Trigonometric limits Please help!

Question

lim (x--> p/3) sin(x-3)/(4cos²-1)

Answer 1

sin(x-3) is a subtraction of sine angles.  This becomes     sin(x)cos(3) - cos(x)sin(3)

 

 

lim (x--> π/3)     sin(x-3)/(4cos²x-1) =

 

lim (x--> π/3)    [sin(x)cos(3) - cos(x)sin(3)] / [(2cosx-1)(2cosx+1)] =

 

Substitute π/3 into this limit.

 

lim (x--> π/3)     [sin(π/3)cos(3) - cos(π/3)sin(3)] / [(2cos(π/3)-1)(2cos(π/3)+1)] =

 

lim (x--> π/3)     [(√3cos(3))/2 - (sin3)/2] / [((2*1/2)-1)*(2*1/2)+1))] =

 

lim (x--> π/3)     [(√3cos(3))/2 - (sin3)/2] / (0*2) =

 

 

Since the denominator will be zero, this makes the limit undefined.  Therefore we must use L'Hospital Rule.  We take the derivative of the numerator divided by the derivative of the denominator.  Afterwards, substitute  π/3 and evaluate limit.