How's g(x) = sin(x)/x removal discontinuity?

Question

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??

Answer 1

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 $f(x)=\sin(x)$.

Answer 2

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).