Finding Limit: Stuck on Calculus Exam Problem

[Redoctober]

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 please tell me or i ll be screwed in my upcoming Calculus exam :S . Thanks in advance

 

[HallsofIvy]

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.

 

[Redoctober]

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.

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 please post it or atleast an advice :D !

 

[lurflurf]

tan(pi/3)=sqrt(3) so recognize tan'(pi/3)
\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}