Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    [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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook