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Finding a Trig Limit by hand, no L'Hopitals!

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\stackrel{lim}{x\rightarrow 0}\frac{cos^2x-1}{2xsinx}[/itex]


    2. Relevant equations

    [itex]\stackrel{lim}{x\rightarrow 0}\frac{1-cosx}{x}=0[/itex]

    [itex]\stackrel{lim}{x\rightarrow 0}\frac{sinx}{x}=1[/itex]

    3. The attempt at a solution

    I found this problem online (and can't remember where). It was in a limits section, so it can be solved without using l'Hopitals rule. I gave it to my class today, and then we all got stuck. We know the answer is -.5 but can't get it algebraically. Any advice?
     
  2. jcsd
  3. Mar 28, 2012 #2
    Remember that 1 = cos^2(x) + sin^2(x).

    Now substitute and simplify and you can use one of your known limits.
     
  4. Mar 28, 2012 #3
    Thank you! I got thrown into this class half way through the year, and while I have re-learned the calc, my trig is horrible. Thank you!
     
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