Proving that $\sin(1/x)$ is not continuous at 0 Let
$f(x) = \begin{cases}
0 &\text{ if $x=0$}\\
\sin(1/x) &\text{ otherwise}
\end{cases}$. Prove that $f$ is discontinuous at $0$
My proof goes like this: for the function to be continuous at 0, the following limit:
$\lim_{x\to 0}(\sin(1/x))$ needs to exist and be equal to 0.
Let $1/x=k$, I rewrite the limit expression as: $\lim_{k\to\infty}(\sin(k))$.
And since this limit oscillates, the limit does not exist. Therefore f(x) is not continuous at 0.
Am I correct?
 A: I think you could be more explicit. By writing for example, as $k \to \infty$,
$$
\frac1x=(4k+1)\frac \pi2 \implies \sin \left( (4k+1)\frac \pi2\right)=1\neq f(0)=0
$$ it is clearer.
A: Here a full other proof using the negation Cauchy-continuous function definition. As I am a student too i hope that it is correct and rigourous and that it will help other students.
Motivation
The negation of the Cauchy continuous function states that if a function $f(x)$ is not continuous at a point $x_0$ it exists at least one cauchy sequence $x'_n$ that converges to $x_0$ but the sequence $f(x'_n)$ is not a Cauchy sequence.
First let note that: $sin(a)=+-1$ if $a_n=\pi (\frac{1}{2}+n) \;, n \in \mathbb{N}$.
Why did I writte that? because in this case the sequence $y_n=sin(a_n)$ won't be obviously a cauchy sequence.
Prove
So because we want:$a_n=1/x_n$ hence we choose the following sequence $x_n=\frac{1}{\pi (\frac{1}{2}+n)}$. Obviously too $x_n$ is a Cauchy sequence as $x_n\underset{n\rightarrow \infty}{\rightarrow}0$ ($x_n$ converges).
We finally obtain: $y_n=f(x_n)=sin(\frac{1}{x_n})$ that is equal to $1$ for even $n$ and to $-1$ for odd $n$. So as we ve allreday explained above $y_n$ doesn't converge at all and especially not to $0$ as $y_n$ is NEVER equal to zero.
Conclusion
And this finish our prove that $sin(1/x)$ is not continuous at $0$. At is exists at least one sequence $x_n$ that converge to $0$ but with $f(x_n)$ that doesn't converge at all and especially in our case not to
$0$ as $y_n$ NEVER takes the value zero.
