# Giving infinity a value

1. Aug 24, 2015

### Einstein's Cat

Is infinity divided by infinity equal to 1? 6 divided by 6 is equal to 1 however as infinity resembles 0 in the sense that 0 dived by 0 is equal to 0, I am uncertain whether infinity divided by infinity would equal 1 or instead, infinity.

2. Aug 24, 2015

### DEvens

Usually you can't do things like multiply or divide by infinity. It is not defined. Similarly, 0 divided by 0 is not 0. It is not defined.

What you can sometimes do is examine a limit. So $\lim_{x->0} \frac{\sin(x)}{x}$ is defined, and is 1. So in this sense, in this case, dividing a zero by a zero gives you 1. But only as the limit.

https://en.wikipedia.org/wiki/L'Hôpital's_rule

3. Aug 24, 2015

### soarce

The uncertainties of the type $\infty \cdot 0$, $\infty/\infty$ or $0/0$ acquire a definite value only as a limit, you can't simply operate with $\infty$ as being a number. So, unless we can go through the limit process it makes no sense to say that some uncertainty is equal to some value.

4. Aug 24, 2015

### AliGh

You are considering infinity as a constant number
Depends on the infinities you are working with quotient of two infinities can be zero or infinity too

5. Aug 24, 2015

### Einstein's Cat

6. Aug 31, 2015

### Staff: Mentor

You have managed to pack a number of things that aren't true into a small number of words.

No.
The indeterminate form $[\frac{\infty}{\infty}]$ shows up in calculus as limits that can literally come out to any number, as well as negative or positive infinity. Here are some simple examples:
1. $\lim_{x \to \infty}\frac{x^2}{x} = \infty$
2. $\lim_{x \to \infty}\frac{x}{x^3} = 0$
3. $\lim_{x \to \infty}\frac{x^2 + 3}{3x^2 - x + 7} = \frac 1 3$

No, not at all.

No.
Division by 0 is not defined. The indeterminate form $[\frac 0 0]$ also shows up in calculus limits, and can come out to any number. Some examples of this:
1. $\lim_{x \to 0}\frac{x^2}{x} = 0$
2. $\lim_{x \to 0}\frac{x}{x^2}$ does not exist
3. $\lim_{x \to 0}\frac{x}{x^3} = \infty$
4. $\lim_{x \to 0}\frac{\sin(2x)}{x} = 2$