Is my proof correct about limit of $\sin\left(\frac{1}{x}\right)$?

Question

Apostol's book Calculus asks to show that there is not a value $A$ such that $f(x)=\sin\left(\frac{1}{x}\right)\to A$ when $x \to 0$. And my proof is:

Suppose for the sake of contradiction that there exists such $A$. If we take $\epsilon=1$, we have two cases:

1. If $A>0$, then $A - \epsilon > -1$, so $-1$ does not satisfy $|-1-A|<\epsilon$ inequality. Now for given $\delta>0$ we can take a natural number $n$ such that $\frac{2-3\pi\delta}{4\pi\delta}<n $ and then $0<\frac{1}{\pi\left( \frac{3}{2}+2n \right)}<\delta$. But $$f\left( \left( \frac{3}{2}\pi+2n\pi \right)^{-1} \right)=\sin \pi\left( \frac{3}{2}+2n \right)=-1. $$

2. If $A\le0$, then $A+\epsilon\le1$ so $1$ does not satisfy $|1-A|<\epsilon $ inequality. Now for given $\delta>0$ we can take a natural number $n$ such that $\frac{2-\pi\delta}{4\pi\delta}<n $ and then $0<\frac{1}{\pi\left( \frac{1}{2}+2n \right)}<\delta$. But $$f\left( \left( \frac{1}{2}\pi+2n\pi \right)^{-1} \right)=\sin \pi\left( \frac{1}{2}+2n \right)=1.$$

We have proved that with $\epsilon=1$ for every $\delta>0$ there exist $x$ such that $0<x<\delta$ but $|\sin \frac{1}{x} - A|<\epsilon$ does not hold. Therefore we have a contradiction, so such $A$ does not exist.

Is my proof correct? there is a shorter argument?. Thanks in advance.

Answer 1

You are, in essence, showing there exists sequences $y_n,x_n$ such that $\lim\limits_{n\to\infty} y_n=\lim\limits_{n\to\infty} x_n=0$ yet $$\lim\limits_{n\to\infty} \sin y_n^{-1}\neq \lim\limits_{n\to\infty}\sin x_n^{-1}$$

This can be generalized:

PROP A function $f:(a,b)\to\Bbb R$ has limit $\ell$ as $x\to \xi\in (a,b)$ if and only if for every sequence $\langle\, x_n:n\in\Bbb N\,\rangle $ of numbers in $(a,b)$ such that $x_n\neq\xi$ and $x_n\to\xi$, it follows that $\lim\limits_{n\to\infty} f(x_n)=\ell$.

Answer 2

Here is something easier: as x → 0 sin(1/x) takes on the value 1 infinitely many times (when x = 2/mπ -- m = 1,5,9,...) and 0 infinitely many times (when x = 1/nπ -- n any integer ). So you can find a subsequence of values of sin(1/x) whose limit is 1; and another subsequence whose limit is 0.

For the limit to exist, all subsequences must converge to the same number, and that is clearly not the case here.