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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*}


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

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.

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