# I want to make sure my proof is correct using $\epsilon, \delta$ definition of continuity for $\sin\left(\frac{1}{x}\right)$

So, let's consider the function

$$f(x)=\begin{cases} \sin\left(\frac{1}{x}\right) & x\neq 0 \\ 0 & x=0 \end{cases}$$

I want to prove the above function is discontinuous at $x=0$.

Let $\epsilon=\frac{1}{2}$. And let's consider $x_n=\frac{2}{(4n+1)\pi}$. Suppose $\frac{2}{(4n+1)\pi} < \delta$ for $\delta>0$. If the function were continuous we should have $|f(x_n)-0|<\epsilon$. But clearly $f(x_n)=1$ for all n, then we have $|f(x_n)|>\epsilon=\frac{1}{2}$. Thus, the function isn't continous at x=0. Is it correct?

• Yep, that looks good. You'll want to make note that $x_n \rightarrow 0$ so as to be able to find an $n$ with $x_n$ less than an arbitrarily chosen $\delta > 0$. There are other ways to do this as well; for example if we restrict the domain to a compact set like $[0,1]$, then the image should be compact if the function were continuous, but $f([0,1])$ isn't even closed. However, really any other method requires basically as much work. – Kaj Hansen Oct 10 '17 at 19:44

The idea of the proof is correct, but a little bit of wording might be needed. You can't say "Suppose $\frac{2}{(4n+1)\pi} < \delta$, as maybe there exists $\delta \le \frac{2}{(4n+1)\pi}$, which is suitable for our $\epsilon$. Anyway the best way to do it is by contradiction.
Let $\epsilon = \frac 12$ and assume the function is continuous at $x=0$. Then there exists a suitable $\delta > 0$ s.t. $|x| < \delta \implies |f(x)| < \epsilon$. Now using the Archimedian Principle we have that there is an integer $n$, s.t. $n > \frac{1}{2\delta \pi} - \frac 14$. Hence we have that $\frac{2}{(4n+1)\pi} < \delta$ and so we must have: $\left|f\left(\frac{2}{(4n+1)\pi}\right)\right| \le \frac 12$, but this is wrong as shown. Hence $f$ is discontinuous at $0$.