I have this function: $f(x) = \begin{cases} \sin \frac {1}{x} & x \ne 0\\ 0 & x = 0\end{cases}$
I have to establish if $f(x)$ is continuous or discontinuous somewhere and, in this last case, the kind of discontinuity.
$f(0)=0$ and $\not\exists \lim_{x\rightarrow 0} sin \frac {1}{x}$ $x=0$ is a point of discontinuity of third kind. Is it right?
I don't know what of the third kind means but I can show you it is discontinuous another way. A function is continuous if $\lim\limits_{n\to\infty}f(a_n) = f(\lim\limits_{n\to\infty}a_n)$ for convergent sequences $a_n$ in the domain of $f$.
Let's consider $b_n = \frac{1}{\pi n}$. It is obvious that the sequence converges to $0$ as $n\to\infty$. If $f$ was continuous then we would have,
$$0 =f(\lim\limits_{n\to\infty}b_n)=\lim\limits_{n\to\infty}f(b_n)$$
However, $\lim\limits_{n\to\infty} f(b_n)$ does not exist since it goes $0, 1, 0, -1, 0, 1, 0, -1,\ldots$