# Talk:Limit (mathematics)

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Wikipedia Version 1.0 Editorial Team / Vital

## Copy of removed paragraph

Removed this:

====A Brief Note Regarding Division by Zero====

In general, but not in all cases, should u directly substitute c for x (into f(x)) and obtain an illegal fraction with division by zero, check to see whether the numerator equals zero. In cases where such substitution results in 0 / 0, a limit probably exists; in other cases (such as 17 / 0) a limit is less likely. For instance; if f(x) = x³ + 1 / x - 1; then, if one substitutes 1 for x, one will obtain 2 / 0; the limit of f(x) (as x approaches 1) does not exist.

I can't be bothered to do the graph offhand, but there will be a limit: either + or - inf. User:Tarquin

oops Pizza Puzzle

Plus and minus infinity are not limits according to the definition in the article. Please make sure that you have some understanding of the article before you go removing bits. -- Oliver P. 15:42 8 Jun 2003 (UTC)

I'm not aware that infinity is a limit; because, infinity is not a real number and my understanding is that limits must be real numbers. Pizza Puzzle

Yes, that's what I just said. I said it in reply to your statement that "there will be a limit: either + or - inf". If you have changed your mind, and are retracting your previous statement, please replace what you removed from the article. -- Oliver P. 16:02 8 Jun 2003 (UTC)

No sir! I did not state that there will be a limit either + or - inf. The user who does not sign his messages stated that. I have added:

which I believe is what u are referring to above. There is now the question of, if the above user was wrong, does that mean I can reinsert my text:

• For instance; if f(x) = x³ + 1 / x - 1; then, if one substitutes 1 for x, one will obtain 2 / 0; the limit of f(x) (as x approaches 1) does not exist.

or would that be a hostile revert? He had initially removed the entire paragraph, which I put most of it back in, but I didnt put the final line back since there was a debate of sorts regarding it.

## Infinite limit

• As x approaches 0, F(x) = 1 / x² is not approaching a limit as it is unbounded; a function which approaches infinity is not approaching a limit. Note that as x approaches infinity, F(x) = 1 / x² does approach a limit of 0.

Pizza Puzzle

Oh, I see! In that case, I apologise unreservedly for having accused you. I'll blame Tarquin for my error, though, since he was the phantom non-signer. ;) There is a problem in that there are different ways of defining what a limit is. I'll give the article some thought, and come back to it later. I wouldn't object to you putting that example back in, although you should leave out the idea of substitution; a limit only depends on the behaviour as you appraoch the point, not at the point itself. -- Oliver P. 16:15 8 Jun 2003 (UTC)

The subsitution point is, IF you substitute, and you get division by zero, if you get 0 / 0, then there is probably a limit, otherwise there probably isn't. Pizza Puzzle

Oh, I'll think about it later. I should be doing work... -- Oliver P. 16:29 8 Jun 2003 (UTC)

Now here, this text says (in so many words): "The limit, L of f(x), as f(x) increases (or decreases) without bound is an infinite limit. Be sure that you see that the equal sign in "L = infinity" does not mean that the limit exists. Rather, this tells you that the limit fails to exist by being boundless."

It would appear, that it is correct to refer to "infinite limits" but one should understand that an "infinite limit" is not a limit. See also: "unbounded limit" Pizza Puzzle

Would it be too much to expect User: AxelBoldt to explain some of his more "major" changes? It appears that a great deal of information was deleted. If he had a problem with it, it would have been more appropriate to discuss it or improve it; rather than merely deleting it. Pizza Puzzle

Too many subsections before the formal definition. I don't think an encyclopedia article should go that way. I will try to rewrite this later. Wshun

I see limits in this way. If the function is continous for all R then at the limit the function will have a definte value. It doesn't matter if you are trying to find the limit at + or - infinity, or the limit of a function as it approaches a certain value c. In both cases you are dealing with an infinte number of values. If there was no definte value at the limit then limits would'nt be of much use in calculus.

## Inconsistent graphic?

At Limit (mathematics) § Limit of a function, the prose discusses a single scenario, and the right side of the graphic purporting to show it almost shows a zoomed-out view of the left side, but not quite. If the two sides are meant to represent the same thing, the left side needs the vertical line intersection with the x-axis at c - δ to be labeled "S". On the right side, f(x) needs to be equal to L + ε at x = c + δ (i.e. the second hump needs to be above the green-highlighted area). —[AlanM1(talk)]— 23:17, 14 June 2013 (UTC)

## Questionable example

The article states that f(x) = x²-1/x-1 is undefined at x=0. I would disagree, since it can easily be simplified to f(x) = x+1. It seems the same as arguing that x²/x would be undefined at 0 (or actually any g(x)*x/x). I can see how the example is convenient in other ways, because the formula is simple, but I would propose to either replace it by sin(x)/x, or at least note that the statement "f(x) is not defined for x=0" is debatable and that the example was chosen for its simplicity. - Jay 84.171.79.63 (talk) 19:11, 28 June 2014 (UTC)

I don't see your point. The function is not defined at x=0. and sin(x)/x = sinc(x), which usually has sinc(0)=1. — Arthur Rubin (talk) 03:40, 2 July 2014 (UTC)
To clarify... the function in the article is f(x) = (x²-1)/(x-1) (rather than f(x) = x²-1/x-1 ) and the article states that it is not defined at x=1 (rather than at x=0). Meters (talk) 22:45, 7 July 2014 (UTC)
I still don't see the IP's point. Just because $\frac {x^2-1} {x-1}$ can be simplified to x+1, doesn't mean that it is simplified. And our article "sinc" does specify that sinc(x) = sin(x)/x when x ≠ 0. One could make a better case that $e^{x^{-2}}$ isn't defined at x = 0, but that isn't quite correct either, when we work on the extended real line. — Arthur Rubin (talk) 16:08, 9 July 2014 (UTC)
I think the IP's point might be that in practice, other than to come up (in a textbook section about limits) with a function with a specific value excluded, no-one would ever define a function like $\frac {x^2-1} {x-1}$. A less trivial and perhaps somehow "better" example would be, for instance, $f(x) = \frac {\ln{x}} {x-1}$ with a hole at x=1, because it cannot be trivially simplified. - DVdm (talk) 16:53, 9 July 2014 (UTC)