What type of discontinuity is $\sin(1/x)$? For those of you familiar with the graph of $\sin(\dfrac{1}{x})$, things get quite 'intense' as $x \to 0$.
Is this a removable discontinuity or an infinite discontinuity or even discontinuous at all?
Lastly, is it differentiable at $x = 0$?
 A: First, the function $$f(x) = \sin\left(\frac{1}{x}\right)$$ isn't defined at $0$. So it is not continuous. For a function to be differentiable it needs to be continuous, so it also isn't differentiable.
Now besides not being defined at $0$, the limit:
$$
\lim_{x\to 0} f(x)
$$
also doesn't exist. So it isn't a removable discontinuity. One might call this an essential discontinuity. As defined in the Wikipedia article, one might also call this an infinite discontinuity (but in my opinion one should keep that term for then one of the sided limits is plus or minus infinity).
A: The function $\sin{\frac{1}{x}}$ Does not attain a limit as $x\rightarrow 0$. 
Let $f(x)=\sin{\frac{1}{x}}$
$f(x)$ attains a limit $l$ then there exists a $\delta>0$ for every $\epsilon >0$ such that 
$\left|f(x)-l\right|<\epsilon$ such that $0<\left|x-0\right|<\delta$.
Considering it from the right we have $0<x<\delta$ or $\infty<\frac{1}{x}<\frac{1}{\delta}$
. Which means There are infinite number of $\frac{1}{x}$ of the form $2n\pi$. Which means the function is oscillating in the interval and  limit does not  exist
A: I don't know what "infinite discontinuity" is, but $\,x=0\,$ is a discontinuity point of the second, and strongest, kind: the one sided limits of the function there don't even exist in the generalized form:
$$x_n:=\frac{1}{2n\pi}\;,\;n\in\Bbb N\Longrightarrow \lim_{n\to\infty}\sin\frac{1}{x_n}=\lim_{n\to\infty}\sin 2n\pi =0$$
and now repeat the above with
$$x_n:=\frac{2}{(4n-1)\pi}\;,\;\;x_n:=\frac{2}{(4n+1)\pi}\,,\,\,n\in\Bbb N\,,\ldots$$
