Is ##\frac{0}{\infty}## equal to 0 or infinity in mathematics?

[Rectifier]

Is $"\frac{0}{\infty}"=0$ ?

 

[phinds]

Is $"\frac{0}{\infty}"=0$ ?

No, it is undefined because you are treating infinity as though it was a number. It isn't. You can't, in most circumstances, use infinity in normal math and expect meaningful results.

 

[.Scott]

Is $"\frac{0}{\infty}"=0$ ?

$\infty$ isn't a number, so you shouldn't be dividing by it.
That said, in most contexts, yes, $\frac{0}{\infty}=0$

 

[phinds]

$\infty$ isn't a number, so you shouldn't be dividing by it.
That said, in most contexts, yes, $\frac{0}{\infty}=0$

No, it's still undefined in most contexts, I think. What IS true is that as n approaches infinity, 0/n approaches 0.

EDIT: I hasten to add, I don't do any math where there IS any meaningful context for 0/infinity, so I could be wrong. What contexts did you have in mind?

 

[.Scott]

No, it's still undefined in most contexts, I think. What IS true is that as n approaches infinity, 0/n approaches 0.

OK, in most meaningful contexts.
As a matter of practicality, if someone ended up with infinity in the denominator, they were probably using shorthand in dealing with a limit. The type of nomenclature is commonly used to explain limits - with the caveat that it is not proper general-use nomenclature. Since the OP put the fraction in quotes, I believe the he already knows about this.

The symbol $\infty$ can be used in other more proper contexts, none of which would allow it to be in a denominator. One is to indicate the result of a limit. Another is to indicate the cardinality of a set, although Aleph is better at that. In cases where the order of Alephs is important, the omega symbol is used. Division can be defined for the omegas, but significant caveats are needed.

 

[Rectifier]

What contexts did you have in mind?

I was trying to calculate following limit:
$\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4 \left( 1 + \frac{\ln x}{x^3}\right) }{ x \left( 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} \right) } = \lim_{x\rightarrow \infty} x^3\frac{1 + \frac{\ln x}{x^3} }{ 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} }$

$\lim_{x\rightarrow \infty} x^3 = \infty$

$\lim_{x\rightarrow \infty}\frac{\ln x}{x^3}=0$

$\lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0$

$\lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x} = "\frac{0}{\infty}"$

HERE-----------------^

 

[Mark44]

I was trying to calculate following limit:
$\lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4 \left( 1 + \frac{\ln x}{x^3}\right) }{ x \left( 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} \right) } = \lim_{x\rightarrow \infty} x^3\frac{1 + \frac{\ln x}{x^3} }{ 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} }$

$\lim_{x\rightarrow \infty} x^3 = \infty$

$\lim_{x\rightarrow \infty}\frac{\ln x}{x^3}=0$

$\lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0$

$\lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x} = "\frac{0}{\infty}"$

That last limit is 0. Don't write it as $\frac 0 {\infty}$, either with or without quotes.

BTW, I am moving this thread to the technical math section, as it is more of a conceptual question than an actual homework-type question.

 

[Rectifier]

I don't understand why the last limit is 0. :,(

 

[Rectifier]

I think I got it.$\lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x} = \lim_{x\rightarrow \infty} \frac{ 1 }{x} \left( \frac{2}{3} \right)^x$

$\lim_{x\rightarrow \infty} \frac{ 1 }{x} = 0$

$\lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0$

 

[wrobel]

Is $"\frac{0}{\infty}"=0$ ?

as for me, this notation is completely clear. If we have two functions $f(x)\to 0$ and $g(x)\to \infty$ as $x\to a$ then $f/g\to 0$

 

[mfb]

I think I got it.$\lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x} = \lim_{x\rightarrow \infty} \frac{ 1 }{x} \left( \frac{2}{3} \right)^x$

$\lim_{x\rightarrow \infty} \frac{ 1 }{x} = 0$

$\lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0$

That approach is much better, because 0*0 is defined in the real numbers.

 

[micromass]

No, it's still undefined in most contexts, I think. What IS true is that as n approaches infinity, 0/n approaches 0.

EDIT: I hasten to add, I don't do any math where there IS any meaningful context for 0/infinity, so I could be wrong. What contexts did you have in mind?

You are wrong. In most contexts where the OP makes sense, it is indeed true that $0/\infty = 0$.

 

[micromass]

That approach is much better, because 0*0 is defined in the real numbers.

0*0 is undefined?

 

[mathman]

0*0 is undefined?

Why do you think so? 0 is a number, and multiplying any number by 0 is 0.

 

[mfb]

0*0 is undefined?

Huh? I said it is defined.

 

[phinds]

You are wrong. In most contexts where the OP makes sense, it is indeed true that $0/\infty = 0$.

micromass, than you for that correction, but I wonder if you could expand on it for me a bit? I'm confused as to how it is that "0/infinity = 0" is not treating infinity like a normal number

 

[mfb]

If you treat infinity like a normal number, you should be allowed to multiply by it, but then you get $0=0\cdot \infty$ which is not well-defined any more.

 

[phinds]

If you treat infinity like a normal number, you should be allowed to multiply by it, but then you get $0=0\cdot \infty$ which is not well-defined any more.

But that's my point. I've always been told that you CAN'T normally treat infinity like a normal number. Is this wrong?

 

[mfb]

I've always been told that you CAN'T normally treat infinity like a normal number.

Correct. You can still perform some limited operations in most contexts, and 0/infinity=0 is one of them (e.g. for limits). That is not treating it as normal number, that is a specific rule for this operation.

 

[phinds]

Correct. You can still perform some limited operations in most contexts, and 0/infinity=0 is one of them (e.g. for limits). That is not treating it as normal number, that is a specific rule for this operation.

OK, thanks.

 

[micromass]

Huh? I said it is defined.

I'm sorry, I must have misread undefined instead of defined.

 

[Daymare]

Why is it that we don't define operations like 0*infinity and 0/infinity etc.They do occur in maths.Is it because defining these things causes inconsistencies?

 

[micromass]

Why is it that we don't define operations like 0*infinity and 0/infinity etc.They do occur in maths.Is it because defining these things causes inconsistencies?

Like I said, $0/\infty$ is defined in math to be $0$. As for $0\cdot \infty$, how would you define it?

 

[Daymare]

You could define it as 0

 

[micromass]

You could define it as 0

Right, and this is done in measure theory, where you rigorously define what an area is. The logic is that a straight line has length $\infty$ and width $0$ and has an area of $0$. So if you define $0\cdot \infty = 0$ then this is consistent.

But why is this not done in the rest of mathematics. Consider $f(n) = 1/n$ and $g(n) = n^2$. Then $f(n)\rightarrow 0$ and $g(n)\rightarrow +\infty$. But $f(n)g(n)\rightarrow +\infty$. So in this sense at least, $0\cdot \infty = \infty$ also makes sense.

And if $f(n) = 1/n$ and $g(n) = n$, then $f(n)g(n)\rightarrow 1$. So $0\cdot \infty = 1$ also makes sense in this sense.

 

[mathman]

Right, and this is done in measure theory, where you rigorously define what an area is. The logic is that a straight line has length $\infty$ and width $0$ and has an area of $0$. So if you define $0\cdot \infty = 0$ then this is consistent.

But why is this not done in the rest of mathematics. Consider $f(n) = 1/n$ and $g(n) = n^2$. Then $f(n)\rightarrow 0$ and $g(n)\rightarrow +\infty$. But $f(n)g(n)\rightarrow +\infty$. So in this sense at least, $0\cdot \infty = \infty$ also makes sense.

And if $f(n) = 1/n$ and $g(n) = n$, then $f(n)g(n)\rightarrow 1$. So $0\cdot \infty = 1$ also makes sense in this sense.

In the first example, the limit is that of a sequence of finite length lines becoming longer without any limit so the result is a limit on 0.n=0 as n becomes infinite. In cases like this you must always consider the underlying sequence that leads to the expression.