On plotting graph for $\frac{\sin x}{x}$ using Wolfram|Alpha and Google, got that : plot of sin(x)/x

also, I can get the value of $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ using squeeze theorem and as illustrated on sources such as MIT.

But I'm not able to understand how the function is defined at $x=0$ and its value came out to be $1$ at that point. As promised in the plot for the function.

Using limits I got the idea about the behavior of the function in the neighborhood of that point but not its value exactly at $x=0$.

  • 6
    $\begingroup$ The function $f(x)=\frac{\sin(x)}{x}$ is not defined in $0$ and so it is not cont. in $0$. However the function can be extended to continuous one which is cont. in zero: $g(x)=\sin(x)/x$ if $x\neq 0$ and $g(x)=1$ if $x=0$ .. $\endgroup$
    – Albert
    Dec 26 '15 at 14:30
  • 4
    $\begingroup$ The function $f(x)=\frac{\sin x}{x}$ isn't defined at $x=0$, but it can be continuously extended to give the value $f(0)=1$. $\endgroup$
    – Harry Reed
    Dec 26 '15 at 14:31
  • 1
    $\begingroup$ we consider the behaviour of a function around a point $x=c$ and its possible value at that point differently. there is lots of explanations about this certain case on the web. $\endgroup$
    – Mikasa
    Dec 26 '15 at 14:33
  • $\begingroup$ Note that a limit does not mean the function exists at the given point. $\endgroup$ Dec 26 '15 at 15:26

A function $f(x)=\frac{\sin x}{x}$ cannot be defined for $x=0$ since division by $0$ is not defined in any numerical field. So this function is not continuous in $x=0$ but, since ( as noted in OP) $$ \lim_{x \to 0}\frac{\sin x}{x}=1 $$ the discontinuity is removable and the function defined as : $$ f: \mathbb{R}\to \mathbb{R} \quad f(x)= \begin {cases}\frac{\sin x}{x} &\text{for }x \ne 0 \\ 1 &\text{for }x=0\end {cases} $$

is continuous in $(-\infty, +\infty)$.

  • 1
    $\begingroup$ What makes this answer "better" than the others? It is interesting to note that there is not concrete explanation on how any of this came to be. $\endgroup$ Dec 26 '15 at 16:14
  • 1
    $\begingroup$ I would not say that $f$ is not continuous at $x=0$. Continuity is only defined for points of the domain. $\endgroup$ Dec 29 '15 at 6:38

The function is not defined at $x=0$. But it can be continuously extended (since the limit from the left is equal to the limit from the right).

So the graph is not technically correct, it just shows the extended function.


From the Maclaurin series expansion of $f(x)=\sin x$,we have that $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$

or, considering $x\not = 0,$ $$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots$$ And hence by using the notion of uniform convergence of a power series, it can be said that $$\lim_\limits{x\to 0} \frac{\sin x}{x} = 1$$

Thus at $x=0$, the function $f(x)=\frac{\sin x}{x}$ has R.H.L.=L.H.L.=$1 \not = f(0)$

So at $x=0$, $f(x)$ has removable discontinuity.

Hence to make the function continuous at $x=0$, $f(0)$ is defined to be $1$.

So the actual function should be like $$f(x)=\begin{cases}\frac{\sin x}{x} & x\not = 0 \\ \, \, \, 1 & x=0\end{cases}$$

For the given function, it can be remarked that it is discontinuous at $x=0$.

  • $\begingroup$ And an explanation for the downvote? $\endgroup$ Dec 26 '15 at 15:33
  • $\begingroup$ You should include the part where you could divide your expansion by $x$, as it is easy to do. I didn't down vote btw. $\endgroup$ Dec 26 '15 at 15:34
  • $\begingroup$ @SimpleArt I didn't get you.. which part are you talking of..? $\endgroup$ Dec 26 '15 at 15:35
  • $\begingroup$ You could divide the expansion by $x$ and substitute $x=0$ in. ;) $\endgroup$ Dec 26 '15 at 15:36
  • $\begingroup$ @SimpleArt But then you have $0$ in the denominator! Division by $0$ is undefined... I don't like it also ;-) $\endgroup$ Dec 26 '15 at 15:37

The graph is absolutely correct, it just misses a big empty circle representing the point of discontinuity. A point is an infinitely small entity, so it cannot be seen. It is just an identity, a position. It has no dimensions. So that point, namely, $x=0,y=1$ is missing.


Consider the Maclaurin expansion of $\frac{sin(x)}{x}$


Dividing by $x$ you get a series that has a value of $1$ at $x=0$

  • 2
    $\begingroup$ At $x=0$, you cannot divide by $x$, since division by $0$ is meaningless. $\endgroup$ Dec 26 '15 at 14:36
  • $\begingroup$ Maybe you can't, but I just did. $\endgroup$
    – StephenG
    Dec 26 '15 at 14:38
  • $\begingroup$ that's hilarious. $\endgroup$
    – Mr. Y
    Dec 28 '15 at 8:56

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