You need to evaluate the given limit such that:

`lim_(x->1/3) (sqrt(27x^3 - 1))/(9x^2 - 1) = (sqrt(27*1/27 - 1))/(9*1/9 - 1) = 0/0`

The indetermination `0/0` allows you to use l'Hospital's theorem, such that:

`lim_(x->1/3) sqrt(27x^3 - 1)/(9x^2 - 1) = lim_(x->1/3) (sqrt(27x^3 - 1)')/((9x^2 - 1)') `

`lim_(x->1/3) sqrt(27x^3 - 1)/(9x^2 - 1) = lim_(x->1/3) ((81x^2)/(2sqrt(27x^3 - 1)))/(18x) `

Reducing duplicate factors yields:

`lim_(x->1/3) sqrt(27x^3 - 1)/(9x^2 - 1) = lim_(x->1/3)(9x)/(4sqrt(27x^3 - 1))`

`lim_(x->1/3)(9x)/(4sqrt(27x^3 - 1)) = 9*(1/3)/(4*0) = 3/0 = oo`

**Hence, evaluating the given limit `lim_(x->1/3) (sqrt(27x^3 - 1))/(9x^2 - 1),` using l'Hospital's theorem, yields `lim_(x->1/3) (sqrt(27x^3 - 1))/(9x^2 - 1) = oo` .**

lim sqrt [( 27x^3 - 1 )/( 9x^2 - 1 )], if x - > 1/3

To calculate the limit, we'll have to substitute x by the indicated value, namely 1/3. We'll check if we'll get an indeterminacy case.

lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = sqrt (27*1/27 - 1)/(9*1/9- 1)

lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = sqrt (1-1)/(1-1)

lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = 0/0, indetermination

To calculate the limit we'll use factorization. We notice that the numerator is a difference of cubes:

27x^3 - 1 = (3x)^3 - (1)^3

We'll apply the formula:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

a = 3x and b = 1

(3x)^3 - (1)^3 = (3x-1)(9x^2 + 3x + 1)

We also notice that the denominator is a difference of squares:

9x^2-1 = (3x)^2 - 1^2

We'll apply the formula:

a^2 - b^2 = (a-b)(a+b)

(3x)^2 - 1^2 = (3x-1)(3x+1)

We'll substitute the differences by their products:

lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = lim sqrt [(3x-1)(9x^2 + 3x + 1)/(3x-1)(3x+1)]

We'll simplify:

lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = lim sqrt [(9x^2 + 3x + 1)/(3x+1)]

Now, we'll substitute x by 1/3:

lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt(9*1/9 + 3*1/3 + 1)/(3*1/3 + 1)

lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt (1+1+1)/(1+1)

lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt (3/2)

lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt6/2