What is the limit of function y=(1-cos^3 x)/(x^2) if x approaches to 0 ?

We recognize to numerator a difference of 2 cubes and we'll
apply the identity:

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

1 - (cos x)^3 = (1 - cos x)[1 + cos x + (cos
x)^2]

We'll re-write th limit:

lim [1 -
(cos x)^3]/x^2 = lim (1 - cos x)[1 + cos x + (cos x)^2]/x^2

lim [1 -
(cos x)^3]/x^2 = lim [(1 - cos x)/x^2]*lim[1 + cos x + (cos
x)^2]

We'll evaluate the limit of each
factor:

We'll substitute the numerator by the half angle
formula:

1 - cos x = 2 [sin(x/2)]^2

Lim
(1-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2

Lim {2 [sin(x/2)]^2}/x^2 =
Lim [sin(x/2)]^2/(x^2/2)

Lim [sin(x/2)]^2/(x^2/2) =
lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)]

Since the elementary limit
is:

lim [sin f(x)]/f(x) = 1, if f(x) -> 0, we'll
get:

lim (1-cosx)/x^2 = 1/2 (1)

We'll
calculate the limit of the next factor:

lim[1 + cos x + (cos x)^2] =
[1 + cos 0 + (cos 0)^2] = (1 + 1 + 1)=3 (2)

We'll multiply (1) by
(2) and we'll get the limit of the function:

lim [1 - (cos x)^3]/x^2
= 3/2

The limit of the given function, if x approaches
to 0, is: lim [1 - (cos x)^3]/x^2 = 3/2.