Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    Why do those limits don't exist?

    limx-->0 1/x2?

    Aren't we here just approaching "x"? And, thus will not get "0" and avoid the division on "0"?

    limx-->0 |x|/x
    Offline

    0
    ReputationRep:
    limx->0- |x|/x = -1
    limx->0+ |x|/x = 1

    The function has two different one-sided limits at 0, so it has no limit at 0.

    limx->0 1/x2 = infinity

    It does exist. You might have confused that with 1/x, whose limit at 0 does not exist for the same reason as above: it has two different one-sided limits (-infinity and infinity).
    Offline

    12
    ReputationRep:
    (Original post by auger)

    It does exist. You might have confused that with 1/x, whose limit at 0 does not exist for the same reason as above: it has two different one-sided limits (-infinity and infinity).
    Infinity and -infinity are not in the real numbers. If it converges to a limit then the limit is finite in the sense of real analysis. You can extend the real numbers or consider the riemann sphere to talk about convergence to infinity, but judging by the level of the questions I'd say that the expected answer is that 1/x^2 does not converge because infinity is not a real number (this is written as diverges to infinity).
    Offline

    0
    ReputationRep:
    (Original post by IrrationalNumber)
    Infinity and -infinity are not in the real numbers. If it converges to a limit then the limit is finite in the sense of real analysis. You can extend the real numbers or consider the riemann sphere to talk about convergence to infinity, but judging by the level of the questions I'd say that the expected answer is that 1/x^2 does not converge because infinity is not a real number (this is written as diverges to infinity).
    It depends on your definition of a limit, I've always used the extended real number set to define limits. But you would be right, the OP seemed to be sure that the function doesn't have a limit, so you could assume that he only meant the real numbers.
    • PS Helper
    Offline

    14
    PS Helper
    As regards the question, it's a bit ambiguous. You need to make the distinction between a limit existing, and a limit lying in the given space. Here, the space is \mathbb{R}; in the first case, the limit exists, but it doesn't lie in the space. In the second case, the limit simply doesn't exist.

    Elaboration (inc. spoiler)
    \displaystyle \lim_{x \to 0} \dfrac{1}{x^2} doesn't converge in \mathbb{R}, but it does have a limit (namely, \infty), in a similar way that \displaystyle \lim_{x \to 0} x doesn't converge in \mathbb{R} \backslash \{ 0 \} but it does have a limit (namely, zero). Conversely, \displaystyle \lim_{x \to 0} \dfrac{1}{x} has no limit (if we consider \infty and -\infty as distinct "points"), and nor does \lim_{x \to 0} H(x), where H(x) is the Heaviside step function (as it tends to 0 from one side and 1 from the other, regardless of whether we define its value at x=0 or not).
    Offline

    12
    ReputationRep:
    I disagree, I think with no context given most mathematicians would agree that the real number line is meant in a case like this, and I don't really agree that there is a difference between the limit existing and lying in a space: if you wanted to do stuff formally, you would first say what set you are working in, then you would pose the question.
    Once you have done that, limits do or don't exist. They don't 'exist but not in the space' because that's the same as saying they don't exist.

    I mean, this is all just pedantry really, but if we permit the second one to have limit existing then a textbook would have to phrase the sum rule with
    Suppose  a_n \to a and  b_n \to b with  a,b finite. Then  da_n+cb_n \to da+cb
    I just feel like having to do this for every theorem in real analysis would make it lose it's elegance and simplicity.

    (Original post by SWEngineer)
    Aren't we here just approaching "x"? And, thus will not get "0" and avoid the division on "0"?
    It's good that you have made this distinction though - it is irrelevant what happens at the point 0. For example, a function which takes the value 42 at 0 and 1 everywhere else has limit 1 at 0. The problem is that the function grows as it approaches 0, so it has no limit (or the limit exists but isn't in the real numbers, depending on whether or not you agree with the above convention)
    Offline

    12
    ReputationRep:
    Better example: stating the completeness axiom as A sequence is cauchy iff it has a limit would no longer hold and have to be replaced by convergent to a number in the space because sequences that diverge to infinity are not cauchy and examples of such sequences exist.
 
 
 
Poll
Cats or dogs?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.