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
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