Prove that sin (1/x) has no limit as x approaches 0

Define the function:

$f(x) = \sin \frac{1}{x} \qquad \text{for } x \neq 0.$

Prove there is no $A \in \mathbb{R}$ such that

$\lim_{x \to 0} f(x) = A.$

Proof. We prove this by contradiction. Suppose there does exist some $A \in \mathbb{R}$ such that $\lim_{x \to 0} f(x) = A$ . From the definition of limit this means that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that

$| f(x) - A | < \varepsilon \qquad \text{whenever} \qquad 0 < |x| < \delta.$

First, we claim that for any such $A$ we must have $|A| \leq 1$ . This must be the case since if $|A| > 1$ then $|A| - 1 > 0$ , so we may choose $\varepsilon$ such that $0 < \varepsilon < (|A|-1)$ . But then, since $|f(x)| \leq 1$ for all $x$ , we have

$|f(x) - A| = |A - f(x)| \geq |A| - |f(x)| \geq |A| - 1 > \varepsilon.$

(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality $|A - f(x) \geq |A| - |f(x)|$ .) This contradicts our choice of $\varepsilon$ , so $|A|$ must be less than or equal to 1.
Next, suppose $|A| \leq 1$ . Then choose $\varepsilon = \frac{1}{2} > 0$ . To obtain our contradiction we must show that there is no $\delta > 0$ such that

$|f(x) - A| < \frac{1}{2}, \qquad \text{whenever} \qquad 0 < |x| < \delta.$

By the Archimedean property of the real numbers we know that for any $x \in \mathbb{R}$ , there exists a positive integer $n$ with $n \equiv 1 \pmod{4}$ such that $0 < \frac{2}{n \pi} < |x|$ . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being $1 \pmod{4}$ .) But,

$0 < \frac{2}{n \pi} < |x| \quad \implies \quad 0 < \frac{2}{(n+2) \pi} < |x|.$

Then, from the definition of $f$ and since $n \equiv 1 \pmod{4}$ and $n+2 \equiv 3 \pmod{4}$ we have,

$\begin{align*} f \left( \frac{2}{n \pi} \right) &= \sin \left( \frac{n \pi}{2} \right) = 1 \\ f \left( \frac{2}{(n+2) \pi} \right) &= \sin \left( \frac{(n+2)\pi}{2} \right) = -1. \end{align*}$

But then,

$\begin{align*} \left| f \left( \frac{n \pi}{2} \right) - A \right| < \frac{1}{2} && \implies && |1 - A| &< \frac{1}{2} \\ && \implies && \frac{1}{2} &< A \leq 1 &(\text{from above } A \leq 1). \end{align*}$

Furthermore,

$\left| f \left( \frac{(n+2) \pi}{2} \right) - A \right| = |-1-A| > \frac{1}{2}.$

This contradicts that if $\lim_{x \to 0} f(x) = A$ then for every $\varepsilon > 0$ we have $|f(x) - A| < \varepsilon$ for all $0 < |x| < \delta$ . In other words, we found an $\varepsilon$ greater than 0 (in particular, we found $\varepsilon = \frac{1}{2}$ ) such that no matter how small we choose $\delta$ there exists an $x$ such that $|x|$ is smaller than $\delta$ , but $|f(x) - A|$ is bigger than $\varepsilon$ . This exactly contradicts the definition of $\lim_{x \to 0} f(x) = A$ . Hence, there can be no such number $A \in \mathbb{R}. \qquad \blacksquare$

This proof shows that there is now way to define $f(0)$ so that $f$ is continuous at 0. We see this since for $f$ to be continuous at 0 we must have $\lim_{x \to 0} f(x) = f(0)$ . But if we try to define $f(0) = A$ for any $A \in \mathbb{R}$ , we will fail to achieve continuity since the proof shows $\lim_{x \to 0} f(x) \neq A$ for any $A$ we might want to choose.

Stumbling Robot

Prove that sin (1/x) has no limit as x approaches 0

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