What is the value of $\frac{\sin x}x$ at $x=0$? On plotting graph for $\frac{\sin x}{x}$ using Wolfram|Alpha and Google, got that :

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 $x=0$.
 A: 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 $(-\infty, +\infty)$.
A: 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.
A: 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 $x=0$.
A: 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.
A: Consider the Maclaurin expansion of $\frac{sin(x)}{x}$
$$sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$
Dividing by $x$ you get a series that has a value of $1$ at $x=0$
