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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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