The Algebraic Limit Theorem
Let \(\lim x_n = x\) and \(\lim x_n = x\). Then, \(\lim (x_ny_n) = xy\).

Strategy

We want to find \(N \in \mathbb{N}\) such that for any \(\epsilon > 0\)

$$ \begin{align*} \lvert x_ny_n - xy \rvert < \epsilon. \end{align*} $$

The trick here to add and subtract a term \(yx_n\) and then use the triangle inequality

$$ \begin{align*} \lvert x_ny_n - xy \rvert &= \lvert x_ny_n - x_ny + x_ny - xy \rvert \\ &\leq \lvert x_ny_n - yx_n \rvert + \lvert x_ny - xy \rvert \\ &= |x_n| \lvert y_n - y \rvert + |y|\lvert x_n - x \rvert. \end{align*} $$

We can make \(\lvert a_n - a \rvert\) be as small as we want and similarly we can make \(\lvert y_n - y \rvert\) 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 \(\lim y_n = y\), then we know that there exists an \(N_1 \in \mathbb{N}\) such that for \(n \geq N_1\),

$$ \begin{align*} \lvert x_n - x \rvert \leq \frac{\epsilon}{2} \frac{1}{|y|}. \\ \end{align*} $$

Similarly, given that \(\lim y_n = y\), then we know that there exists an \(N_2 \in \mathbb{N}\) such that for \(n \geq N_2\),

$$ \begin{align*} \lvert y_n - y \rvert \leq \frac{\epsilon}{2} \frac{1}{|x_n|}. \end{align*} $$

We still need to bound the term \(|x_n|\) since it’s not a constant. But we know since it converges, then it’s bounded and there is an \(M > 0\) such that every term \(|x_n| \leq M\). Therefore, let

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

If we plug in everything back, we get

$$ \begin{align*} \lvert x_ny_n - xy \rvert &\leq |x_n| \lvert y_n - y \rvert + |y|\lvert x_n - x \rvert \\ &< M\big(\frac{1}{M}\frac{\epsilon}{2}\big) + |y|\big(\frac{1}{|y|}\frac{\epsilon}{2}\big). \\ &= \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon. \\ \end{align*} $$

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 given. Since \((x_n)_n\) is convergent, then we know it’s bounded. Therefore, there exists an \(M\) such that \(|x_n| \leq M\) (if \(M = 0\), then let \(M = 1\)). Moreover, since \(\lim x_n = x\), then there is an \(N \in \mathbb{N}\) such that for all \(n \geq N_1\),

$$ \begin{align*} \lvert x_n - x \rvert < \frac{\epsilon}{2|y|} \end{align*} $$

Similarly, since \(\lim y_n = y\), there exists an \(N_2 \in \mathbb{N}\) such that for all \(n \geq N_2\),

$$ \begin{align*} \lvert y_n - y \rvert < \frac{\epsilon}{2|M|} \end{align*} $$

Therefore, choose a natural number \(N\) such that \(N = \max(N_1,N_2)\). Assume \(n \geq N\). Then,

$$ \begin{align*} \lvert x_ny_n - xy \rvert &= \lvert x_ny_n - x_ny + x_ny - xy \rvert \\ &\leq \lvert x_ny_n - yx_n \rvert + \lvert x_ny - xy \rvert \\ &= |x_n| \lvert y_n - y \rvert + |y|\lvert x_n - x \rvert \\ &< M\big(\frac{1}{M}\frac{\epsilon}{2}\big) + |y|\big(\frac{1}{|y|}\frac{\epsilon}{2}\big). \\ &= \frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon. \\ \end{align*} $$

Thus, since \(\lvert x_ny_n - xy \rvert < \epsilon\) for all \(n \geq N\), then \(\lim (x_ny_n) = xy\) as we wanted to show. \(\ \blacksquare\)

References