Solution:

Given: lim x--->π/6. √3sinx-cosx/x-π/6.

We need to evaluate the given limit.

Method:

To evaluate the given limit, we can use the standard trigonometric limit:

lim x-->0 [(sin x)/x] = 1

We can also use the following trigonometric identities:

sin (π/6) = 1/2
cos (π/6) = √3/2

Using these identities, we can simplify the given limit as follows:

√3sinx-cosx/x-π/6 = [(√3sinx-cosx)/(x-π/6)] * [(x-π/6)/(sin x - π/6)]

Now, we can apply the standard trigonometric limit to the first factor:

lim x-->π/6 [(√3sinx-cosx)/(x-π/6)] = lim x-->π/6 [(√3sinx-cosx)/(sin x - π/6)] * [(sin x - π/6)/(x-π/6)]

= √3 * lim x-->π/6 [(sin x)/(x-π/6)] - lim x-->π/6 [(cos x)/(x-π/6)]

= √3 * lim x-->0 [(sin (x+π/6))/(x+π/6)] - lim x-->0 [(cos (x+π/6))/(x+π/6)]

= √3 * lim x-->0 [(sin x cos π/6 + cos x sin π/6)/(x+π/6)] - lim x-->0 [(cos x cos π/6 - sin x sin π/6)/(x+π/6)]

= √3 * lim x-->0 [(sin x + √3 cos x)/(x+π/6)] - lim x-->0 [(cos x - √3 sin x)/(x+π/6)]

= √3 * lim x-->0 [(sin x/x)/(1/√3 + π/6x)] - lim x-->0 [(cos x/x)/(1 - √3π/6x)]

= √3 * 1/(1/√3) - 1/1

= 2√3 - 1

Next, we can apply L'Hopital's rule to the second factor:

lim x-->π/6 [(x-π/6)/(sin x - π/6)] = lim x-->π/6 [1/(cos x)] = 2

Therefore, the given limit can be evaluated as follows:

lim x--->π/6. √3sinx-cosx/x-π/6 = (2√3 - 1) * 2 = 4√3 - 2

Answer:

lim x--->π/6. √3sinx-cosx/x-π/6 = 4√3 - 2.