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 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 :
$$
f: \mathbb{R}\to \mathbb{R} \quad f(x)= \begin {cases}\frac{\sin x}{x} &amp;\text{for }x \ne 0 \\ 1 &amp;\text{for }x=0\end {cases}
$$

is continuous in $(-\infty, +\infty)$.
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 $x=0$