# Is $\frac 1 0$ undefined or equal to $\tilde{\infty}$? [duplicate]

Is $\displaystyle\frac 1 0$ undefined or equal to $\tilde{\infty}$?

I know that $\displaystyle\lim_{x\to0}\frac 1 x=\tilde{\infty}$, how about $\displaystyle\frac 1 0$?

Thank you.

p.s. $\tilde{\infty}$ denots the complex infinity.

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## marked as duplicate by J. M., Start wearing purple, TMM, Lord_Farin, AmzotiJun 2 '13 at 14:01

What should I think about videos like this one stating that 1/0= infinity? youtu.be/AJ4zlvqOtE8?t=4m43s – user8005 Jun 2 '13 at 15:43
I want to answer this, please reopen. – Anixx Dec 17 '14 at 17:05

It is undefined.

By the way, the limit $\displaystyle \lim_{x\to 0} \frac 1 x$ does not exist, since the limit from the right ($\infty$) does not equal the limit from the left ($-\infty$).

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Note that the question was amended after this answer was posted; the answer made good sense before the amendment, but less so afterwards. – Harald Hanche-Olsen Jun 2 '13 at 12:44
The limit exists on the affine real line/complex plane. – Anixx Dec 17 '14 at 17:06

If you know what "complex infinity" means, then you probably mean to ask this question in the complex projective numbers (i.e. the arithmetic of the Riemann sphere), in which case the answer is, of course, $\infty$. The same is true in the real projective numbers.

In the arithmetic of real numbers, of course, the answer is undefined. The same is true for the extended real numbers.

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Let us think about WHY it is undefined.

Can we multiply a number with $0$ to get $1$? No. Every number we multiply with $0$ will result in $0$ itself. Hence this is undefined.

However, let $\epsilon^{+}\to 0$. This number is not $0$, but tends to $0$. If we multiply $\epsilon$ with $\frac{1}{\epsilon}$, then we can indeed get $1$! If $\epsilon^{+}\to 0$, then $\frac{1}{\epsilon}\to\infty$.

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It depends on context. Normally, $1/0$ is considered to be undefined, but in the present context, with a complex infinity present, it makes perfectly good sense to define $1/0$ to be that infinity. It makes sense because it makes the function $x\mapsto 1/x$ continuous at $x=0$. However, if you do so, you must throw some rules out the window: For example, the algebraic rule $x\cdot(1/x)=1$ will not hold at $x=0$. There is frequently such a price for extending a notion beyond its original domain. But sometimes, the price is worth it.

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It will if one defines 0/0=0 which is perfectly sane. – Anixx Dec 17 '14 at 17:08

If you are doing complex analysis, then it might make sense, in some context, to say what you did. If you compactify $\mathbb{C}$ by adding $\infty$ to get the Riemann sphere $\hat{\mathbb{C}}$, then indeed it makes sense to put $1/0 = \infty$. For instancance, if $h : \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is holomorphic, and $g(z) = h(1/z)$, and you ask what is $g(0)$, then it won't be bad to say $g(z) = h(\infty)$ because $1/0 = \infty$.
If you are doing real analysis, you have at least two reasonable ways to compactify $\mathbb{R}$: by adding one "unsigned" infinity, or two "signed" infinities $+\infty, - \infty$. It is strange to speak of complex infinity in this context. If you accept one $\infty$, then perhaps $1/0 = \infty$; else it is probably undefined.
If you are doing algebra, then $1/0$ makes no sense. After all, $x = 1/0$ should have the property that $x \cdot 0 = 1$. But this never happens because $x \cdot 0 = 0 \neq 1$, end of story.