You need to factor out 2 to denominator such that:

`lim_(x->pi/6) sin(x - pi/6)/(2(sqrt3/2 - cos x))`

You should remember that `cos(pi/6) = sqrt3/2` , hence, you may substitute `cos(pi/6)` for `sqrt3/2` such that:

`lim_(x->pi/6) sin(x - pi/6)/(2(cos(pi/6) - cos x))`

You need to convert the difference of cosines into...

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You need to factor out 2 to denominator such that:

`lim_(x->pi/6) sin(x - pi/6)/(2(sqrt3/2 - cos x))`

You should remember that `cos(pi/6) = sqrt3/2` , hence, you may substitute `cos(pi/6)` for `sqrt3/2` such that:

`lim_(x->pi/6) sin(x - pi/6)/(2(cos(pi/6) - cos x))`

You need to convert the difference of cosines into a product using the formula `cos a - cos b = 2 sin ((a+b)/2) sin((b-a)/2)` such that:

`cos(pi/6) - cos x = 2sin ((pi/6+x)/2) sin((x-pi/6)/2)`

You may write the numerator such that:

`sin(x - pi/6) = sin 2((x - pi/6)/2) = 2 sin((x - pi/6)/2)cos((x - pi/6)/2)`

`lim_(x->pi/6) (2 sin((x - pi/6)/2)cos((x - pi/6)/2))/(4sin((x +pi/6)/2)sin((x - pi/6)/2))`

Reducing by `2sin((x - pi/6)/2))` yields:

`(1/2)lim_(x->pi/6) cos((x - pi/6)/2)/sin((x - pi/6)/2) = (1/2)lim_(x->pi/6) cot ((x - pi/6)/2) = (1/2) cot 0 = oo`

**Hence, evaluating the given limit, avoiding l'Hospital's theorem, yields `lim_(x->pi/6) sin(x - pi/6)/(2(sqrt3/2 - cos x)) = oo.` **