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# Explain this watch

1. (x-1)/(x^2-1) = (x-1)/(x-1).(x+1) = 1/(x+1)

If I set x = 1, then why does (x-1)/(x^2-1) = 0/0

but 1/(x+1) = 1/2
2. (x-1)/(x^2-1) isn't defined at x=1 but the limit as x tends to 1 exists and is 1/2
3. mind = blown
4. (Original post by SsEe)
(x-1)/(x^2-1) isn't defined at x=1 but the limit as x tends to 1 exists and is 1/2
so x=1 isnt defined at (x-1)/(x^2-1)

but x =1 is defined at 1/(x+1)

despite the fact that (x-1)/(x^2-1) = 1/(x+1)
so x=1 isnt defined at (x-1)/(x^2-1)

but x =1 is defined at 1/(x+1)

despite the fact that (x-1)/(x^2-1) = 1/(x+1)
If you are talking about graphs, and geometrically representing these as functions, then I don't think you can say that

While elementary algebra gives an apparent equivalence between the two, geometrical and algebraic forms are still separate representations, and care must be taken converting from one form to the other.

If you want one way of thinking about it, the algebraic manipulation you have gone through is simply a way of 'ignoring' the presence of the asymptote. Try plotting both graphs and comparing their shapes, maybe.
6. (Original post by james.h)
If you are talking about graphs, and geometrically representing these as functions, then I don't think you can say that

While elementary algebra gives an apparent equivalence between the two, geometrical and algebraic forms are still separate representations, and care must be taken converting from one form to the other.

If you want one way of thinking about it, the algebraic manipulation you have gone through is simply a way of 'ignoring' the presence of the asymptote. Try plotting both graphs and comparing their shapes, maybe.
They have identical shapes...
so x=1 isnt defined at (x-1)/(x^2-1)

but x =1 is defined at 1/(x+1)

despite the fact that (x-1)/(x^2-1) = 1/(x+1)
Both are defined and equal everywhere except x=1 and x=-1. At x=-1, neither are defined. At x=1, the former is not defined (but has a limit as x tends to 1) and the latter is defined. So really, they're not strictly identical.
8. You have kind of divided by (x-1), which when x=1 is 0, which isn't allowed. Naughty
9. (Original post by MathsHamster)
You have kind of divided by (x-1), which when x=1 is 0, which isn't allowed. Naughty
Ha! Can't believe I missed that.
10. (Original post by MathsHamster)
You have kind of divided by (x-1), which when x=1 is 0, which isn't allowed. Naughty
(x-1)/(x^2-1) = 1/(x+1)
This isn't true. It's "almost" true, in various very specific and technical senses, but it's not actually true, because (as you said yourself) the left hand side isn't quite defined at x = 1 (it is trivially easy to work out how you'd like to define it there, but it is not defined there by default, because the actual theory behind showing you can do so isn't easy), and the right hand side is. They behave identically everywhere else.

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