# Continuity and simplification of a function

Suppose a function $$f(x) = \frac{x^2 - x}{ x - 1},$$ we can simplify this function to be $f(x) = x$. Yet, we say that this function is discontinuous at $x = 1$ but after the simplification, we say that the function $f(x)$ is continuous.

Which one is correct? The fact that $f$ is discontinuous or continuous?

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What is correct depends on what the person grading will mark as correct. It is safest to say that the function is not continuous at $x=1$ because it is not defined there. But some people (and programs) will not even notice a removable discontinuity. –  André Nicolas Apr 21 '14 at 8:00

The original function $$f\left(x\right)=\frac{x^{2}-x}{x-1}$$ has $\mathbb{R}\backslash\left\{ 1\right\}$ as (maximal) domain and is continuous. It is not defined on $\left\{ 1\right\}$ and consequently statements like '$f$ is (dis)continuous at $1$' don't make sense. It can only be (dis)continuous at points that belong to its domain.

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Exactly! Saying $f$ is discontinuous at $1$ is senseless. It makes as much sense as saying it is discontinuous at $\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}$. –  Git Gud Apr 21 '14 at 8:22
@GitGud Actually, I wonder if this is true. I've never seen discontinuity defined, other than as the negation of continuity. Since it is not continuous at $1$, it must be discontinuous. Similarly the function $f$ is also discontinuous at $\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}$ and $\operatorname{Spec}(\Bbb Z)$ and everything else too. –  Mario Carneiro Apr 21 '14 at 16:10
@MarioCarneiro Given $a\in \mathbb R$, assuming $a\in \text{dom}(f)$, $a$ is continuous if, and only if, $\text{usual statement}$. The $a$ is given before you define continuity. If you replace $a$ by $1$ you get: assuming $1\in \text{dom(f)} \ldots$. Whatever comes next won't mean anything. Granted the conditional statement is true, but that says nothing of the consequent which is what concerns us. –  Git Gud Apr 21 '14 at 16:24
@GitGud Ugh, conditional definitions are annoying and nonsensical. Propositional formulas are true or false; they are never "undefined", whatever that means. A predicate is an abbreviation for a propositional formula, and as such always has meaning. I can write down the expression "$f$ is continuous at $1$" down as a formula, and it will be either true or false. Since $1\notin {\rm dom}(f)$, it will be false. Thus the function is not continuous at $1$. –  Mario Carneiro Apr 21 '14 at 16:48
@MarioCarneiro "I can write down the expression "$f$ is continuous at $1$" down as a formula, and it will be either true or false." It will not be false, part of the formula is $f(1)$, this doesn't make sense. I agree with the first part of your comment, but I claim that what has meaning is the formula for "$1\in \text{dom}(f)\implies f\text{ is continuous}$". This will either be true or false. Saying $f$ is discontinuous at $1$ is saying the consequent of this assertion is false. Saying it is continuous is saying the consequent is true. –  Git Gud Apr 21 '14 at 16:56

The original function has domain $x: x \neq 1$ while the second function has domain all real numbers. So they are two different functions, and they are not equal. $f$ is discontinuous at $x = 1$. Here $f(x) = \dfrac{x^2 - x}{x - 1}$

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Beat me to it... –  Mon Kee Poo Apr 21 '14 at 7:54
I must be an idiot. Haha I totally had a brain fart. Thanks –  Chris Myung Hyun Kim Apr 21 '14 at 7:58
If $f$ is not defined at $x=1$ then it cannot be stated that $f$ is discontinuous at $x=1$. –  drhab Apr 21 '14 at 7:58

The "two" functions you are talking about are exactly equal (as far as analysis is concerned) as long as you want their domains equal (when you specify a function in the form $f(x)=...$ one really should qualify x for completeness). Take the limits from the left and right at 1 and you'll get 1 so this is a perfectly legitimate function over all of $\mathbb R$ if you define it to be such.

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