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.
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.
