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Simple trig limit with cos

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limits that exist:

    lim x->0 of [tex]\frac{1-cos4x}{9x^{2}}[/tex]

    2. Relevant equations
    lim x->0 [tex]\frac{1-cosax}{ax}[/tex] = 0

    3. The attempt at a solution
    so far I've got this
    The second part has a limit of 0 but I don't know what to do about the first fraction. To me it looks undefined but according to some java applet I found on the internet the limit should be 8/9. What's the right answer?
  2. jcsd
  3. Sep 30, 2007 #2


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    The limit has the form 0/0. That suggests you use L'Hopital's rule. Alternatively, if you know the power series expansion of cos(x)=1-x^2/2!+x^4/3!-..., you could substitute x->4x and use that.
    Last edited: Sep 30, 2007
  4. Sep 30, 2007 #3
    Thank you. So far we are in Ch 2. so we haven't done L'Hopital's rule or power series. I guess saying it is undefined is enough for now.
  5. Sep 30, 2007 #4


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    The trouble with that is that is IS defined. If you can't use those, then you have to use tricks. Try a double angle formula like cos(2x)=1-2*sin(x)^2. Use that to express cos(4x). Then use lim x->0 sin(ax)/(ax)=1. I'm guessing you do know that from the relevant equation you posted.
  6. Sep 30, 2007 #5
    oops, I got confused a bit...
    Anyways, I used your suggestion and solved it.
    Thanks again.
  7. Sep 30, 2007 #6


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    Very sneaky little problem: the limit law it looks the most obvious to use is the wrong one!. Keep these kinds of manipulations in mind on exam problems...
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