How to make crazy work:
Say \(x\) is a number arbitrarily close to \(10\) but not exactly \(10\). We need to find \(\lim_{x\to 10} \frac{x^2-100}{x-10}=\lim_{x\to 10} x+10=20\)

Adapting this idea, you could arrive at any numerical answer you want. Hence, limits would be unable to resolve this issue, as it is highly dependent on the path that you arrive at the limit.

Note: A similar problem is to define \(0^0\) though limits. "Most of the time" we get 1, though if you take \( \lim_{x \rightarrow 0} 0^x\), you will get 0.

Don't u know when we write (x^2-y^2)/(x-y)=(x+y) ,it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =(x+y) as result,..here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong

answer is 20!
may sound crazy. But as I go thru discussions, the fact x + y could be the answer is presented.

whenever you encounter 'zero upon zero' situation, it is of course 'indeterminate' - but it also means that is not the end. There is a real answer different from 'indeterminate'. You may arrive at it by suitable algebraic simplification / trigonometry simplification etc.

There are even a few theorems to work out the real answers

The question is not (x^2-100)/(x-10) with x tending to zero. We can interpret (100-100)/(10-10) in different ways and get different answers as Calvin said. So the answer does not necessarily have to be 20.
For eg. the question can be interpreted as lim x->10 [(x^3-9x^2-100)/(x-10)] which gives the answer as 120. So I think answer is undefined.

Don't u know when we write [a(x-y)/(x-y)=a] it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =a as result...here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong...

Undefined. The division of 0 is not allowed in Maths since you can get a variety of answers. Since anything mutiplied by 0 is 0, 0 divided by 0 is everything. Like wise, nothing multiplied by 0 is not 0, hence any other number divided by 0 is just not possible

Hey you can't divide by zero ..... there are plethora of mathematical contradictions and absurdities based on division by zero ....... If you divide by zero serious flaws creep into your logic leading to absolute fallacies

We can only cancel the term taking the supposition that it is not zero. So, if you want to cancel a term you will have to assume that it is not zero. So, in this case you cannot cancel the term my friend. so, your answer is totally wrong.

Don't u know when we write (x^2-y^2)/(x-y)=(x+y) ,it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =(x+y) as result,..here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong

hi
it's not possible, you cannot divide any real by 0, so dividing 0 by it's self is one of the craziest thing I saw, since a/0 might me greater than infinit, because that infinit times 0 is still 0, but I cannot guess what might be 0/0, since it could be \(0^{23-5}=0\) , \(0^{1-23}=1/0\) or even \(0^{0}\) or 0^ any number ,we can accept the strictly positive ones, but not the the others , so this is particularly weird I think

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## Comments

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TopNewestNot exactly "undefined"; a more fitting term would be "indeterminate", because any number \(x\) multiplied by zero is zero.

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How to makeSay \(x\) is a number arbitrarily close to \(10\) but not exactly \(10\). We need to find \(\lim_{x\to 10} \frac{x^2-100}{x-10}=\lim_{x\to 10} x+10=20\)crazywork:Log in to reply

The main problem with this is that your expression is 'completely arbitrary'. For example, I could say that

\[ \lim_{x \rightarrow 10} \frac { \frac {1}{2} x^2 + 5x - 100}{ x-10} = \lim_{x \rightarrow 10} \frac {1}{2} x + 10 = 15.\]

Adapting this idea, you could arrive at any numerical answer you want. Hence, limits would be unable to resolve this issue, as it is highly dependent on the path that you arrive at the limit.

Note: A similar problem is to define \(0^0\) though limits. "Most of the time" we get 1, though if you take \( \lim_{x \rightarrow 0} 0^x\), you will get 0.

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as from what i learned the real and specific term for 0/0 is INDETERMINATE not undefined

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UNDEFINED

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i wish mathematics had something better to offer :/

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Comment deleted Feb 18, 2013

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Don't u know when we write (x^2-y^2)/(x-y)=(x+y) ,it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =(x+y) as result,..here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong

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there is nothing wrong with the solution. study about "limits" bro, it helps sometimes.

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why is sayan C spamming :\

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answer is 20! may sound crazy. But as I go thru discussions, the fact x + y could be the answer is presented.

whenever you encounter 'zero upon zero' situation, it is of course 'indeterminate' - but it also means that is not the end. There is a real answer different from 'indeterminate'. You may arrive at it by suitable algebraic simplification / trigonometry simplification etc.

There are even a few theorems to work out the real answers

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The question is not (x^2-100)/(x-10) with x tending to zero. We can interpret (100-100)/(10-10) in different ways and get different answers as Calvin said. So the answer does not necessarily have to be 20. For eg. the question can be interpreted as lim x->10 [(x^3-9x^2-100)/(x-10)] which gives the answer as 120. So I think answer is undefined.

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would u please state your ' a few theorem'...i do not know it...

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remember ....it is the problem only (100-100)/(10-10)=?...& not anything should be thought or taken into account....

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many possible answers are there.....like 10,20,undefuned,etc... depending on the person who solve it...

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u can not divide anything by zero.

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Though on the

{If that was what u meant by "interesting", \((100-100)/(10-10)\) = \(10(10-10)/(10-10)\) = \(10\) }*crazy side*Log in to reply

no cancellation rule applies for zero

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in my comment??*"crazy side"*Log in to reply

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Don't u know when we write [a(x-y)/(x-y)=a] it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =a as result...here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong...

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side"*crazy*Log in to reply

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cant be determined

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indeterminate

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It is Undetermined or Undefined.

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and UNDEFINED tooo

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(100-100)/(10-10) is nothing but indeterminate form ....which cannot be solved...we may solve it using limits

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0/0

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UNDEFINED

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Undefined. The division of 0 is not allowed in Maths since you can get a variety of answers. Since anything mutiplied by 0 is 0, 0 divided by 0 is everything. Like wise, nothing multiplied by 0 is not 0, hence any other number divided by 0 is just not possible

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Hey you can't divide by zero ..... there are plethora of mathematical contradictions and absurdities based on division by zero ....... If you divide by zero serious flaws creep into your logic leading to absolute fallacies

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it can be (10^{2}-10^{2})/(10-10) =(10+10)(10-10)/(10-10) =20

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how can you cancel zero by zero?????

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i agree with that point

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We can only cancel the term taking the supposition that it is not zero. So, if you want to cancel a term you will have to assume that it is not zero. So, in this case you cannot cancel the term my friend. so, your answer is totally wrong.

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It

be 20 :canhttps://brilliant.org/discussions/thread/an-interesting-question/#comment-a3f52161e92

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Don't u know when we write (x^2-y^2)/(x-y)=(x+y) ,it is given or we assume that, x is not equal to y...or (x-y)is not equal to zero....and only then we divide both the numerator and denominator by (x-y)...and we get =(x+y) as result,..here we are seeing clearly (10-10)=0 so we cannot proceed with such a operation...and the result we would get be absolute wrong

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hi it's not possible, you cannot divide any real by 0, so dividing 0 by it's self is one of the craziest thing I saw, since a/0 might me greater than infinit, because that infinit times 0 is still 0, but I cannot guess what might be 0/0, since it could be \(0^{23-5}=0\) , \(0^{1-23}=1/0\) or even \(0^{0}\) or 0^ any number ,we can accept the strictly positive ones, but not the the others , so this is particularly weird I think

but I guess you know all this stuff already ^^

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use limit bro

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0

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10(10-10)/(10-10) = 10

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