It's a commonly known fact that one does not simply divide by zero. There are many mathematical proofs on the Internet and on Brilliant that are flawed due to dividing by zero, such as proving that 1+1=1. But if we aren't allowed to divide by 0, shouldn't we not be allowed to divide by infinity?

When you type in \(0\times \infty\) on wolfram alpha, contrary to popular belief, the answer isn't 0, but rather it's indeterminate.

let me show what I mean

\(\dfrac{1}{\infty}=0\)

\(1=\infty \times 0\)

Or

\(\dfrac{2}{\infty}=0\)

\(2=\infty \times 0\)

What... There are two results, but how can this be.

How about this:

\(1=\infty \times 0\)

\(\dfrac{1}{0}=\infty\)

Or

\(2=\infty \times 0\)

\(\dfrac{2}{0}=\infty\)

So my question to you is "why is dividing by \(\infty\) allowed but dividing by 0 is not)

## Comments

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TopNewestThere are 7 indeterminant forms namely,

\(\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, \infty \times 0, \infty^{0}, 0^{0}, 1^{\infty} \)

here \(0\), \(1\) and \(\infty \) are tending \(0\),tending \(\infty\) and tending 1But

\(\frac{exact 0}{exact 0}, \frac{tending0}{exact 0}, \infty + \infty, \infty \times \infty, \infty^{\infty}, \pm \infty\)

Are undefined

Let me explain further

\(exact 1^{\infty} = 1\)

But

\(tending 1 ^{\infty}\) is indeterminant

Another example

\(\frac{exact 0}{tending 0} = 0\)

But

\(\frac{tending 0}{tending 0}\) is indeterminant

\(\displaystyle exact0 \times \infty = 0 \)

But

\(\displaystyle tending0 \times \infty\) is indeterminant

Ask anything if you don't understand because I copied the comment of mine when I explained same thing to other – Krishna Sharma · 2 years, 11 months ago

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– Trevor Arashiro · 2 years, 11 months ago

This is actually a great explaintion. If I understand, it's like using limits in a sense.Log in to reply

Exactly. Using limit sin x 0 value equals 0. Using limit e^x , value equals 1. Using limit e^1/x value equals infinite – Swarupendra Nath Chakraborty · 2 years, 2 months ago

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One should understand the difference between these two situations:

\(\Large a.\) When a number is \(\textit {exactly}\) equal to some value &

\(\Large b.\) When it \(\textit {approaches from left or right} \) to some value.

In another way \(0×\infty=0\) but \(\textbf {(something infinitesimally small)×(something infinitely big)=undefined!! }\) – Sanjeet Raria · 2 years, 11 months ago

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