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- Thread starter Helicobacter
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StatusX

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Are you asking if a function approaches infinity at a point from both directions (like 1/x^{2} at x=0, but not like 1/x), is its limit at this point infinity? I would say yes, although some people may say the limit does not exist (because infinity is not a real number).

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StatusX said:Are you asking if a function approaches infinity at a point from both directions (like 1/x^{2}at x=0, but not like 1/x), is its limit at this point infinity? I would say yes, although some people may say the limit does not exist (because infinity is not a real number).

Actually, I was thinking about something similar to -(1/x

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StatusX

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Hurkyl

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One of the drawbacks of having definite answers is that math requires precisely stated problems.

In particular, you haven't specified in what number system you are taking the limit.

If you're working over the reals, then the limit doesn't exist.

If you're working over the extended reals, then that limit is equal to positive infinity. (the extended reals add two points "at infinity" to the real line, which we name positive infinity and negative infinity)

If you're working over the projective reals, then that limit is equal to infinity. (the projective reals have only one point "at infinity")

Your function doesn't have to be exactly symmetrical. For example, the following function is a**continuous** extended-real-valued function that is everywhere defined on the extended reals

[tex]

f(x) := \begin{cases}

-\frac{1}{x} & x < 0 \\

+\infty & x = 0 \\

14 + \frac{3}{x^2} & x > 0

[/tex]

(note that [itex]f(-\infty) = 0[/itex] and [itex]f(+\infty) = 14[/itex])

In particular, you haven't specified in what number system you are taking the limit.

If you're working over the reals, then the limit doesn't exist.

If you're working over the extended reals, then that limit is equal to positive infinity. (the extended reals add two points "at infinity" to the real line, which we name positive infinity and negative infinity)

If you're working over the projective reals, then that limit is equal to infinity. (the projective reals have only one point "at infinity")

Your function doesn't have to be exactly symmetrical. For example, the following function is a

[tex]

f(x) := \begin{cases}

-\frac{1}{x} & x < 0 \\

+\infty & x = 0 \\

14 + \frac{3}{x^2} & x > 0

[/tex]

(note that [itex]f(-\infty) = 0[/itex] and [itex]f(+\infty) = 14[/itex])

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StatusX

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Maybe a mathematician can tell you what they call it, but like I said, it doesn't really matter. It's just a name. If this is a step in a larger problem, what you call your result at this step doesn't matter, as long as you apply it correctly to the next one. If this is your final answer to a problem, say, on a test, I think either choice should be accepted. When talking with other people about a problem, you should try to be extra clear in these circumstances, as I don't think everyone would use the same convention. Perhaps there is a preference to call a series that has an infinite limit a "divergent series" rather than saying it "converges to infinity", where as it is more likely that function like 1/x^2 would be said to have a limit of infinity at x=0.

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