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Having trouble finding this limit

  1. Oct 10, 2007 #1
    Hi all,

    My girlfriend asked me this question just now, however I have no idea how I should approach to solve it, I highly appreciate if anyone could shed lights on this:

    lim x^2 times cos^2(x^-2)

    I tried using the squeeze theorem:

    -1 < cos(1/x) < 1
    -x^2 < x^2 (cos(1/x)) < x^2
    -x^2 cos(1/x) < x^2 (cos^2(x^-2)) < x^2 cos(1/x)
    Therefore, as x->0 in the middle, the two sides also approach 0.

    But I don't think it makes any sense... since 1/x as x-> 0 cannot really be used as part of the intervals representing -1 and 1.

    I also tried rearranging cos^2 (x^-2), but I don't think it's any use.

    Please enlighten on this, thanks in advance.
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 11, 2007 #2
    calc 1 or 2?

    is your problem

  4. Oct 11, 2007 #3
    -1<cos(theta) <1 for any value of theta

    so x^2 * cos (theta)->0 as x -> 0

    What's wrong with that?
  5. Oct 11, 2007 #4


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    [itex]-1\le cos(\theta)\le 1[/itex] even when [itex]\theta= 1/x^2[/itex]! That's christianjb's point.
  6. Oct 11, 2007 #5
    Thank you for your replies, guys.

    Yes, that's the correct output of the question. I haven't done calculus for a long time, but I believe it's calc 1.

    christianjib & HallsofIvy:
    But if it's cos^2 (x^-2), can you still apply the same rule?
  7. Oct 11, 2007 #6


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    Actually, in that case, it's easier: [itex]0\le cos^2(\theta)\le 1[/itex] for all [itex]\theta[/itex]!
  8. Oct 14, 2007 #7
    Wow, never thought of that, thanks so much HallsofIvy. :D
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