Given the function $ f(n) = \begin{cases} \sin (1/x), & x \neq 0 \\ 0, & x=0 \end{cases} $
describe the interval on which the function is continuous.
I know that the function is continuous at all real values except $x=0$, because $y=1/x$ is continuous at all points except for $x=0$, and $\sin (1/x)$ is a composite function.
But what about the bottom half of the function? It states that there exists a point at $(0,0)$, so why wouldn't it be continuous there as well?
Note that $\lim_{x\to 0} \sin \frac{1}{x} $ doesn't exist, indeed for
$x_n=\frac1{2\pi n}\to 0\implies \sin\frac1{x_n}=\sin 2\pi n=0$
$x_n=\frac2{\pi (4n-3)}\to 0\implies \sin\frac1{x_n}=\sin \frac{\pi (4n-3)}2 =1$
then recall the definition of continuity.