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

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