[The Algebraic Limit Theorem] Let \(\lim a_n = a\) and \(\lim b_n = b\). Then,
  1. \(\lim (ca_n) = ca\) for all \(c \in \mathbb{R}\);
  2. \(\lim (a_n + b_n) = a + b\);
  3. \(\lim (a_nb_n) = ab\);
  4. \(\lim (a_n/b_n) = a\);

Previous Definitions:


Problem Discussion

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 \(\lim a_n = a\), 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:

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

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. Since \(\lim a_n = a\) and \(\lim b_n = b\), both sequences are convergent and hence bounded by the convergent sequences theorem. Therefore, there exists a constant \(M\) such that \(|a_n| \leq M\) and \(|b_n| \leq M\) for all \(n\). Now, since \(\lim a_n = a\), there exists \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\),

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

where \(M\) is a constant such that \(|a_n| \leq M\) for all \(n\). Similarly, since \(\lim b_n = b\), there exists \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\),

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

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,

$$ \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{(\(|ab| = |a||b|\))} \\ &< M\big(\frac{\epsilon}{2M}\big) + |a|\big(\frac{\epsilon}{2|a|}\big). \\ &= \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon \\ \end{align*} $$

as we wanted to show. \(\blacksquare\)


References