If you have a function $f(x)=\frac{x^2}{x}$, then is the function continuous at x=0? If you have a function $f(x)=\dfrac{x^2}{x}$, then is the function continuous at $x=0$?
On one hand, if you simplify it and end up with $f(x)=x$, it is continuous at $0$, but if you keep it in its original form, at $x=0$ the function is not defined.
 A: By definition, a function is continuous at a point $a$ if $f(a)$ is defined and
$$\lim_{x\to a}f(x)=f(a).$$
As you already noted, your function is not defined at $x=0$, and is therefore not continuous.
It is however possible, as pointed out in a comment, to extend the function to a continuous one by simply defining $f(0)=0$.
A: If you talk about the expression $\frac{x^2}{x}$ then you are right in saying that it cannot be evaluated in that form at 0. In such cases when an algebraic expression of a function is not applicable you look for extensions that are either continuous, $\mathcal{C}^1$ or $\mathcal{C}^2$ or whatever property you want to preserve.  
In this case you merely cancel out the $x$ from the denominator and the expression $f(x)=\frac{x^2}{x}$ is equivalent to $f(x)=x$. Hence $f(x)=x$ is an extension of $f(x)=\frac{x^2}{x}$ to 0. 
A: It is not by the definition of continuity.
$\lim \limits_{x \to 0} \frac{x^2}{x}$ is not defined at the absolute value of $x=0$ even though its $RHL$ and $LHL$ do exist at that point both being equal to $0$. It's a kind of removable continuity where by certain adjustments the function can be made continuous.
