The Algebraic Limit Theorem
Let \(\lim\limits_{n \to \infty} x_n = x\) and \(\lim\limits_{n \to \infty} y_n = y\). Then, \(\lim\limits_{n \to \infty} \frac{x_n}{y_n} = \frac{x}{y}\) if \(y \neq 0\).

Strategy

Let \(\epsilon > 0\) be given. First we need the following lemma to establish a lower bound on \(y_n\)

Lemma
If \(\lim\limits_{n \to \infty} y_n = y\) and \(y \neq 0\), then there exists an \(N \in \mathbb{N}\) such that for all \(n \geq N\), we have \(|y_n| > \frac{|y|}{2}\).

Recall the reverse triangle inequality, for any \(a,b \in \mathbb{R}\), we have

$$ \begin{align*} |a| - |b| &\leq |a - b| \\ |b| - |a| &\leq |a - b| \end{align*} $$

Now, since we’re given that \(\lim\limits_{n \to \infty} y_n = y\), then choose \(\epsilon = \frac{|y|}{2}\). There exists an \(N \in \mathbb{N}\) such that for all \(n \geq N\)

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

Using the reverse inequality, then

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

Therefore,

$$ \begin{align*} |y| - |y_n| < \frac{|y|}{2} \\ \frac{|y|}{2} < |y_n| \\ \end{align*} $$

Formal Proof

Let \(\epsilon > 0\) be given. Since \(\lim x_n = x\), then there is an \(N_1 \in \mathbb{N}\) such that for all \(n \geq N_1\),

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

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

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

Moreover, by the lemma, there exists an \(N_3 \in \mathbb{N}\) such that for all \(n \geq N_3\),

$$ \begin{align*} \frac{|y|}{2} < |y_n| \end{align*} $$

Therefore, choose a natural number \(N\) such that \(N = \max(N_1,N_2,N_3)\). Assume \(n \geq N\). Then,

$$ \begin{align*} \left\lvert \frac{x_n}{y_n} - \frac{x}{y} \right\rvert &= \left\lvert \frac{x_ny - y_nx}{y_ny} \right\rvert \\ &= \left\lvert \frac{x_ny -xy + xy - y_nx}{y_ny} \right\rvert \\ &= \left\lvert \frac{(x_ny -xy) + (xy - y_nx)}{y_ny} \right\rvert \\ &\leq \frac{| (x_ny -xy) | + | (xy - y_nx) | }{|y_ny|} \quad \text{(Triangle Inequality)} \\ &\leq \frac{|y||(x_n -x) | + |x||(y - y_n) | }{|y_n||y|} \\ &< \frac{|y||(x_n -x) | + |x||(y - y_n) | }{\frac{|y|}{2}|y|} \quad \text{(since $|y_n|>|y|/2$)} \\ &< \frac{|y| \frac{\epsilon}{|y|} + |x| \frac{\epsilon}{|x|} }{2}\\ &< \frac{2\epsilon }{2} = \epsilon\\ \end{align*} $$

Thus, since \(\lvert \frac{x_n}{y_n} - \frac{x}{y} \rvert < \epsilon\) for all \(n \geq N\), then \(\lim \frac{x_n}{y_n} = \frac{x}{y}\) as we wanted to show. \(\ \blacksquare\)


References