What is the limit of the function (1-cosx*cos2x*...*cosnx)/x^2, x approaches to zero?

First, we'll substitute x by the value of accumulation
point:

lim (1-cosx*cos2x*...*cosnx)/x^2 = (1-cos 0*cos
2*0*....cos n*0)/0^2 = 0/0

Since we've get an
indetermination, we'll apply L'Hospital rule:

lim
(1-cosx*cos2x*...*cosnx)/x^2 = lim
(1-cosx*cos2x*...*cosnx)'/(x^2)'

We'll apply product rule
for the second term of the numerator:

lim
(1-cosx*cos2x*...*cosnx)'/(x^2)'=lim(sin x*cos 2x*...*cos nx + 2sin 2x*cos x*...*cos nx
+ n*sin nx*cos x*...*cos(n-1)*x)/2x

Since we'll get an
indetermination again, we'll apply L'Hospital rule one more time. We'll use the
remarcable limit:

lim k*sin kx/k = k^2, k
integer

lim (1-cosx*cos2x*...*cosnx)'/(x^2)'=
(1^2+2^2+...+n^2)/2

The sum of the squares
is:

1^2+2^2+...+n^2 =
n*(n+1)*(2n+1)/6

lim (1-cosx*cos2x*...*cosnx)'/(x^2)'=
n*(n+1)*(2n+1)/6

The limit of the given
function, for x approaches to 0, is: lim (1-cosx*cos2x*...*cosnx)/x^2
=  n*(n+1)*(2n+1)/6