rvert = \big\lvert c \big\rvert \big\lvert a_n - a \big\rvert < \big\lvert c \big\rvert \frac{\epsilon}{|c|}. \\ \end{align*} $$

This means that

$$ \begin{align*} \big\lvert ca_n - ca \big\rvert < \epsilon. \\ \end{align*} $$

as required. $\blacksquare$

Notes

I’m not too comfortable yet with this proof. I don’t know why it was okay to keep $|a_n - a|$ as is?

References:

Understanding Analysis by Stephen Abbott