# How is $\frac{\sin(x)}{x} = 1$ at $x = 0$ [duplicate]

I have a function: $$\text{sinc}(x) = \frac{\sin(x)}{x}$$ and the example says that: $\text{sinc}(0) = 1$, How is it true?

I know that $\lim\limits_{x \to 0} \frac{\sin(x)}{x} = 1$, But the graph of the function $\text{sinc}(x)$ shows that it's continuous at $x = 0$ and that doesn't make sense.

## marked as duplicate by Simon S, Mark Fantini, Daniel W. Farlow, marwalix, user147263 May 1 '15 at 20:35

• what is $sinc(x)$? – imranfat May 1 '15 at 15:36
• $\sin x \approx x \quad \forall x \rightarrow 0$ – Autolatry May 1 '15 at 15:38
• Sorry for this, I edited the question. – Farouk Sabry May 1 '15 at 15:39
• @Autolatry, you meant the constraint $x \simeq 0$ – abel May 1 '15 at 15:40
• @imranfat: it's the cardinal sine, one of the special functions. – Bernard May 1 '15 at 15:41

In an elementary book, they should define $\mathrm{sinc}$ like this $$\mathrm{sinc}\; x = \begin{cases} \frac{\sin x}{x}\qquad x \ne 0 \\ 1\qquad x=0 \end{cases}$$ and then immediately prove that it is continuous at $0$.
In a slightly more advanced book, they will just say $$\mathrm{sinc}\;x = \frac{\sin x}{x}$$ and the reader will understand that removable singularities should be removed.
The function $\operatorname{sinc}$ is defined as $$\operatorname{sinc}\colon x\in\mathbb{R} \mapsto \begin{cases} \frac{\sin x}{x} & \text{ if } x\neq 0\\ 1 & \text{ if } x = 0 \end{cases}$$ (note that you cannot write $\frac{\sin x}{x}$ for the case $x=0$). It is continuous on $\mathbb{R}$, because for $x\neq 0$ $\operatorname{sinc}(x) = \frac{\sin x}{x} \xrightarrow[x\to 0]{} 1 = \operatorname{sinc}(0)$.
The exact definition of $sinc$ function is $$sinc(x)= \begin{cases} \frac{\sin(x)}x, & \text{iff x\neq0}\\ 1, & \text{iff x=0} \end{cases}$$
Did you expect it to go to $\infty?$ In such cases L' Hospital's Rule can be used to find the proper limit. The graph of the function $\text{sinc}(x)$ shows that it is continuous at $x = 0$ and that is why it makes sense.