Is the graph $y=\frac{x^2}{x}$ continuous?

Question

I randomly came across the thought today that since dividing by zero is undefined, will the graph $y=\frac{x^2}{x}$ be discontinuous? Or would it be considered reducible and therefore continuous and identical to the graph $y=x$. Thanks everyone.

Answer 1

To begin with, a graph of a function $f$ is just a set $\{\langle x, f(x)\rangle\}$. You probably wanted to ask whether the function $f(x)=\frac{x^2}{x}$ is continuous.

The answer is yes: $f$ is continuous for all $x$ where it is defined. Note that $f(0)$ is not defined because of division by zero.

However, $f$ admits a continuous extension $g$, defined by $g(x)=\lim\limits_{x' \rightarrow x}f(x')$. In this case, $g(x)=x$. This corresponds to the graph of $g$ being the closure of the graph of $f$.

Answer 2

You can never divide by zero.

Setting aside concerns of domain and range (my guess is OP hasn't taken real analysis), if you just write $y =\dfrac{x^2}{x}$, it is implicit in the definition that you're only saying this applies for $x \neq 0$, because division by zero is undefined.

However, if you write $$y = \begin{cases} \dfrac{x^2}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$$ this is in fact equivalent to saying that $y = x$.

Answer 3

$y=\frac{x^2}{x}$ is clearly continuous everywhere aside from $x=0$, as it can be reduced to $y=x$.

Continuity at $x=0$ depends on how $y(0)$ is defined. If $y(0)=0$ then $y(x)$ is continuous at $x=0$. If $y(0) \ne 0$ then $y(x)$ is discontinuous at $x=0$.

Answer 4

An important thing to note is that two functions $f(x)$ and $g(x)$ are not identical for only their output values (the formula). The domain is another important factor in whether two functions are equivalent.

$$f(x) = \frac{x^2}{x} \quad g(x) = x$$

These two functions are not the same unless you add some sort of specification for the domain. This is because the first function is defined for all $x \neq 0$ while the second function is defined for all $x \in \mathbb{R}$. Hence, if you simply leave the two functions in that form, they can’t be considered the same since the first function has a gap at $(0, 0)$ (even though they are completely equivalent for all other points).

You can, however, add a specification which could make the two functions equivalent.

$$f(0) = \lim_{x \to 0}g(x) = 0$$

From here, let $f(x) = x$ for $x \neq 0$ and $f(x) = 0$ for $x = 0$. Now, they’re equivalent. (This is known as a removable discontinuity, because the gap can be removed.)