How to determine the limit of the function y=(1-cos^2x)/x^2, if x goes to 0? (don't use l'Hospital)

We'll substitute the difference of the squares from
numerator by the product: (1 - cos x)(1 + cos x)

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

We'll use the half angle
identity:

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

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

We'll create
the elementary limit:

lim sin x/x =
1

lim [sin(x/2)/x] = lim [sin(x/2)/2*x/2] =
(1/2)*lim[sin(x/2)/(x/2)]

lim [sin(x/2)/x] =
1/2

The limit will become:

lim
(1 - cos x)(1 + cos x)/x^2 = 2*(1/2)*(1/2)*(1+cos 0)

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

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

The requested limit of the
function is: lim (1 - cos x)(1 + cos x)/x^2 = 1.