# (x/0) / (x/0) = 1

Hurkyl
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it shows that it is 0.
No, it doesn't show that either.

your right it doesnt show that x/0 / x/0 =0 it shows that the limit is zero, therfore it can be said that it is zero

We all agree that L'Hopital's Rule works, but it just doesn't apply to the original topic.

mathmike said:
your right it doesnt show that x/0 / x/0 =0 it shows that the limit is zero, therfore it can be said that it is zero
That is not true. Just because a limit exists, it does not mean that the point that the limit approaches exists. We see this all the time.

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

But the point (0,1) does not exist on that graph.

AKG said:
I'm guessing you don't know what undefined means. Tell me, is a;kljdfa;lkjasdf greater than ;kljasdf;kljsdf? Does that previous question even make sense?.
but can't you say that "a;kljdfa;lkjasdf" / "a;kljdfa;lkjasdf" equals one, since "a;kljdfa;lkjasdf" = "a;kljdfa;lkjasdf" ?

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Hurkyl
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No, you cannot. "a;kljdfa;lkjasdf" is undefined, and thus so is any expression involving it.

You also cannot say "a;kljdfa;lkjasdf = a;kljdfa;lkjasdf"

Then, how can an undefined entity differ from itself? If they are the same, comparing sames doesn't equal one?

but the implication of [x / 0] / [x / 0] = 1 is perposterous. but you are right in saying that it can be manipulated to get a numerical result

also, can't you apply geomtric reasoning, since we DO know what happens to C/X when X approaches 0?

Hurkyl
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Then, how can an undefined entity differ from itself?
It is just as invalid to say "a;kljdfa;lkjasdf $\neq$ a;kljdfa;lkjasdf" as it is to say "a;kljdfa;lkjasdf = a;kljdfa;lkjasdf".

HallsofIvy
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The point is, you can't write any valid equation involving something that doesn't exist!

mathmike said:
but the implication of [x / 0] / [x / 0] = 1 is perposterous. but you are right in saying that it can be manipulated to get a numerical result
To say that it equals zero is even more preposterous. I think we've reached a conclusion for this thread: abstract concepts cannot follow all of the same rules as concrete numbers.

Gokul43201
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No matter how "abstract" a object is (in mathematics), it is still defined "as rigorously" as something more "concrete". Different objects obey different rules, but some objects don't have laxer rules making them less well-defined.

Agreed. My only point was that you cannot apply all of the rules of algebra to infinity and division by zero.

yes but i can show that the limit is zero when x approches 0 but it cannot be shown in any manner that it is 1. so saying it is zero is not perposterous, in fact it follows l'hopitals therom. can you show that the limit is in any way 1.

Hurkyl
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The limit of the particular function you mentioned is zero. Limits of the form inf/inf generally are not zero.

actually more often than not they are zero.

LeonhardEuler
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What do you mean by "more often than not"? You took the example of f(x)=x/e^x, but one could just as easily take f(x)=(e^x)/x, which clearly approaches infinity as x approaches infinity. And in any case, niether example shows what (1/0)/(1/0) is equal to, they simply show how some particular functions behave as they approach this.