Is the function $\frac{\sin{x}}{x}$ continuous at $x=0$? Here the function is $$f(x) =\frac{\sin{x}}{x}.$$
We see that the right hand limit equals the left hand limit but does $f(0)$ exist?
 A: Indeed, the given function $$f(x) := \frac{\sin{x}}{x}$$ is not defined at $x=0$. Because $f(0)$ isn't defined (at this stage), it does not really make sense to ask if $f$ is continuous at the point $x=0$. This is because continuity at $x=0$ requires that
$$
\lim_{x \to 0} f(x) = f(0).
$$
However, this is meaningless if we don't know what $f(0)$ is!
Now, as you've observed, the right and left limits agree, i.e.
$$
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-}  f(x) = 1.
$$
This means that the function $f$ can be made continuous at $x=0$ by assigning it the value $f(0) := 1$. Alternatively, this means that the function
$$
f(x) := \begin{cases} \frac{\sin{x}}{x} & \text{if }x \neq 0,\\
1 & \text{if }x = 0
\end{cases}
$$
is continuous everywhere. 
A: As it is $f(0)$ is not defined and $f$ is continuous on $\mathbb R \setminus \{0\}$.  If you extend the definition of $f$ by defining $f(0)$ as $1$ then it becomes continuous. 
A: $f(0)$ is undefined since $\sin(0)/0$ is undefined. However, the limit of $f(x)$ as $x$ approaches zero, as you noted, is one. If you define $f(0)=1$, then $f$ is continuous (everywhere) due to this.
