To find the lt (x^3-27)/(x^2-9) as x-->3.

If we put x = 3, then (x^3-27)/(x^2-9) becomes (3^3-27)/(3^2-9) = (27-27)/(9-9) = 0/0 form.

Since x - 3 makes both numerator and denominator zero, by reaminder theorem x-3 is a factor of both numerator and denominator. So we divide both numerator and denominator by (x-3).

Numerator: x^3-27 = (x^3-3^3)= (x-3)(x^2+3x+3^2) , as a^3-b^3 =(a-b)(a^2+ab+b)

Denominator (x^2-9) = (x^2-3^2), as a^2-b^2 = (a-b)(a+b)

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

Therfore Lt (x^3-27)/(x^2-9) = (x^2+9x+9)(x+3) as x--> 3 = (3^2+3*3+9)/(3+3) = 27/6 = 4.5

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

Let's see:

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

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

We'll re-write the numerator using the formula of difference of cubes:

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

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)

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

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

We'll re-write the limit:

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

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

We have to evaluate the limit (x^3-27)/(x^2-9) x-->3

Now for x=3,

(x^3 - 27) / (x^2 - 9) = (3^3 - 27) / (3^2 - 9) = 0 / 0, which cannot be determined.

Now we know that a^3 - b^3 = (a-b) (a^2 + b^2 + ab)

=> (x^3-27) = (x - 3) ( x^2 + 3^2 + 3a)

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

Dividing the two:

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

=> (x - 3) ( x^2 + 3^2 + 3x) / (x- 3) ( x+3 )

=> ( x^2 + 3^2 + 3x) / ( x+3 )

Now substitute x = 3

=> (9 + 9 + 9) / (3+3)

=> 27 / 6

=> 9/2

=> 4.5

**Therefore limit (x^3-27)/(x^2-9) x-->3 is 9/2**