Finding limit :s

  • Thread starter Redoctober
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #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.
 
  • #3
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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.
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 !
 
  • #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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