Math question

[DyslexicHobo]

Originally posted by: f95toli
You need to remember that lim x->0 is NOT the same thing as saying thay x ever IS zero. Hence, you can't ever set x=0 as you do.
Also, in general you need to keep track of which direction you are approacing the limit from.
If you e.g. have a step function which is 1 for X>0 and 0 for x<0 you obviously have that lim x->0 from the positive side is equal to 1. and from the negative side equal to 0.

But x=0 is undefined and therefore you can't say anything about the value of the function at x=0.
(You CAN still differentiate, however. The result is the Dirac delta function)

I understand that saying x=0 is not the same thing as when x approaches 0. However, it still seems as though there is contradiction in the expression.

I'm still not exactly sure why people are talking about differentiating the function. Where does this come into play?


Anyway, back to the anomaly: If f(x) = (1/x)*x, at x=0, it is undefined. However, f(x) = g(x) if g(x) = 1. Therefore g(0) MUST equal f(0). Therefore g(0) = f(0) = 1.
Am I breaking any laws of math by taking this backdoor route to the solution? Obviously two different solutions are found depending which path you choose.

 

[polarmystery]

It is undefined...

Simply put

(x/0)*0

Order of operation is within the parenthesis... so it is undefined before it can be multiplied, hence it is undefined regardless of how you try to solve it.

 

[PCTC2]

Originally posted by: eaj0010
It is undefined...

Simply put

(x/0)*0

Order of operation is within the parenthesis... so it is undefined before it can be multiplied, hence it is undefined regardless of how you try to solve it.

and if you reorganize that to x*(0/0), it is still undefined because 0/0 is also undefined.

but if you want a silly definition of x/0, it is called nullity (the symbol is an I superimposed over a 0). so 0/0=Nullity
it is found here:Nullity

 

[Matthias99]

Originally posted by: DyslexicHobo
I understand that saying x=0 is not the same thing as when x approaches 0. However, it still seems as though there is contradiction in the expression.

I'm still not exactly sure why people are talking about differentiating the function. Where does this come into play?

Some people were talking about L'Hopital's Rule earlier, which has to do with differentiation of functions (but won't help you with the original question, since the derivatives are 0).

Anyway, back to the anomaly: If f(x) = (1/x)*x, at x=0, it is undefined. However, f(x) = g(x) if g(x) = 1. Therefore g(0) MUST equal f(0). Therefore g(0) = f(0) = 1.
Am I breaking any laws of math by taking this backdoor route to the solution? Obviously two different solutions are found depending which path you choose.

Edit: er, that didn't make a lot of sense.

If you define g(x) = 1, f(x) equals g(x) when x != 0. The limit of f(x) as x goes to 0 is 1, but f(0) is still undefined and is not equal to 1.

 

[MagnusTheBrewer]

I was always taught that "anything" * 0 = 0 not because of the value of the product but, because it is a stated identity. Therefore, x/0 * 0 = 0 and i * 0 = 0

 

[BrownTown]

you were taught worng.

 

[darkhorror]

Remember when using the limit that the value never actually equals where you are looking. so it's not acutally equal to the value at x = what ever.

"I was always taught that "anything" * 0 = 0 not because of the value of the product but, because it is a stated identity. Therefore, x/0 * 0 = 0 and i * 0 = 0"

your anything still needs to be a number, x/0 is not a number.

 

[Matthias99]

Originally posted by: MagnusTheBrewer
I was always taught that "anything" * 0 = 0 not because of the value of the product but, because it is a stated identity. Therefore, x/0 * 0 = 0 and i * 0 = 0

The problem is that there are two incompatible "rules":

1) Any real number multiplied by zero is equal to zero.

2) The result of dividing a real number by zero is undefined.

What you are missing is another (usually implicit) rule:

0) Undefined values are not real numbers.

This "rule" means you have to ensure that the values you work with are real numbers before you can apply the "any real number multiplied by zero is equal to zero" rule.

(x/0 * 0) then can't be simplified with rule 1 (since (x/0) is undefined and thus not a real number). If you change things around with order of operations, you can get ((x*0)/0), but then that simplifies to (0/0), which is also undefined.

i is a complex number, and if you are working with complex numbers, i * 0 = 0. However, if you limit yourself to the reals, sqrt(x) is undefined when (x < 0).

 

[MagnusTheBrewer]

Originally posted by: Matthias99

Originally posted by: MagnusTheBrewer
I was always taught that "anything" * 0 = 0 not because of the value of the product but, because it is a stated identity. Therefore, x/0 * 0 = 0 and i * 0 = 0

The problem is that there are two incompatible "rules":

1) Any real number multiplied by zero is equal to zero.

2) The result of dividing a real number by zero is undefined.

What you are missing is another (usually implicit) rule:

0) Undefined values are not real numbers.

This "rule" means you have to ensure that the values you work with are real numbers before you can apply the "any real number multiplied by zero is equal to zero" rule.

(x/0 * 0) then can't be simplified with rule 1 (since (x/0) is undefined and thus not a real number). If you change things around with order of operations, you can get ((x*0)/0), but then that simplifies to (0/0), which is also undefined.

i is a complex number, and if you are working with complex numbers, i * 0 = 0. However, if you limit yourself to the reals, sqrt(x) is undefined when (x < 0).

Thank you for a clear explanation. That makes sense to me.

 

[thesurge]

Originally posted by: DyslexicHobo

Originally posted by: thesurge
lim_y-->0(x*(y/y))=x if the 0/0 formed from the function y/y.

Basically, you have a removable discontinuity (I believe) at y=0 (if you let x be constant and iterate y). However, if the 0/0 formed in a different way (as eLiu pointed out), you obviously can have different solutions.

But obviously we cannot have different solutions for the same expression! This would lead us to believe that 1=2. Err... 1 = all reals. Actually... 1=all reals and non-reals.

As previously stated, this would be a pretty boring and useless number system.



So what I've reasoned from this topic so far... if left as-is, it would come to be undefined because of divide by zero and whatnot. However... (I believe Cyclo might have mentioned this) what if we let f(x) = (1/x)*x. At x=0, the limit is going to equal 1, but the actual answer is undefined. Is this correct? We can also state that g(x) = f(x) if g(x) = 1. Am I correct in saying that?

So if g(1) = 1, then f(1) = 1. This proves that (1/0)*0 = 1.

Actually I kind of made up my rule to equate f(x) and g(x), and now that I think about it, it's not correct. But the logic makes sense using algebra, doesn't it? Or is this not a valid proof. Why?

I think it's pretty obvious that this expression is undefined. One of the reasons, aforementioned, is that it in fact can take on more than one real value. The value is computed from a limit and therefore it doesn't render our real number system "useless".

 

[DyslexicHobo]

Originally posted by: thesurge

Originally posted by: DyslexicHobo

Originally posted by: thesurge
lim_y-->0(x*(y/y))=x if the 0/0 formed from the function y/y.

Basically, you have a removable discontinuity (I believe) at y=0 (if you let x be constant and iterate y). However, if the 0/0 formed in a different way (as eLiu pointed out), you obviously can have different solutions.

But obviously we cannot have different solutions for the same expression! This would lead us to believe that 1=2. Err... 1 = all reals. Actually... 1=all reals and non-reals.

As previously stated, this would be a pretty boring and useless number system.



So what I've reasoned from this topic so far... if left as-is, it would come to be undefined because of divide by zero and whatnot. However... (I believe Cyclo might have mentioned this) what if we let f(x) = (1/x)*x. At x=0, the limit is going to equal 1, but the actual answer is undefined. Is this correct? We can also state that g(x) = f(x) if g(x) = 1. Am I correct in saying that?

So if g(1) = 1, then f(1) = 1. This proves that (1/0)*0 = 1.

Actually I kind of made up my rule to equate f(x) and g(x), and now that I think about it, it's not correct. But the logic makes sense using algebra, doesn't it? Or is this not a valid proof. Why?

I think it's pretty obvious that this expression is undefined. One of the reasons, aforementioned, is that it in fact can take on more than one real value. The value is computed from a limit and therefore it doesn't render our real number system "useless".

There are many equations that can rear more than one real solution. Take the equation y = x^2. y = x and y = -x

 

[CycloWizard]

Originally posted by: Matthias99
0) Undefined values are not real numbers.

I think this is the crux of the issue. Funny that I should say this in two threads in one day, but +/- infinity aren't in the set of all real numbers.

 

[DrPizza]

Originally posted by: BrownTown
L'Hospital's rule obviously wouldn't apply in this situation as the variables in question are constants (at 0) so their derivitives are also 0 and you are still boned. As I remember we used it mroe in the case of infinity/infinity, but obviously you can do 0/0 too although I'm not sure exactly how mand usefull situations that would apply too. Just off the top of my head I guess you could do a trivial case like x/(2x) which evealuated to 0/0 at 0, but is really 1/2 by applying L'Hospital's rule (or just canceling the x terms / using common sense).

lim (as x->0) sinx/x

(There's one example of a 0/0 case for L'Hospital's rule)

However, even in the infinity over infinity case, it's really no different than the 0/0 case
(I'll do it the other way around though):

lim (as x->0) (1/x)/(1/sinx)
My first 0/0 type is now an infinity/infinity type. So, of course, you can switch most infinity over infinity types to a 0/0 type using this trick.

 

[BrownTown]

Umm, I don't see why people feel the need to keep explaining random L'Hospital's rule examples here, we all know what the rule is and how to use it. However the OP did not ask about what the sinc function equated to at 0 or what (1/x)/(1/sinx) equals at 0. What the OP asked was a single question which was x/0*0 in the ONE EXAMPLE L'Hospital's rule does not apply because 0 is a CONSTANT and the derivitive of a cosntant is 0 again.

 

[CycloWizard]

Originally posted by: BrownTown
Umm, I don't see why people feel the need to keep explaining random L'Hospital's rule examples here, we all know what the rule is and how to use it. However the OP did not ask about what the sinc function equated to at 0 or what (1/x)/(1/sinx) equals at 0. What the OP asked was a single question which was x/0*0 in the ONE EXAMPLE L'Hospital's rule does not apply because 0 is a CONSTANT and the derivitive of a cosntant is 0 again.

The OP's question is meaningless as stated since the answer depends on all of the other factors that people are bringing up. Zero may be manifested as a constant, but that is only one way that it appears. We are simply bringing up the other things that could bring about this situation and trying to give further insight based on these possibilities. Why is this so offensive to you?

 

[DrPizza]

Originally posted by: BrownTown
Umm, I don't see why people feel the need to keep explaining random L'Hospital's rule examples here, we all know what the rule is and how to use it. However the OP did not ask about what the sinc function equated to at 0 or what (1/x)/(1/sinx) equals at 0. What the OP asked was a single question which was x/0*0 in the ONE EXAMPLE L'Hospital's rule does not apply because 0 is a CONSTANT and the derivitive of a cosntant is 0 again.

Sorry, I didn't mean to imply that L'hospital's rule had anything to do with the OP's question. I figured his question was answered pretty quickly. I was simply pointing out that...

As I remember we used it mroe in the case of infinity/infinity, but obviously you can do 0/0 too although I'm not sure exactly how mand usefull situations that would apply too

any (almost any?) cases where a limit - if you were using L'Hopital's rule, which has absolutely nothing to do with the OP's question - evaluates to infinity over infinity, it can be rewritten to evaluate to 0 over 0; i.e. there's no sense differentiating the two types of indeterminancy.

 

[thesurge]

Originally posted by: DyslexicHobo

Originally posted by: thesurge

Originally posted by: DyslexicHobo

Originally posted by: thesurge
lim_y-->0(x*(y/y))=x if the 0/0 formed from the function y/y.

Basically, you have a removable discontinuity (I believe) at y=0 (if you let x be constant and iterate y). However, if the 0/0 formed in a different way (as eLiu pointed out), you obviously can have different solutions.

But obviously we cannot have different solutions for the same expression! This would lead us to believe that 1=2. Err... 1 = all reals. Actually... 1=all reals and non-reals.

As previously stated, this would be a pretty boring and useless number system.



So what I've reasoned from this topic so far... if left as-is, it would come to be undefined because of divide by zero and whatnot. However... (I believe Cyclo might have mentioned this) what if we let f(x) = (1/x)*x. At x=0, the limit is going to equal 1, but the actual answer is undefined. Is this correct? We can also state that g(x) = f(x) if g(x) = 1. Am I correct in saying that?

So if g(1) = 1, then f(1) = 1. This proves that (1/0)*0 = 1.

Actually I kind of made up my rule to equate f(x) and g(x), and now that I think about it, it's not correct. But the logic makes sense using algebra, doesn't it? Or is this not a valid proof. Why?

I think it's pretty obvious that this expression is undefined. One of the reasons, aforementioned, is that it in fact can take on more than one real value. The value is computed from a limit and therefore it doesn't render our real number system "useless".

There are many equations that can rear more than one real solution. Take the equation y = x^2. y = x and y = -x

All I see are three constants.

 

[manowar821]

DON'T DIVIDE BY ZERO, EVER.

 

[rocadelpunk]

It's seemingly natural if you have two functions say a and b

you want to show that these are equal so you say a = b and algebraically manipulate them until one looks like the other and say aha!

However, this isn't technically correct from a proof standpoint, as your first step begins by assuming exactly what you're trying to prove.

From a brief glance I think this is what you were trying to do in one ex op.

 

[piasabird]

if a=b then a squared -ab = 0
2*0 = 1*0
0=0

 

[DyslexicHobo]

Originally posted by: piasabird
if a=b then a squared -ab = 0
2*0 = 1*0
0=0

Err... what?

 

[AgentJean]

You can not divide by zero.
If you do bad things happen.