[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

We’ll follow a similar approach to proving (i) and (ii) in here and here For (iii) we want to show that $\lim (a_nb_n) = ab$. So we want to find $N$ such that for any $\epsilon > 0$,

$$ \begin{align*} \lvert a_nb_n - ab \rvert < \epsilon. \end{align*} $$

The trick here to add and subtract a term $ab_n$ and then use the triangle inequality:

$$ \begin{align*} \lvert a_nb_n - ab \rvert &= \lvert a_nb_n + ab_n - ab_n - ab \rvert \\ &\leq \lvert a_nb_n - ab_n \rvert + \lvert ab_n - ab \rvert \\ &= |b_n| \lvert a_n - a \rvert + |a|\lvert b_n - b \rvert. \end{align*} $$

Using the same trick from (ii), we can make $\lvert a_n - a \rvert$ be as small as we want and similarly we can make $\lvert b_n - b \rvert$ be as small as we want. The idea is to bound these two terms to values such that the total will be exactly $\epsilon$ which is what we want to prove. So, given that $\lim b_n = b$, then we know that for some $n \geq N_1$,

$$ \begin{align*} |a|\lvert b_n - b \rvert \leq \frac{\epsilon}{2}. \\ \lvert b_n - b \rvert \leq \frac{\epsilon}{2} \frac{1}{|a|}. \\ \end{align*} $$

Similarly, given that $\lim b_n = b$, then we know that for some $n \geq N_2$,

$$ \begin{align*} |b_n|\lvert a_n - a \rvert \leq \frac{\epsilon}{2} \\ \lvert a_n - a \rvert \leq \frac{\epsilon}{2} \frac{1}{|b_n|}. \end{align*} $$

We still need to bound the term $|b_n|$ since it’s not a constant. One trick is to use the convergent sequences theorem which says that all convergent sequences are bounded by some term $M$. This means that we can replace the bound above with the following:

$$ \begin{align*} \lvert a_n - a \rvert < \frac{\epsilon}{2} \frac{1}{M}. \end{align*} $$

If we plug in everything back, we get

$$ \begin{align*} \lvert a_nb_n - ab \rvert &= \lvert a_nb_n + ab_n - ab_n - ab \rvert \\ &= \lvert a(b_n - b) + b_n(a_n - a) \rvert \\ &\leq \lvert a(b_n - b) \rvert + \lvert b_n(a_n - a) \rvert \quad \quad \text{(Triangle Inequality)}\\ &= |a|\lvert b_n - b \rvert + |b_n|\lvert a_n - a \rvert \quad \quad \text{($|ab| = |a||b|$)} \\ &< |a|\big(\frac{1}{|a|}\frac{\epsilon}{2}\big) + M\big(\frac{1}{M}\frac{\epsilon}{2}\big). \\ &= \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon. \\ \end{align*} $$

Finally we can choose $N$ to be the maximum of $N_1$ and $N_2$ to guarantee the above bounds.

Formal Proof

Let $\epsilon > 0$ be arbitrary. We are given that $\lim a_n = a$. This means that for some $n \geq N_1$ and for any choice of $\epsilon$ , the following holds:

$$ \begin{align*} \lvert a_n - a \rvert \leq \epsilon. \end{align*} $$

Since this holds for any $\epsilon$, let $\epsilon = \big(\frac{1}{|a|}\frac{\epsilon}{2}\big)$ so,

$$ \begin{align*} \lvert a_n - a \rvert \leq \frac{1}{|a|}\frac{\epsilon}{2}. \end{align*} $$

Similarly, given that $\lim b_n = b$. This means that for some $n \geq N_2$, the following holds:

$$ \begin{align*} \lvert a_n - a \rvert \leq \frac{1}{M}\frac{\epsilon}{2}. \end{align*} $$

$M$ is an upper bound on the sequence $(b_n)$ guaranteed by convergent sequences theorem. Therefore, choose a natural number $N$ satisfying $\max(N_1,N_2)$. We now verify that this choice has the desired property. Let $n \geq N$. Then,

as we wanted to show. $\blacksquare$

References:

Understanding Analysis by Stephen Abbott