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A function $f$ is said to be continuous if $\lim\limits_{x \to a} f(x)=f(a)$

Now when we evaluate a function like $\frac{x^2-1}{x-1}$ as $x \to 1$ we find that the limit is 2.

$$\lim\limits_{x \to 1}\frac{x^2-1}{x-1} =\lim\limits_{x \to 1}\frac{(x-1)(x+1)}{(x-1)}=\lim\limits_{x \to 1}(x+1)=2$$

but the function $\frac{x^2-1}{x-1}$ itself is not defined at $x=1$.

Does this mean this function is not continuous at $x=1$?

Orpheus
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    Yes, it does. For a function to be continuous at a point it must exist at said point. – Kyan Cheung Nov 29 '21 at 10:17
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    However, it can be extended by continuity at $x=0$. – PinkyWay Nov 29 '21 at 10:18
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    See also: https://math.stackexchange.com/q/1476310, https://math.stackexchange.com/q/462199, https://math.stackexchange.com/q/1998842 – all found with Approach0 – Martin R Nov 29 '21 at 10:20
  • @Kyky when you say "yes, it does", note the question in the title is the opposite of the final question in the text. Presumably you intended "yes, this means the function is not continuous at $x=1$" – Henry Nov 29 '21 at 10:27
  • I intended to reply to the question at the end of the question, good catch. – Kyan Cheung Nov 29 '21 at 10:43
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    @Invisible You mean at $x=1$ – Peter Nov 29 '21 at 11:54
  • @MartinR The duplicate in the close vote has a slight difference to the case here, namely that the limit does not exist (or is it a typo and it should be $x-2$ in the denominator ?) As written, the function in the duplicate cannot even be made continous by adding a point. – Peter Nov 29 '21 at 11:58
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    @Peter: Yes, I had noticed that (too late) and suggested in CURED to chose one of the questions from my other comment as a duplicate target. Both https://math.stackexchange.com/q/1476310 and https://math.stackexchange.com/q/1998842 are the identical question. – Martin R Nov 29 '21 at 12:04

2 Answers2

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Does this mean this function is not continuous at $x=1$?

Yes, and you give the correct reason, that is, $f(1)$ does not exist.

However, $f$ can be extended in order to make it continuous:

$$f(x)=\begin{cases} \frac{x^2-1}{x-1} &\text{ if } x\neq 1 \\ 2 &\text{ if } x=1\end{cases}$$

is a continuous function and is equivalent to the function $f(x) = x+1\ (\forall x\in\mathbb{R})$.

Adam Rubinson
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  • So you mean the way..... I defined $f(x)$ was not correct....but – Orpheus Nov 29 '21 at 10:29
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    Well, it would have been better if you stated the domain of your original function. As you wrote it, I assume that the input is all of the real numbers that the function can take, $\ \mathbb{R} \setminus {1},\ $ in which case of course the function is well-defined. But it is not continuous at $\ x=1.\ $ In fact, you shouldn't even been talking about $\ x=1\ $ because $\ x=1\ $ is not a part of the domain. See: https://en.wikipedia.org/wiki/Domain_of_a_function#:~:text=In%20mathematics%2C%20the%20domain%20or%20set%20of%20departure,X%20%E2%86%92%20Y%2C%20and%20is%20alternatively%20denoted%20as – Adam Rubinson Nov 29 '21 at 10:34
  • Can I call this a removable discontinuity? – Orpheus Nov 29 '21 at 15:23
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No, your function is not continuous at 1. It doesn't satisfy your definition for continuity at some a. Although the limit exists, $f(1)$ doesn't.

sato
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