Turn on thread page Beta
    Offline

    2
    ReputationRep:
    (Original post by RichE)
    Maybe I shouldn't lecture him until he's in Oxford :rolleyes: :eek:
    LOL!
    • Thread Starter
    Offline

    15
    ReputationRep:
    Nope
    Lol well do my employers? It's a Thursday and i'm supposedly contracted to work :eek:

    Maybe I shouldn't lecture him until he's in Oxford
    Offline

    0
    ReputationRep:
    (Original post by Phil23)
    do you use your holidays what they are used for :rolleyes: ...sleeping..resting....sleeping some more?!?!?

    STOP STUDYING:eek:, lol
    why did i get so much rep for this:eek: - can't anyone on TSR appreciate a joke now and then - why is everyone so serious :mad:

    my positve 4 rep points was a record whilst it lasted, for just over 24 hours :cool: - can anyone TOP me up please...dont really like the red diamond :rolleyes:

    :tsr:
    :cheers: :party:
    Offline

    0
    ReputationRep:
    (Original post by RichE)
    Well maybe I'm being too hard on Phil. Maybe I shouldn't lecture him until he's in Oxford :rolleyes: :eek:
    i'd prefer it if you didn't lecture me when and if i get to oxford :rolleyes:
    Offline

    15
    ReputationRep:
    (Original post by Phil23)
    i'd prefer it if you didn't lecture me when and if i get to oxford :rolleyes:
    I also but mysterious ways and all that...
    Offline

    0
    ReputationRep:
    (Original post by RichE)
    I also but mysterious ways and all that...
    in english please :rolleyes:
    • Thread Starter
    Offline

    15
    ReputationRep:
    I have another question. This is concerned with functions rather than series but is still analysis.

    The book states:
    Theorem 4.1.11: Suppose that f(x) and g(x) are defined for all x<a and satisfy the inequality g(x)\geqf(x) for all x<a, where a is some given real number. If g(x)\rightarrow \infty as x\rightarrow -\infty, then f(x)\rightarrow \infty as x\rightarrow -\infty

    I suspect that there is a typo in this theorem however.
    g(x) \rightarrow \infty as x \rightarrow -\infty implies:
    g(x)&gt;A \forall x&lt;X (X&lt;a)
    But g(x) \geq f(x) does not imply that f(x)&gt;A \forall x&lt;X and so we have not guaranteed f(x) \rightarrow \infty as x \rightarrow -\infty.

    Should it instead be worded:
    Theorem 4.1.11: Suppose that f(x) and g(x) are defined for all x<a and satisfy the inequality g(x)\leqf(x) for all x<a, where a is some given real number. If g(x)\rightarrow \infty as x\rightarrow -\infty, then f(x)\rightarrow \infty as x\rightarrow -\infty

    Here note:
    A&lt;g(x)\leq f(x) \forall x&lt;X and so f(x)&gt;A \forall x&lt;X so f(x) \rightarrow \infty as x \rightarrow -\infty

    Thanks.
    Offline

    15
    ReputationRep:
    In my copy of Hart it is actually printed as you suggest it should. So maybe you have an earlier edition and the typo was noticed later - certainly it isn't true as you first present it.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by RichE)
    In my copy of Hart it is actually printed as you suggest it should. So maybe you have an earlier edition and the typo was noticed later - certainly it isn't true as you first present it.
    I bought my copy (second edition) at the start of this month. Perhaps it's specific to a few copies.
    In any case, thanks for the confirmation that there's a typo in my copy - thankfully it's not an error in my understanding.
    • Thread Starter
    Offline

    15
    ReputationRep:
    Am I right in thinking that \lim_{ x \rightarrow 1} \frac{cosx}{1-x} does not exist as \lim_{ x \rightarrow 1-} \frac{cosx}{1-x} \neq \lim_{ x \rightarrow 1+} \frac{cosx}{1-x}?
    I think the author of the book has used l'Hopital's rule inappropriately as cos1 \neq 0 but they have given the answer \lim_{ x \rightarrow 1} \frac{cosx}{1-x} = sin1
    Offline

    15
    ReputationRep:
    (Original post by Gaz031)
    Am I right in thinking that \lim_{ x \rightarrow 1} \frac{cosx}{1-x} does not exist as \lim_{ x \rightarrow 1-} \frac{cosx}{1-x} \neq \lim_{ x \rightarrow 1+} \frac{cosx}{1-x}?
    I think the author of the book has used l'Hopital's rule inappropriately as cos1 \neq 0 but they have given the answer \lim_{ x \rightarrow 1} \frac{cosx}{1-x} = sin1
    It doesn't exist from either side unless you include plus or minus infinity in your definition of a limit. Yes the book seems wrong. Is this Hart? (again!?)
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by RichE)
    It doesn't exist from either side unless you include plus or minus infinity in your definition of a limit.
    I usually count \pm \infty as limits and see things like \lim_{x \rightarrow \infty} sinx as not having a limit.

    Yes the book seems wrong. Is this Hart? (again!?)
    Yes, not to worry. Thanks for the confirmation.
 
 
 
Turn on thread page Beta
Updated: July 26, 2005

3,836

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
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.