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Question: What is the domain of $f(x)=\frac{1}{x}-\frac{1}{x}$? Does the function have a removable discontinuity at $x=0$?

My attempt: My first intuition told me that it was $\mathbb R$, since we just have $f(x)=0$. However, when $x=0$, we get $f(x)= \mathrm{undefined} - \mathrm{undefined}=\mathrm{undefined}$.

So I think that it is rather $\mathbb R_{\neq0}$. If that is correct, can we call $x=0$ a removable discontinuity?

wythagoras
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    In the same idea, what about the domain of $f(x)=\ln(x)-\ln(x)$? – Surb Jul 28 '15 at 18:06
  • Correct. Yes we can. – David C. Ullrich Jul 28 '15 at 18:07
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    @Surb That one is even more interesting regarding the question about the removable discontinuity, because for the removable discontinuity we need to have $\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{-}}f(x)$, and the negative variant doesn't exist while the positive variant is 0. – wythagoras Jul 28 '15 at 18:12
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    Actually things like these are used a lot! For example if you want to plot a part of the curve $y=\sin{(x)}$ from $[0,2\pi]$ then you can use the equation $y=\sin{(x)} + 0 \sqrt{x} \sqrt{2\pi - x}$ !! This is hence used in restricting domain (while plotting only the Real part)! – NeilRoy Jul 28 '15 at 18:14
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    @NeilRoy: With the $\sqrt{x}$ restricting us to $\left[0, +\infty\right)$ and the $\sqrt{2\pi-x}$ restricting us to $\left(-\infty, 2\pi\right]$, we then take the intersection otherwise $f$ won't be defined outside, right? – Khallil Jul 28 '15 at 18:15
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    Right. $a-a = 0$, for all real numbers $a$. If $a$ isn't a real number, all bets are off. – chharvey Jul 28 '15 at 19:25
  • @Khallil Yep...that's right! And we multiply by "$0$" so that $\sqrt{x}$ and $\sqrt{2\pi -x}$ do not take part in the equation but just change the domain...like controllers! – NeilRoy Aug 03 '15 at 02:52

3 Answers3

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Well as you mentioned you get the domain of $f(x)$ as $\Bbb{R}-\{0\}$ and the range for the given function will just be zero.

So the domain will be

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The range will be

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Go for domain before simplifying . for example : $y=f(x) :\mathbb{R}\rightarrow \mathbb{R}$ $$y=(\sqrt{x})^2\\ $$if you simplify $y=x , \mathrm{domain}=\mathbb{R}$

but it's wrong $$\mathrm{domain}=\left \{ x|x \geq 0 \right \}$$ because of $\sqrt{x}$.

Khosrotash
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  • $(\sqrt{x})^2$ does not equal $x$. It is ok to simplify, as long as you simplify right, and during simplification say something about domain. In your example (which is not what the OP asked) you cannot simplify unless you say that yourt simplification is only valid when $x\ge0$ – Mirko Jul 28 '15 at 18:26
  • @Mirko Intuitively, it does. That is what he probably meant. Also, $\frac1x-\frac1x$ doesn't equal 0 when $x=0$. – wythagoras Jul 28 '15 at 18:28
  • @wythagoras $(\sqrt{-3})^2$ is not $-3$, I do not know what they meant, they did not say it right, and what they say is more confusing than helping. – Mirko Jul 28 '15 at 18:31
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The domain of a function needs to be specified as part of the definition of the function. A formula for a function does not define the domain of the function, unless we specify the domain in some way from the formula: for example, we might choose the specification

"the domain of $f$ is the set of real numbers $x$ such that the given formula for $f(x)$ is well defined".

But this is not a definition of "the domain of $f$"; it is just one of many possibilities for the domain. Thus your question is well defined only if we add to it some specification of the domain. If we chose the specification above, then the answer is $\Bbb R\setminus\{0\}$ or, in your excellent notation, $\Bbb R_{\neq0}$. Zero is indeed a removable discontinuity in this case.

John Bentin
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