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Limit question

  1. Jul 7, 2006 #1
    Lets say you have a function that is constant except at interval [-2,2] where it drains down to infinity. The whole function is reflected symetrically at x=0. Is the limit of this function, as x approaches 0, negative infinity?
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  3. Jul 7, 2006 #2


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    Are you asking if a function approaches infinity at a point from both directions (like 1/x2 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).
    Last edited: Jul 7, 2006
  4. Jul 7, 2006 #3
    Actually, I was thinking about something similar to -(1/x2), but it doesn't matter. Is the limit undefined or infinity?
    Last edited: Jul 7, 2006
  5. Jul 7, 2006 #4


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    Infinity isn't a real number, so you might argue it can't be the result of any limit, and you could just say that a sequence or function with an infinite limit diverges, or has no limit. On the other hand, if you're interested in exactly how it diverges (positive or negative infinity), or if you want to distinguish functions that diverge nicely (like 1/x^2) from those whose limit is truly undefined (like 1/x or sin(1/x)), then you might say it has an infinite limit. But it's just a matter of convention. I don't think there is a universal consensus one way or the other (it varies from problem to problem and mathematician to mathematician), but it doesn't really matter.
  6. Jul 7, 2006 #5
    I generally thought that mathematics is one of the few areas where answers are always definite. I've read that infinity does not equal infinity. To my logic I agree that when you deal with infinity, limited quantities become sometimes undiscernable, but in this specific function both sides have identical values if reflected at x=0..
  7. Jul 7, 2006 #6


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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

    f(x) := \begin{cases}
    -\frac{1}{x} & x < 0 \\
    +\infty & x = 0 \\
    14 + \frac{3}{x^2} & x > 0

    (note that [itex]f(-\infty) = 0[/itex] and [itex]f(+\infty) = 14[/itex])
    Last edited: Jul 7, 2006
  8. Jul 7, 2006 #7


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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.
    Last edited: Jul 7, 2006
  9. Jul 7, 2006 #8
    well...is it...i can't picture it. you mean it goes constant and it starts falling at x=-2 and has a vertical asymptote at x=2? like a...parabola that is opened diagonally towards the third quadrant? if so it depends on the fall. If from x=-2 to x=2 it keeps on falling...it depends HOW it falls.
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