-1
$\begingroup$

Because when we draw a graph there have no break point and breakthrough where we can notice that function jumped from his path or removed?? https://i.stack.imgur.com/XGkQ8.jpg

$\endgroup$
3
  • 2
    $\begingroup$ A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case). $\endgroup$ Aug 3, 2018 at 14:12
  • $\begingroup$ In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry. $\endgroup$
    – Arthur
    Aug 3, 2018 at 14:15
  • $\begingroup$ What is your definition of a removable discontinuity? $\endgroup$
    – user88319
    Aug 3, 2018 at 14:19

2 Answers 2

7
$\begingroup$

The function $f : \mathbb R\setminus \{0\}\to \mathbb R$ defined by the formula $f(x) = \frac{\sin x}{x}$ is not discontinuous at $x=0$, because it is not even defined there. In most texts you'll find out there, continuity is defined in such a way that functions cannot be continuous or discontinuous at points outside their domain.

However you know from a geometric argument (or Taylor series) that $$\lim_{x\to 0} \frac{\sin x}{x} = 1, $$ so you may define a continuous extension $ g : \mathbb R \to \mathbb R$ of your function, $$g(x) = \begin{cases} \dfrac{ \sin x }{x} & x\neq 0, \\ 1 & x=0\end{cases} $$ so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.

This you can do whenever a function is not defined at a point but has finite limit at that point.

Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)

In this sense you may say thay your function has a removable singularity at $x=0$ (instead of a discontinuity).

$\endgroup$
14
  • 1
    $\begingroup$ Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point. $\endgroup$
    – Mark
    Aug 3, 2018 at 14:27
  • $\begingroup$ @Mark See my addendum. $\endgroup$
    – giobrach
    Aug 3, 2018 at 14:29
  • 1
    $\begingroup$ That's fine. But for me it's still a point at which the function is not continuous. $\endgroup$
    – Mark
    Aug 3, 2018 at 14:40
  • 1
    $\begingroup$ If you really want to do things properly you shouldn't start with "The function $f(x)=\frac{\sin x}{x}$..." because $f(x)=\frac{\sin x}{x}$ is not a function. $\endgroup$ Aug 3, 2018 at 14:55
  • 1
    $\begingroup$ Thanks ! It helped a lot. giobrach $\endgroup$ Aug 3, 2018 at 14:58
0
$\begingroup$

Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let \begin{align*} g(x) :=\frac{f(x)}{x} \text{ for } x\neq 0. \end{align*} The following formula is a direct consequence of l'Hopital's rule: \begin{align*} \lim_{x\rightarrow 0} \frac{f(x)}{x}=f'(0) \end{align*}

Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.

This is directly applicable to $f(x)=\sin(x)$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .