Discontinuity of $\sin(\frac{1}{x})$ I’ve heard and read in many books that the function 
$$\sin\left(\frac{1}{x}\right)$$
is discontinuous at $x=0$ since as $x$ tends to zero the function ‘oscillates’ rapidly that is , for numbers very close to each other the number takes valued such as $-1$ and $1$ hence we cannot define a limit. But I’ve also read that the continuity of a function is defined only over its domain. Then why do we define the continuity of $\sin(\frac{1}{x})$ if $x=0$ does not lie in its domain ?
 A: Actually, what you may have heard is that the function:
$$f(x)=\left\{\begin{array}\sin\left(\frac{1}{x}\right) & x\neq0\\
0 & x=0\end{array}\right.$$
is not continuous at $x=0$. Moreover, this function is not continuous for every number $a$ one may replace $f(0)$ with.
Even if it may be inaccurate, many are used to name the above function just as $\sin\frac{1}{x}$ without referring to $f(0)$. It is more accurate to say the following:

The function $f(x)=\sin\frac{1}{x}$, $x\neq0$ cannot be extended to a continuous function on $\mathbb{R}$

The above means that you cannot find a number $a\in\mathbb{R}$ such that letting $\tilde{f}(0)=a$ would make the function:
$$\tilde{f}(x)=\left\{\begin{array}\sin\left(\frac{1}{x}\right) & x\neq0\\
a & x=0\end{array}\right.$$
be continuous on $\mathbb{R}$.
A: This function is indeed continuous within its domain, and in some contexts it is reasonable to say simply that it is continuous and leave it at that.
However, the number $0$ is a limit point of the domain, and the question therefore arises, whether it is possible to extend this function to a function that is continuous at that limit point. If it were not a limit point of the domain, then it would be vacuously true that it can be so extended.
The function $x\mapsto\dfrac {\sin x} x $ is undefined at $0,$ except that in some contexts it is taken to have the value $1$ at $0$ because that is the only way to extend it to a continuous function at that limit point of its domain. (And that extended continuous function is not only continuous, but very well behaved, in that it is an entire function.)
A: The phrase "(dis)continuous at $x$" is used in different ways in calculus books and in mathematics. This is maybe not surprising, because in fact the word "function" itself is used differently! In a typical  calculus book a function is not a set of ordered pairs, a function is maybe defined as something specified by some "rule", and then in practice "rule" seems to mean "formula".
This may or may not be a bad thing; students have enough trouble with the more naive notion of "function". And of course history is on the side of the calculus books - a "function" for Euler and Fourier was certainly not a set of ordered pairs.
It gets worse. I've seen chapters on the Laplace transform in differential equations texts where the function $f$ defined by $$f(t)=\begin{cases}1,&(t<1),
\\t,&(t\ge1).\end{cases}$$is called a "discontinuous" function, because of the discontinuity in the formula defining $f$.
A: We say that a function $f(x)$ is discontinuos at an isolated point $x=a$ if


*

*$\lim_{x\to a} f(x)$ exists but $\lim_{x\to a} f(x)\neq f(a)$ since f(a) is different by the limit or since $f$ is not defined in $x=a$, in this case we define that a removable discontinuity

*$\lim_{x\to a^+} f(x)\neq \lim_{x\to a^-} f(x)$, in this case we refer to a jump discontinuity

*$\lim_{x\to a} f(x)=\pm \infty$, one or both side, in this case we refer to a infinite discontinuity
In all the other cases we define the discontinuity as an essential discontinuity that is exactly the case for $\sin\left(\frac{1}{x}\right)$ since the limit $x\to 0$ doesn't exist at all.
