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A limits problem

  1. May 20, 2010 #1
    Find the limit algebraically (if it exists).
    If the limit DNE (Does Not Exist), explain why.
    (hint: use the properties of limits and the limit identities)

    ..... cos x-1
    lim --------- =
    x->0 2x^2

    Help??

    *** note: the "...." infront of the cos is just so nothing. I only did it to evenly proportion the fraction
     
  2. jcsd
  3. May 20, 2010 #2

    CompuChip

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    Are you allowed to use L'Hopitals rule?
     
  4. May 20, 2010 #3
    Wrong forum.
     
  5. May 20, 2010 #4

    thrill3rnit3

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    I would assume not.
     
  6. May 20, 2010 #5
    0, multiply it by x/x, cosx - 1/x is 0
     
  7. May 21, 2010 #6

    CompuChip

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    Sorry bignum, your post doesn't make sense.
    First of all, cos(x) - 1/x is not 0. In fact, it's not zero for any (real) value of x.
    If you multiply the top by x it gives x cos(x) - x, and the bottom gives 2x3.
    If you multiply by (1/x) / (1/x), then you would still be left with 2x in the denominator, and the numerator would not become cos(x) - 1/x but cos(x)/x - 1/x.

    I wonder if Edin is still going to respond, or if he found the answer elsewhere.

    The reason I asked about L'Hopital, is because you mentioned you should use "limit identities." One of the most "famous" ones is the standard limit
    [tex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/tex].

    Otherwise, you may want to use a Taylor series to expand cos(x) as 1 + a x2 + O(x4), such that [cos(x) - 1] / x2 becomes a + O(x2) when you take x to zero.
     
  8. May 21, 2010 #7

    Mark44

    Staff: Mentor

    Another identity (including that the one that Compuchip gave) that is very useful in this problem is sin2(x) + cos2(x) = 1.

    You don't need L'Hopital's Rule or Taylor series.
     
  9. May 21, 2010 #8

    CompuChip

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    Are you thinking about (cos2(x) - 1) / 2x2 here, or am I just missing something?

    [edit]
    I am probably missing something, namely that
    [tex]\lim_{x \to a} f(x) g(x) = L_f L_g [/tex]
    if f has limit Lf and g has limit Lg.
    Right?
    [/edit]
     
  10. May 22, 2010 #9
    Guys I don't really understand this it's why I posted it. So if you're going to give an answer can you please show the work so I can try and figure it out?? For the user who said something about Lhopital, I don't really know what that is but judging form your post it seemed you were making a reference to the special limit properties. Ex:

    1-cos x
    -------= 0
    .....x
     
  11. May 22, 2010 #10

    gabbagabbahey

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    Hi, Edin. Welcome to PF!:smile:

    We don't do the work for you here, and we don't give out answers to homework problems. We only give you pointers on how to proceed (so that you can do the work yourself) and point out errors.

    As a hint on how to begin with this problem, I'd start by multiplying both the numerator and denominator by [itex](1+\cos(x))[/itex]....what do you get when you do that?
     
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