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Finding limit :s

  1. Nov 7, 2011 #1
    I i am fed up with these question. some are easy to solve , some solve themselfs along and some seem to have no answer in the far horizon like this question -.-

    lim (tanx - sqrt3)/(3x-pie)
    x->(pie/3)

    I did it until i reached (sin^2(x) - 3cos^2(x))/(cos(x).(3x-pie).(sinx+sqrt3.cosx) where i got stuck !

    Help ! .
    If you have any advice about how to approach such questions plz tell me or i ll be screwed in my upcoming Calculus exam :S . Thanks in advance
     
  2. jcsd
  3. Nov 7, 2011 #2

    HallsofIvy

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    Do you know "L'Hopital's rule"? That makes this problem easy. If not, then it would probably be best to use a trig identity: write tan(x- pi/3) in terms of trig functions of x and pi/3 separately.
     
  4. Nov 7, 2011 #3
    Thanks
    btw L'Hopital's rule is not allowed , I guess i ll try using tan(x-pi/3) = ( tanx - tan(pi/3))/( 1+tanxtan(pi/3))

    If anyone found a solution for this problem plz post it or atleast an advice :D !
     
  5. Nov 7, 2011 #4

    lurflurf

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    tan(pi/3)=sqrt(3) so recognize tan'(pi/3)
    [tex]\lim_{x\rightarrow\pi/3}\frac{\tan(x)-\sqrt{3}}{3x-\pi}=\frac{1}{3}\lim_{x\rightarrow\pi/3}\frac{\tan(x)-\tan(\pi/3)}{x-\pi/3}=\frac{\tan'(\pi/3)}{3}[/tex]
     
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