# Determine the limit of the function y = sqrt( 27x^3 - 1 )/( 9x^2 - 1 ) x --> 1/3

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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**

To find limit of the function y = sqrt( 27x^3 - 1 )/( 9x^2 - 1 ) x --> 1/3.

Solution:

Lt x--> 1/3 y = Lt x--> 1/3 {sqrt (27x^3 -1)/(9x^2-1)} .

If we put x = 1/3 both numerator and denominator become zero., giving us the form of 0/0 indetermation.

Since both numerator and denominators vanish for x = 1/3, x-1/3 or 3x-1 is a facor of both numerator and denominators.

Therefore y = sqrt {(3x-1)(9x^2+3x+1)}/{(3x-1)(3x+1)}

Therefore y = sqrt {9x^2+3x+1}/{(3x+1)sqrt(3x-1)}

Now taking limits , we get:

Lt x--> 1/3 y = sqrt {9(1/3)^2+3*(1/3)+1}/{((3/3) +1)(3/3 - 1)}

Lt x--> 1/3 y = (sqrt3)/{2*0} = infinity.

Therefore Lt x--> 1/3 1/y = Lt x --> 1/3 sqrt(27x^3-1}/(9x^2-1} = infinity.