4

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 ?

Michael Hardy
  • 1
Aditi
  • 1,389
  • Closely related: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark Mar 19 '18 at 19:51

4 Answers4

4

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 $\mathbb{R}$.

Vassilis Markos
  • 4,554
  • Oh ! So you mean that although we define continuity of a function on its domain , here we are trying to show that the function will not be continuous on points outside of its domain such as $x=0$ – Aditi Mar 19 '18 at 17:34
  • 1
    In a way. We mean that there is no function $g$ defined on $\mathbb{R}$ such that $g(x)=f(x)$ for every $x\neq0$ and $g$ is also continuous. – Vassilis Markos Mar 19 '18 at 19:24
2

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.)

Michael Hardy
  • 1
1

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 $f$.

David C. Ullrich
  • 92,839
0

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.

Michael Hardy
  • 1
user
  • 162,563
  • Yes you’re right ! But in all the cases as you mentioned , we define the discontinuity at points that lie within the domain of the function isn’t it ? Then why do we do the same for $x=0$ in $sin(\frac{1}{x})$? – Aditi Mar 19 '18 at 17:37
  • 1
    @Aditi For discontinuity It is also a matter of definition but note that not always in the cases we necessarly deal with points whithn the domain; for example we can say that $1/x$ is discontinuos at $x=0$ with an infinte discontinuity. And also in a jump discontinuity the jump point may lie or not whitin the domain. – user Mar 19 '18 at 17:45
  • Oh okay I get it now . Thank you for helping ! – Aditi Mar 19 '18 at 17:47
  • @Aditi You are welcome! Bye – user Mar 19 '18 at 17:50