I’m assuming because of the delta sign, you aren’t allowed to use L’Hopitals.

Let’s expand those brackets:

[sin(pi/6) * cos(x) + sin(x) * cos(pi/6) - 1/2] / x

[1/2 * cos(x) + sqrt(3)/2 * sin(x) - 1/2] / x

[cos(x) + sqrt(3) * sin(x) - 1] / 2x

[cos(x) - 1] / 2x + sqrt(3) * sin(x) / 2x

We know that sin(x) ≈ x when it approaches zero, thus that part of the second fraction will go to 1:

[cos(x) - 1] / 2x + sqrt(3) / 2

For the first one, rewrite cos(x) as 1 - 2sin²(x/2):

[1 - 2sin²(x/2) - 1] / 2x

-sin²(x/2) / x

Let’s use a substitution:

x/2 = t

x = 2t

As x approaches zero, so will t:

-sin²(t) / 2t

= -1/2 * sin(t) / t * sin(t)

Again we know that sin(t) ≈ t as t approaches zero, leaving you with:

-1/2 * 1 * 0 = 0

Thus the limit is:

0 + sqrt(3) / 2

= sqrt(3) / 2