Zero

What happens when you divide by zero? It must be something important because you even have to find NPVs (non-permissible values) so that you will know if a denominator will end up being zero.

E.g. (x+2)/ (y^2-7) NPVs=sqrt7

Originally posted by prosoccer
What happens when you divide by zero? It must be something important because you even have to find NPVs (non-permissible values) so that you will know if a denominator will end up being zero.

E.g. (x+2)/ (y^2-7) NPVs=sqrt7

Divding by zero can result in two things. Something undefined, or infinity.

Divisions can be rewritten as a multiplication:

15 : 3 = x, then 15 = x * 3

15 : 0 = x, then 15 = x * 0, wich is clearly not something to define.

Rewriting it in another way:

15 : 3 = x, then 15 : x = 3

15 : 0 = x, then 15 : x = 0, wich is a reason to call x infinity.

Unless you're in college, just don't divide by zero. My teacher used to say "Delen door nul is flauweful", wich boils down to "Dividing by zero is nonsense".

Originally posted by TheMaster37
Unless you're in college, just don't divide by zero.

You can't divide by zero, not even in college.

Originally posted by FabianFnas
You can't divide by zero, not even in college.

What's the limit of (sin x)/x as x approaches 0?

Originally posted by XanthosNZ
What's the limit of (sin x)/x as x approaches 0?

The Taylor series expansion of sin x gives

sin x = x - x^3/3 + x^5/5! - x^7/7! + ...+ (-1)^r.x^(2r+1)/(2r+1)! + ...
= x (1 - x^2/3 + ... + (-1)^r.x^2r/(2r+1)! +...)

Hence (sin x)/x = 1 - x^2/3 + ... + (-1)^r.x^2r/(2r+1)! +...
= 1 - x(x/3+...+ (-1)^r.x^(2r-1)/(2r+1)!+...)
Therefore, lim as x tends to 0 of (sinx)/x = 1.

Originally posted by XanthosNZ
What's the limit of (sin x)/x as x approaches 0?

Wou write correctly when x approaches zero, not, when x has reached zero.
The answer is of course 1, not needing to divide by zero.
L'Hôpital rule is well known.

Originally posted by FabianFnas
You can't divide by zero, not even in college.

I don't, physisists do.

Quote from a college of quantummechanics:

"So s*t is greater equal to h/2, but what if we know the position of a particle exactly? In that case s=0. In that case we also have t=infinite, and luckily 0*infinity is just a little over h/2."

What is the limit of x approaching 0 of 1/x?

Correct answer; it has no limit. Though often one says that the limit is infinity. If you calculate 1/x for small x, you'll never get to infinity.

Depending on the function, division by zero could result in several potential situations.

1) A non-zero number divided by zero could approach positive infinity from both sides.

2) It may also approach negative infinity from both sides.

3) It may split on both sides, with one approaching positive inifinity, and the other approaching negative infinity.

4) A function which ends up in the form of 0/0 may have an infinite value.

5) A function which ends up in the form of 0/0 may have a finite value.

6) A function which ends up in the form of 0/0 may be indeterminite having several logical answers (or a range of answers). Since they all cannot be true, none can be determined to be true.

if i have 5 apples and you take away 0 groups of my apples you have 0 apples.

Originally posted by TheMaster37
I don't, physisists do.

Physicists can do anything with numbers. So can economists. Not to mention politicians. But mathematicians can't. (Dare I talk about lawyers?)

One of the misconceptions about division by zero is that the infinity is not a part of R. And when you go outside R unsuspected things can happen. Example: 2/0 is 1/0 doubled. Not true.

Another misconception is that 0/0 is not a number. But a limit may approach 0/0 but the denominator can never reach 0, only be approaching. So the limit sin(x)/x approaches 1 when x approaches 0 doesn’t mean that you really divide by zero. Try the same thing with cos(x)/x and don't go outside R and you will see.

One cannot, ever, divide by zero. Not even physicists, if they do proper math.

Originally posted by prosoccer
What happens when you divide by zero? It must be something important because you even have to find NPVs (non-permissible values) so that you will know if a denominator will end up being zero.

E.g. (x+2)/ (y^2-7) NPVs=sqrt7

My lecturer used to say "My mother warned me of two things; ladies of a certain virtue, and dividing by zero!"

The whole subject of analysis is based around convergence to zero. It's tough, but interesting and necessary for everything from advanced physics to simple* integration and differentiation.

*although upon further analysis (hehehe) of the integration and differentiation we find that they are actually very nasty...but you learn them in school so they are seen as simple 'cause everybody knows that integration is the inverse of differentiation...but actually, intuatively, they shouldn't be!...they are nasty!...

What is the limit of x approaching 0 of 1/x?

Correct answer; it has no limit. Though often one says that the limit is infinity. If you calculate 1/x for small x, you'll never get to infinity.

Although this is true, strictly speaking, you can give the answer as infinity. It's not technically true, but because 1/x can be made arbitrarily large with very small values of x, we pretend that it's infinity.

Originally posted by Fwack

What is the limit of x approaching 0 of 1/x?

Correct answer; it has no limit. Though often one says that the limit is infinity. If you calculate 1/x for small x, you'll never get to infinity.

Although this is true, strictly speaking, you can give the answer as infinity. It's not technically true, but because 1/x can be made arbitrarily large with very small values of x, we pretend that it's infinity.

If the right handed limit is not the same as the left handed limit in 1/x when x approaches zero is not the same, there is no limit, even if the two values belongs to R.

In the case of lim 1/abs(x) the two limits are the same but nevertheless not within R, there isn't any limit (within R).

One should never consider the infinity within R, because it isn't. If we define R as one, R*, where there is an infinit value, oo, we get very mysterious results, like oo+1=oo, oo*2=oo, oo^2=oo, even oo^oo=oo, very confusing. It seems that R* demands arithmetic properties you can't use in ordinary R.

Every time anyone says that 0/0 has a value, that you acutally perform a division with zero in sin(x)/x = 1, than you show that you have not any deeper insights in mathematics.

You are absolutely right, Fwack, I just expand the issue a little.

Originally posted by EcstremeVenom
if i have 5 apples and you take away 0 groups of my apples you have 0 apples.

Please go back to school. Here's a quick sample to show what you did wrong...

5/2 .... You have 5 apples and you want to divide them up into groups of 2, how many groups do you form and how many are left over (two and one respectively).

Now if you have 0 groups all you are saying is 5/x where x > 5.

Originally posted by geepamoogle
Depending on the function, division by zero could result in several potential situations.

1) A non-zero number divided by zero could approach positive infinity from both sides.

2) It may also approach negative infinity from both sides.

3) It may split on both sides, with one approaching positive inifinity, and the other approaching negative infi ...[text shortened]... wers (or a range of answers). Since they all cannot be true, none can be determined to be true.

This is rediculous Because you are attributing properties of limits to Funtions.

F(x) = x^x Gives F(0) = DOES NOT EXIST
Limit F(x) as x approaces 1 = 1

Funtions and Limits Have different properties.