KindiakMath

The Limit of Quotients

Throughout this post, let \mathbb K be an ordered field.

Problem 1. Let \{x_n\} be a \mathbb K-sequence. Suppose x_n \to x, y_n \to y for some x,y \in \mathbb K, and y_n \neq 0 for any n. If y \neq 0, prove that x_n/y_n \to x/y.

Proof. Suppose we have proven that y_n \to 1 implies 1/y_n \to 1. Then y_n \to y \neq 0 implies

\displaystyle \frac 1{y_n} = \frac 1y \cdot \frac{1}{\frac 1y \cdot y_n} \to \frac 1y \cdot \frac 1{\frac 1y \cdot y} = \frac 1y,

where all the limits are defined. Then if x_n \to x,

\displaystyle \frac{x_n}{y_n} = x_n \cdot \frac{1}{y_n} \to x \cdot \frac 1y = \frac xy.

Thus, we prove the special case y_n \to 1 \Rightarrow 1/y_n \to 1. Fix \epsilon > 0. Since y_n \to 1, for any k > 0, there exists N(k) \in \mathbb N such that

n > N(k) \quad \Rightarrow \quad |y_n - 1| < k \epsilon \quad \Rightarrow \quad y_n > 1 - k\epsilon.

If we selected k such that 1-k\epsilon > 0, then y_n > 0 \Rightarrow |y_n| > 0 implies

\displaystyle n > N(k) \quad \Rightarrow \quad |y_n - 1| < k\epsilon \quad \Rightarrow \quad \frac{1}{|y_n|} < \frac{1}{1-k\epsilon}.

Select k_1 = 1/2, k_2 = 1/(2\epsilon) so that 1-k_2\epsilon=1/2. By calculations, for n > \max\{N(k_1),N(k_2)\},

\begin{aligned} \left| \frac{1}{y_n} - 1 \right| &= \frac{|1 - y_n|}{|y_n|} = \frac{|y_n - 1|}{|y_n|} < \frac{k_1 \epsilon}{1 - k_2 \epsilon} = \frac{\epsilon/2}{1 / 2} = \epsilon,\end{aligned}

as required.

Problem 2. Let \{x_n\} be a \mathbb K-sequence. Suppose x_n \to 0. Prove that for any M > 0, there exists N \in \mathbb N such that

n > N\quad \Rightarrow \quad |1/x_n| > M.

In fact this is what we mean by an infinite limit.

Proof. Fix M > 0 so that 1/M > 0. Since x \to 0, there exists N \in \mathbb N such that

n> N \quad \Rightarrow \quad |x_n| < 1/M \quad \Rightarrow \quad |1/x_n| =1/|x_n| > M.

Definition 1. Let \{x_n\} be a \mathbb K-sequence. We write x_n\to \infty to mean

\forall M > 0 \quad \exists N \in \mathbb N:\quad n> N\quad \Rightarrow \quad x_n > M.

Similarly, we write x_n\to -\infty to mean

\forall M > 0 \quad \exists N \in \mathbb N:\quad n> N\quad \Rightarrow \quad x_n < -M.

Corollary 1. Let \{x_n\} be a \mathbb K-sequence. Then x_n \to 0 \iff 1/|x_n| \to \infty.

—Joel Kindiak, 19 Dec 24, 2153H

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