# What is the limit of the function  (x^3-27)/(x^2-9) if x goes to 3?

justaguide | Certified Educator

We have to find the value of: lim x-->3[(x^3-27)/(x^2-9]

lim x-->3[(x^3-27)/(x^2-9]

=> lim x-->3[(x - 3)(x^2 + 3x + 9)/(x - 3)(x + 3)]

=> lim x-->3[(x^2 + 3x + 9)/(x + 3)]

substitute x = 3

=> (3^2 + 3*3 + 9)/6

=> (9 + 9 + 9)/6

=> 27/6

=> 9/2

The required value of lim x-->3[(x^3-27)/(x^2-9] = 9/2

giorgiana1976 | Student

We'll replace x by 3 in the expression of the limit and we'll get the indetermination case: 0/0

(3^3-27)/(3^2-9) = (27-27)/(9-9)

(27-27)/(9-9) = 0/0

We'll notice that the numerator is a difference of cubes:

(a^3-b^3) = (a-b)(a^2 + ab + b^2)

Let a^3 = x^3 and b^3 = 27

x^3 - 27 = (x-3)(x^2 + 3x + 9)

We'll re-write the denominator using the formula of difference of squares:

(a^2-b^2) = (a-b)(a+b)

Let a^2 = x^2 and b^2 = 29

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

The limit will become:

lim (x^3 - 27)/(x^2 - 29)=lim (x-3)(x^2 + 3x + 9)/(x-3)(x+3)

We'll simplify and we'll get:

lim (x-3)(x^2 + 3x + 9)/(x-3)(x+3)=lim (x^2 + 3x + 9)/(x+3)

We'll substitute x by 3:

lim (x^2 + 3x + 9)/(x+3) = (3^2 + 3*3 + 9)/(3+3)

lim (x^3 - 27)/(x^2 - 29) = 27/6

The requested limit, if x approaches to 3, is: lim (x^3 - 27)/(x^2 - 9) = 9/2