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I am reading through one of my maths textbooks at the moment and the following example is given.

Determine whether the following functions are continuous at $x=2$

$f(x) = \frac{x^2 - 4}{x - 2}$

It then goes on to say

The function $f(x)$ is undefined at $x = 2$ and hence is not continuous at $x = 2$

I am very confused by this statement. When they define $f(x)$ is it referring to strictly that expression and asking me to simply replace all $x$ with $2$ without simplifying?

If $g(x) = x + 2$ then is $f(x)$ not equivalent to $g(x)$ and as $g(x)$ is continuous at $x = 2$ then $f(x)$ is also?

Nasica
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  • I suppose the crux of my question is are $f(x)$ and $g(x)$ the same function or simply functions with the same locus? Does $f(2)$ strictly mean replace all instances of $x$ with $2$ and don't simplify? – Nasica Jun 04 '20 at 08:50
  • The domain of $g$ is the whole of the real number line. The domain of $f$ is not - the domain excludes the point $x = 2$. For $f$ and $g$ to be equal, one of the conditions is that their domains are the same (the others are that the codomains are the same and that $f(x) = g(x)$ for all $x$ in their common domain). It is that missing point in the domain of $f$ which specifically indicates that $f$ and $g$ are different - but for that point they would be the same. – Prime Mover Jun 04 '20 at 09:08
  • I kind of understand your trouble with that question. Personally, I would say that the question does not make sense: we may only say that a function is or isn't continuous at a point where it is defined since, by definition, $f$ is continuous at $2$ if $\lim_{x \to 2} f(x) = f(2)$. So, in this case the right hand side is undefined. – Jakub Opršal Jun 04 '20 at 09:34
  • You may also find (Question 1525054) useful, particularly Will R's answer "It comes down to what we mean by a function. In specifying the function, we need to specify not just the "rule", but also the domain, i.e., what are you allowed to put into the function?...". – Jam Jun 04 '20 at 10:47

2 Answers2

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At $x=2$, $g$ is continuous and $f$ is undefined. This is all you can say.

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It is true the function $f$ is not continuous, because at $x=2$ the denominator vanishes and makes the expression of $f$ meaningless (if $f$ was continuous you could evaluate $f(2)$). For $x \neq 2$ you can simplify the fraction and get $x+2$, but this is not the function $g(x)=x+2$ defined over the reals. All you can say is that $f$ is the function $x+2$ defined on $\mathbb R \setminus \{2\}$.

Gibbs
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