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Calculus Limits

  1. Mar 25, 2004 #1
    I am not sure what the answer to this is or how to go about getting an answer for this:

    Find:

    lim (1 - cos(a*x))/x^2
    x->0

    My teacher said the answer would have the term 'a' in it.

    How do I go about getting the answer to a limit problem like this? Thank you.
     
  2. jcsd
  3. Mar 25, 2004 #2
    [tex]\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2} = \frac{1 - \cos 0}{0^2} = \infty ^+[/tex]
    I might be wrong though, but I think the limit is (positive) infinity and that the function has an asymptote at [tex]x = 0[/tex].
     
    Last edited: Mar 25, 2004
  4. Mar 25, 2004 #3
    Sorry to say that Chen, but you are wrong. To solve this, look at the power series of cos(ax). Like this you will have [tex]1/2*a^2x^2+O(x^4)[/tex] on top of the line and another square term below. So the limit is...

    You can also use L'Hôpital (twice), if you know it.
     
    Last edited: Mar 25, 2004
  5. Mar 25, 2004 #4
    It's trivial with l'Hopital's rule, but your teacher may or may not want you to use that ;)

    lim (x -> 0) (1 - cos(a*x)) / x^2 =
    lim (x -> 0) a * sin(ax) / (2x) =
    lim (x -> 0) aa * cos(ax) / 2 =
    a^2 * 1/2 =
    a^2/2
     
  6. Mar 25, 2004 #5
    I figured it was too simple to be true. :wink:
     
  7. Mar 25, 2004 #6
    I'm not supposed to know about L'Hopital's rule yet. How could I come up with that forumla? I have to show some kind of evidence. I don't know how to come up with that formula. Thanks.
     
  8. Mar 25, 2004 #7

    Zurtex

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    Hmm, [tex]\lim _{x \rightarrow 0} \frac{1 - \cos ax}{x^2}[/tex]

    Using the maclaurin series:

    [tex]\cos ax \approx 1 - \frac{(ax)^2}{2}[/tex]

    Try that
     
  9. Mar 25, 2004 #8
    I do not think I am supposed to know that either. Is there an easy way to prove that the limit is (ax^2)/2?
     
  10. Mar 25, 2004 #9
    You could always graph it on a calculator and point.

    cookiemonster
     
  11. Mar 25, 2004 #10
    Yeah I know I did that for when a = 1, 1.5, and 2. But I don't know how to find the forumla for finding the limit if a was anything.
     
  12. Mar 25, 2004 #11
    Anyone know? Thanks.
     
  13. Mar 25, 2004 #12
    Er, I guess you could always try delta-epsilon proofs, but I don't think you want to go there.

    Why not just plot it for a bunch of values of a and look for a pattern? If you have 20 values of a of varying value supporting you, it's pretty solid for your purposes.

    cookiemonster
     
  14. Mar 26, 2004 #13

    arildno

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    Suppose you are allowed to know the limits: lim(x->0)(sin(x)/x)=1,lim(x->0)(cos(x)=1),
    im(x->0)(a(x)*b(x))=lim(x->0)a(x))*(lim(x->0)b(x),
    lim(x->0)(a(x)/b(x))=(lim(x->0)a(x))/(lim(x->0)b(x))
    Then you may do without MacLaurin series/L'Hopital's Rule in your proof
     
  15. Mar 26, 2004 #14
    now that's very rigorous
     
  16. Mar 26, 2004 #15

    Zurtex

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    Then would you mind explaining what areas you have covered around this sort of area?
     
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