You say you can remove h from the denominator. The reason number cancel out is because a number divided by itself is equal to 1, which is fair enough, unless that number is 0. You can not can 0/0, because 0/0 is not equal to 1.

That is another reason your argument does not work when h = 0.

By the triangle I mean: (f(x + h) - f(x)) / h, where h forms the bottom side and (f(x + h) - f(x)) forms the vertical side, and the hypotenuse is the estimate of the tangent. That equation(f''(x) = (f(x + h) - f(x)) / h) holds only when h is not 0, which is why we use the limit as h->0, as we never have to divide by 0.

Infinity is simply large enough. How large, exactly, depends on your specific problem at hand and the amount of allowable error. The same general idea goes for finding limits, but instead of being large enough, you''re going for close enough. While a function may not be defined at a certain point, we can never actually be at that exact point (for a certainty). We can get close enough so that the error is within an acceptable range, but we never actually have to be there.

Here''s one to chew on: How close does x have to be to 4 for x + x to equal 8? x = 4.00...01? (Then again, how close to 8 is 8?)

karg

You cannot divide by 0, that is why we remove the possiblity to get the 0 in the denominator. We do not divide 0 by 0 but h with h.

Also, it has nothing to do with a triangle, it's slope we cope with.

EDIT: This post was for higherspeed


| Enselic's Corner |

[edited by - Enselic on April 25, 2003 1:02:52 PM]

Your argument is fine, except for the setting of h = 0 at the end as h can''t be 0 anywhere further from the definition at the beginning. However you can say that as h -> 0, 8x + h -> 8x, therefore f''(x) = 8x.

If you set h = 0 at the end, then earlier you have divided 0 by 0.

quote: Original post by higherspeed
However you can say that as h -> 0, 8x + h -> 8x, therefore f''(x) = 8x.

That''s a good way to put it.



| Enselic''s Corner |

quote: Original post by Cold_Steel
Off topic- Hmmm, if we live in a closed universe, would infinity and negative infinity possibly be the same anyway? Food for thought.

This is the Riemann Sphere. Imagine the complex plane "rolled" into a sphere, each and every complex infinity would be the same. ( There are better topological explanations ).

quote: Original post by Anonymous Poster

Clearly, something cannot be defined to be valued both -oo, +oo and 1 at the same time.

You should ask Schroedinger about that ;-)




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Geez...
I read about a page and a half of this and it seems that only 1 or 2 people are saying anything meaningful at all. You are all just making stuff up off the top of your head!
Think about it, using the rules of COMPLEX NUMBER ALGEBRA, you cannot consistently define z / 0 to be a single value because division is supposed to be the inverse of multiplication. If z / 0 = w, then w * 0 = z, which implies z = 0. But this equation would then be true for all complex numbers w. Like, the anomynous poster said, if you define division by 0 (at least as a mapping from the complex plane to the complex plane - including infinity,) you would be able to prove that 1 = 2 (because w can be 1 or 2.) Hence the result of division by 0 could not be defined as a complex number. If you defined it as something else, then we could no longer say that the complex numbers formed a field - which is a generization of REAL NUMBER ALGEBRA. Very sad. As for this:

quote: Original post by CoffeeMug
1 / 0 = x
0x = 1
0 = 1

I''m not sure what this is trying to do, but the rules of COMPLEX NUMBER ALGEBRA do not define any of the operations performed here. E.g., you are ASSUMING that you can "cancel" 0 / 0. This should read 1 / 0 = x. => 0 * 1 / 0 = x * 0. Now, who told you that 0 * 1 / 0 = 1? In fact, you are also saying that x = 1 / 0, but x * 0 = 0, make up your mind! This is pointless.
The reason sqrt (-1) is defined is because it CAN be, division by 0 cannot - at least not on complex numbers.

I think that perhaps this topic should be laid to rest now. Also Gauler, although I agree with you that what most of what has been said is rubbish, I think you have to remember that most people have only studied applied maths. I''ve only done a small amount of pure maths at A-Level involving groups. I''ll be doing plenty of algebra next year at Uni though. But most people don''t have the benefit of any course, so don''t seem surprised when they don''t understand what exactly they''re doing.

the day we figure out what infinity is is the day we find god, the day we find the universal equation the day we find... etc etc.

stupid question imo ;/ sorry

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