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Basic doubt in limits

  1. Sep 11, 2013 #1
    When the limit of a function turns out to be infinity ,how should we interpret it ?

    For ex. we have [itex]\lim_{x \to 0}\frac {1}{x^2} = \infty[/itex]

    What does this mean ?

    Does it mean that the limit is ∞ or does it mean that the limit does not exist as ∞ is not a number ?

    I would appreciate if someone could clarify this doubt .
  2. jcsd
  3. Sep 11, 2013 #2
    In your example the limit as x approaches 0 is ∞, that's what the limit equals. If you were to plug in 0 into the equation you would get undefined since you can't divide by 0. The limit and the actual value are two different things. What it's really saying is that the limit is not approaching any particular value which makes sense since the value it's approaching would give you an undefined value. Hope this clears up something?
  4. Sep 11, 2013 #3
    Do we say that the limit exists and is equal to infinity ?
  5. Sep 11, 2013 #4
    I guess this is a better way of explaining it and I realized that I was misleading in my first message, sorry. A limit only exists if there is a continuous function. Since your equation is not continuous at that point then a limit does not exist.
  6. Sep 11, 2013 #5


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    By definition, a limit which tends to infinity, plus or minus, does not exist. A limit exists only when a finite number, however large or small in magnitude, is obtained.
  7. Sep 11, 2013 #6
    Thanks SteamKing...

    But then what is the difference between

    1. [itex]\lim_{x \to 0}\frac {1}{x^2} = \infty[/itex]

    2. [itex]\lim_{x \to 0}\frac {1}{x}[/itex] = does not exist

    I understand how the second one has different right and left limits (one infinity other -infinity),thereby limit not existing.

    But why dont we write the limit as does not exist in the first case and instead write the limit as infinity.(Afterall infinity means limit does not exist)
  8. Sep 11, 2013 #7


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    The difference can be summed up as follows:

    1. is a nice form of non-existence, 2. is an unnice form of non-existence. :smile:
  9. Sep 11, 2013 #8


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    If a limit equals infinity, it technically does not exist. We say it equals infinity though because it is a good description of the behavior of the function (which is the purpose of calculating limits, see what the function does as x gets close to that point or gets really big).

    The distinction is very unimportant when someone just asks you to calculate a limit and say whether it exists, unless they are being a pendantic grader or something. However, there are many theorems about existence of limits where you have to be careful. For example,
    [tex] \lim_{x\to 0} f(x) + g(x) = \lim_{x\to 0} f(x) + \lim_{x\to 0} g(x) [/tex]
    is true if both limits on the right hand side exist. In particular, they cannot be infinity. So if you write down
    [tex] 0 = \lim_{x\to 0} \frac{1}{x^2} - \frac{1}{x^2} = \lim_{x\to 0} \frac{1}{x^2} - \lim_{x\to 0} \frac{1}{x^2} = \infty - \infty [/tex]
    then what you have makes no sense at all.

    On the other hand knowing the limit is infinity can be useful in certain situations. For example, if f(x) is continuous then
    [tex] \lim_{x\to a} f(x) = f(a) [/tex]
    Composing two functions gives us
    [tex] \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) [/tex]
    as long as that inside limit exists (i.e. is a number). But even if it doesn't exist, if it's infinity we still might be able to get a result. For example

    [tex] \lim_{x\to 0} e^{-1/x^2} = e^{-\lim_{x\to 0} 1/x^2} = 0[/tex]
    Even though the inside limit doesnt exist, knowing it's "equal" to infinity lets me figure out what this bigger limit is equal to
  10. Sep 11, 2013 #9


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    No, if a function is continuous at a point, then by definition of continuity, it has a limit there.
    But a limit can exist even though it is not continuous there.
  11. Sep 12, 2013 #10
    Brilliant explaination!!! Thanks Office_Shredder :smile:
  12. Sep 12, 2013 #11
    the simplest and most straightforward reason why [itex]\lim_{x \to 0}\frac {1}{x}[/itex] does not exist is the very same reason you understood [itex]\lim_{x \to 0}\frac {1}{x^2}[/itex] to be [itex]\infty[/itex]. [itex]\lim_{x \to 0}\frac {1}{x}[/itex] = DNE b/c [itex]\lim_{x+ \to 0}\frac {1}{x}[/itex] = [itex]\infty[/itex] and [itex]\lim_{x- \to 0}\frac {1}{x}[/itex] = [itex]-\infty[/itex]. but [itex]\lim_{x \to 0}\frac {1}{x^2}[/itex] = positive [itex]\infty[/itex] whether you approach 0 from the left or the right. since the limit of this function is the same from both a left approach and a right approach, the limit is defined as its calculated value (even if [itex]\infty[/itex] is not a number or a well defined limit).
  13. Sep 12, 2013 #12

    Stephen Tashi

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    Read the relevant definitions in your course materials. There is more than one kind of limit - just as there is more than one kind of "trig function". The term "trig function" is ambiguous; it doesn't refer to one particular function. Likewise the notation [itex] \lim_{x \rightarrow a} f(x) [/itex] doesn't refer to one particular kind of limit.

    If you are using mathematically respectable study materials, there is a definition for [itex] \lim_{x \rightarrow a} f(x) = L [/itex] when [itex] L [/itex] is a number and there is a different definition for [itex] \lim_{x \rightarrow a} f(x) = \infty [/itex].

    When [itex] lim_{x \rightarrow a} f(x) = \infty [/itex] is satisifed by [itex]f [/itex] and [itex] a [/itex] in the sense of the second definition then [itex] lim_{x \rightarrow a} f(x) [/itex] does not exist in the sense of the first definition.

    As you pointed out, It is possible to have situation where neither definition is satisfied.
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