find the limit :

Question

lim of (cos(x)-sqrt(3)sin(x))/(6x-pi)

 

as x approaches pi/6

Answer 1

By direct substitution the given limit has the indeterminate form 0/0.

 

So, using L'Hopital's Rule, the given limit is equivalent to:

 

    lim_x→π/6 [(-sinx - √3 cosx) / 6] = (-sin(π/6) - √3 cos(π/6))/6

 

                                                  = (-1/2 - 3/2) / 6 = -1/3