0/0

Euler

L'Hopitals Rule :rolleyes:

yeah, that's the one :biggrin:

Reply 21

how come z can hold any value when x=0 and y=0?

Reply 22

RichE

You can't apply L'Hopital's rule to x/y as LHR works for one variable. x/y hasn't a limit as i said.

But if you know LHR then you could apply to say to the following limits as x tends to 0 (all of the 0/0 variety)

sin(x)/x, tanx/x^2, (1-cosx)/x^2,

and get

1, infinity, 1/2

as your answers - which surely convinces that 0/0 cannot be sensibly assigned a single value in all circumstances.

I'm not sure on this but couldn't you simply differentiate it implicitly? :confused:

Reply 23

darkenergy

how come z can hold any value when x=0 and y=0?

'cause it's a vertical line perpendicular to the xy plane

oh yeah......

You can't make any sense of 0/0 as a real number. Like El Stevo says it's called an indeterminate.

That makes sense. You can't really apply normal rules of algebra here as there are infinitely many values.

1^infinity, infinity/infinity, infinity - infinity, infinity^0

Aren't 1^infinity and infinity^0 just 1? Perhaps the others can be indeterminate though if you say there are an infinite number of possible relative 'sizes' of infinity.

Reply 26

El Stevo

Gaz031

What's so indeterminate about:
1^infinity - isn't it just 1?
infinity/infinity - isn't it just 1?
infinity-infinity - isn't it just 0?
infinity^0 - isn't it just 1?

there are different infinities, all of which i presume will give different 'answers'

Reply 27

but from your logic:
if infinity/infinity = 1, why shouldnt 0/0 be 1?

if infinity^0 is 1, surely 0^0 = 1?

Lastly 0/0 is same as 0^0.

Reply 28

Gaz031

Okay i'm not sure sure about inf^0 but surely 1^inf is 1?
It's a very interesting topic but arguably more philosophical than most areas of mathematics.
Do we have a set of theorems to do with the laws concerning infinity or do we rely on intuition?

Reply 29

1^infinity is definite 1. Unless repetitive multiplication of 1 changes its property :biggrin:

Reply 30

i'm pretty sure inf^0 is 1

Reply 31

darkenergy

but from your logic:
if infinity/infinity = 1, why shouldnt 0/0 be 1?

if infinity^0 is 1, surely 0^0 = 1?

Lastly 0/0 is same as 0^0.

how is 0/0 = 0^0?

Reply 32

well:

2 / 2 = 2 ^ 0

so similarly

0 / 0 = 0 ^ 0

Reply 33

coldfusion

I'm not sure on this but couldn't you simply differentiate it implicitly? :confused:

No - because x and y vary independently - y doesn't depend on x.

Reply 34

Gaz031

Aren't 1^infinity and infinity^0 just 1? Perhaps the others can be indeterminate though if you say there are an infinite number of possible relative 'sizes' of infinity.

No 1^infinity can take any positive value as well.

A famous example is (1+1/x)^x which as x->inf gives e.

(1+1/x^2)^x -> 1 as x->inf

(1+1/x)^(x^2) -> inf as x -> inf

On the infinity^0 front consider

(a^x)^(1/x) which gives a as x -> infinity

Reply 35

El Stevo

there are different infinities, all of which i presume will give different 'answers'

Senor El Stevo

Hmm not really this is a different idea to the different infinities you see in set theory.

There are two infinities in real analysis normally - just the +inf and -inf notions.

Reply 36

darkenergy

so similarly

0 / 0 = 0 ^ 0

How does this help - they're both indeterminate :frown:

Reply 37

RichE

How does this help - they're both indeterminate :frown:

yes but coldfusion asked, "how is 0/0 = 0^0?"

Reply 38

Gaz031

RichE

No 1^infinity can take any positive value as well.

A famous example is (1+1/x)^x which as x->inf gives e.

On the infinity^0 front consider

(a^x)^(1/x) which gives a as x -> infinity

Interesting examples, thanks. Surely in the first example though you never actually reach '1' on the inside of the fraction? Doesn't it just tend to 1 but the infestimal part remaining means the value inside is slightly greater than 1 and so it can only then be multiplied to give a number above 1?

Reply 39