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Limits and their algebra

  1. Dec 26, 2011 #1
    I have the following doubts. Please help me get through them.....

    (1) Is there any difference between the terms 'limit does not exist' and 'limit is not defined'?

    (2) For a limit to exist does it have to be a finite number?

    (3) When we do the algebra of limits as lim (x->a) [f(x) + g(x)] = [lim (x->a) f(x)] + [lim (x->a) g(x)], is it the necessary condition for both the limits to be finite?

    (4) When can we write 1/x as 0, as x->0. (This part is very confusing. Some insight or examples would be great.)

    (5)What is the basis of the L'Hospital's Rule???

    (6) Are there any 'if s' to the Taylor series. If Yes, what are they???
  2. jcsd
  3. Dec 26, 2011 #2
    I can help you with a few, but for (5) and (6) I suggest you try to read your calculus book a little more.

    1 - Yes, there is a difference. To say that a limit "does not exist" means that (in the 2 dimensional sense) the function doesn't behave in the same way when approached from the left or the right, and also that those two limits might not be the same as the value of the function at that point. For a limit to exist: the limit on the right (as you approach from the right side), and limit on the left, and the value of the function at that point (f(a)) must all be equal. This isn't the rigorous definition, but it's what is normally taught to first year calc students.

    To say that a limit" is undefined means that you take the limit doesn't make sense. For example, take the function f(x) = √(1-x), and consider the limit as x approaches 2 of f(x). This is undefined (for now) as f(x) is not defined (in the real numbers) when there's a negative number under the square root. You can make it defined by removing either the left-sided limit or the right-sided limit (depending on the function).

    2 - No, it does not have to be a finite number. A limit could very well be infinity. Consider the function :
    f(x) = \left |{\frac{1}{x}} \right |
    The limit as x approaches 0 from the left and the right is positive infinity. Check this from the graph as well:

    3 - No, it is not necessary, as they are now just limits on their own.
    4 - The limit of 1/x as x approaches 0 is not 0. From the left it approaches negative infinity, and from the right it approaches positive infinity. Here's a graph to help you see what I'm talking about:
  4. Dec 26, 2011 #3


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    Depends. If you're working in the real number system, yes. If you're working in the extended real number system, no.

    Not necessarily. However, it is possible to run across the indeterminate form:
    ∞ - ∞

    Never. However:

    [tex]\lim_{x \to 0} \frac{1}{|x|} = \infty[/tex]

    I'm not sure why you think it's 0?

    That's a little vague. What do you mean?
  5. Dec 26, 2011 #4
    I disagree with (2) and (3).

    I always learned that for a limit to exist, it had to be finite. So if the limit is infinite, then it is said to not exist. However, this depends on the book you use.

    For (3), it IS crucial that the limits are finite! If the limit is infinite or doesn't exist, then the sum might not be well-defined!!!! You must be EXTREMELY careful when you split the limit of a sum in two terms!!
  6. Dec 26, 2011 #5


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    This is not quite right. If we are working withing the extended real number or the projective real numbers, then there are infinite limits. If we are working within the real numbers however, then a limit cannot be infinity (since infinity is not a part of our number system).

    If we are working within ℝ, people often use the shorthand limx→0|x-1| = +∞, but this really means that the limit does not exist in a particular way. It does not mean that the limit is actually infinity.

    Edit: So many replies in such a short time, haha
  7. Dec 26, 2011 #6

    Stephen Tashi

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    Ashu2912's confusion about the algebra of limits appears to justified!

    I don't know whether modern calculus texts provide any formal axioms for "the extended real numbers". The one's I have (from 20 years ago) do not. They merely had a set of verbal statements that said when particular algebraic rules applied and in some rules they allowed terms to be limits equal to [itex] \infty [/itex] or [itex] - \infty [/itex]. You had to read the fine print to see what was going on. You couldn't memorize any simple set of guidelines and assume they applied to all the rules.

    In mathematics you have to read the fine print. I suspect most people don't memorize the fine print about the laws of limits. Instead, they form a good intuition about how limits behave and use that in place of remembering what their text said years ago.

    There are formal definitions for [itex] \lim_{x \rightarrow a }f(x) = \infty [/itex] and [itex] \lim_{x \rightarrow \infty} f(x) = \infty [/itex] and so forth. These don't entail all the ways that a limit can fail to exist as a finite limit. A function can oscillate "infinitely fast". Or a function can simply be undefined in an entire interval around where you are trying to take the limit. For example, you can define a function f(x) to be equal to 6 on all real numbers where |x| > 1 and leave it undefined elsewhere. Then [itex] lim_{x \rightarrow 0} f(x) [/itex] doesn't exist.

    My old calculus books often used phrases like "if the limits exist or are infinite" to describe the restrictions that applied to a rule. They considered an "infinite" limit to be a special case of a limit not existing.

    As to l'Hospital's rule, I remember hunting down proofs for all the cases and reading them. There are a few cases of it that are hard to prove and weren't easy to find in books. Now that you have the web, I think you should have no trouble. However, don't expect to find a proof that treats each case ( [itex] 0/0, \infty/\infty,...[/itex] etc.in the same fashion. The cases are fundamentally different situations.
  8. Dec 26, 2011 #7


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    ^ this.

    depending on what meaning you assign to the symbol "∞" you can have one of the following 3 outcomes:

    (a) [itex]\lim_{x \to 0} \frac{1}{x} \text{ is undefined.}[/itex]

    (b) [itex]\lim_{x \to 0} \frac{1}{x} \text{ does not exist.}[/itex]

    (c) [itex]\lim_{x \to 0} \frac{1}{x} = \infty[/itex].

    i believe that is not useful, for a "first look" at calculus, to allow "infinite limits". yes, we can extend the real number system to include non-finite quantities, but doing so "muddies the waters". R U {-∞,∞} is no longer a field, and R U {∞} (the projective reals) is no longer an ordered field (or even an ordered ring).

    it is still possible to assign meanings to:

    [itex]\lim_{x \to a} = \infty[/itex] and [itex]\lim_{x \to \infty} = L[/itex], but for such statements to be meaningful within the context of the real ordered field of real numbers, we have to state them SOLELY in terms of real numbers. in this view of things "∞" is not, and does not become, a real number, or any "extension" of a real number, but a SHORTHAND, for certain statements about real numbers.

    why is such caution prudent? because "intutive" notions about "infinity" (as conceived by inductive reasoning from finite numbers) simply do not hold. while it is certainly possible to create an axiom system that allows for "extending the real numbers to include infinity", that axiom system is certainly more complicated that the axioms for an ordered field, and one has to separate out the "infinite cases from the non-infinite cases" (this becomes particularly important with the projective reals and any theorem involving inequalities).

    that is, before one starts messing around with "improper limits", one should have a firm grasp of "proper limits", that is, when a limit is a real number, AT a real number.
  9. Dec 27, 2011 #8
    Hey friends! Thanks for your replies! :smile: I have a few observations...and clarifications :

    Consider : lim(x to [itex]\infty[/itex]) x/|x|, |x| being the modulus function.
    Now lim (x to [itex]\infty[/itex])x = [itex]\infty[/itex] and lim(x to [itex]\infty[/itex]) 1/|x| = 0. However, the actual limit is 1.

    Is the reasoning correct?????

    Sorry I meant When can we write 1/x as 0, as x->[itex]\infty[/itex]. (This part is very confusing. Some insight or examples would be great.)

    I meant that what are the requirements if we want to apply the Taylor series??? Like some restriction on the function or any variable..... I just want the conditions we need to apply it..
  10. Dec 27, 2011 #9
    Being short on time, I overlooked the separation of limits, there are special cases to consider. However, as even micromass pointed out, a limit can be infinity in the reals, it just depends on what book you've learned from.
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