Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Limits with x to infinity

  • 22-07-2011 3:13pm
    #1
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    In this line of a the problem attached

    (taken from the following problem)
    http://www.sosmath.com/calculus/limcon/limcon04/answer4.html

    how is 4 over 2+SqRt4 arrived at?

    If x is going to infinity, and there is an x on top & bottom, doesn't both top & bottom just go to infinity?

    I realise this mustn't be the case, so I'm wondering what I'm doing wrong. Thank you


Comments

  • #2
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    Hi op.

    You need the previous line from that website you posted. If you let x go to infinity in that line you get the answer. For example the 4/x bit. If you let x get very large the fraction 4/x will get so small, it'll go to zero. Apply the same logic to each component with an x in it.


  • #3
    shizz
    Registered Users, Registered Users 2 Posts: 1,506 ✭✭✭shizz


    TheBody wrote: »
    Hi op.

    You need the previous line from that website you posted. If you let x go to infinity in that line you get the answer. For example the 4/x bit. If you let x get very large the fraction 4/x will get so small, it'll go to zero. Apply the same logic to each component with an x in it.

    In the line that you are referring to, how where they able to divide above and below by x but they divided what was in the squared root by x^2 and not x?


  • #4
    shizz
    Registered Users, Registered Users 2 Posts: 1,506 ✭✭✭shizz


    shizz wrote: »
    In the line that you are referring to, how where they able to divide above and below by x but they divided what was in the squared root by x^2 and not x?

    Never mind I just figured out why. x= (squared root)x^2 so the x^2 can then be brought into the squared root to divide the terms in there.


  • #5
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    To divide inside the square root buy x you need to square it to take it inside the square root. It's a bit awkward to write a good explination. :o


  • #6
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    shizz wrote: »
    Never mind I just figured out why. x= (squared root)x^2 so the x^2 can then be brought into the squared root to divide the terms in there.


    Great!! Are you ok now?


  • Advertisement
  • #7
    shizz
    Registered Users, Registered Users 2 Posts: 1,506 ✭✭✭shizz


    TheBody wrote: »
    Great!! Are you ok now?

    Yup I am. Although Im not the OP just in case you thought I was :)


  • #8
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    shizz wrote: »
    Yup I am. Although Im not the OP just in case you thought I was :)

    Sorry, just noticed that now!! My bad!! :o


  • #9
    shizz
    Registered Users, Registered Users 2 Posts: 1,506 ✭✭✭shizz


    Haha no worries. I think those solutions are badly laid out. They shouldn't say one equation is equal to another and then use the first equation in the next line instead of the one they said it was equal to. Just a bit confusing.


  • #10
    TheBody
    Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    Agreed. It took me a few minutes to figure out what they were at.


  • #11
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    In case you're interested, you could do that limit a different way, using the squeeze rule.

    Sorry I'm not a Latex-er, but anyway:

    (Assuming x>0 throughout, since we are interested in x going to infinity)

    Clearly sqrt(4x^2+5) is always greater than sqrt(4x^2) = 2x

    Also, sqrt(4x^2+5) is always less than sqrt(4x^2 + 5 + 25/(16x^2)) = 2x + 5/(4x)
    [Square the right-hand side to see that.]

    So, the given expression is always between 1 and (1 - 5/(4x)).

    [The document makes the rather rash assertion that there isn't an easier way than the one they've given. It's perhaps a matter of opinion as to whether I've disproven that!]

    As x goes to infinity, 5/(4x) goes to 0, so the limit of the given expression is 1.


  • Advertisement
Advertisement