I need help to solve using remarcable limits.limit of f(x) given by f(x)=(1-cosx)/x^2*cos4x, x->0

We'll re-write the numerator of the function f(x) using
the half angle identity.

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

f(x) =
2[sin(x/2)]^2/x^2*cos4x

We'll evaluate the
limit:

lim 2[sin(x/2)]^2/x^2*cos4x =
2lim[sin(x/2)]^2/x^2*lim 1/cos4x

2lim[sin(x/2)]^2/x^2*lim
1/cos4x = 2lim[sin(x/2)]/x*lim sin(x/2)/x*lim 1/cos4x

We
know that the remarcable limit is:

lim sin u(x)/u(x) =
1

We'll create the remarcable
limit:

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

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

The limit of the function
is:

2lim[sin(x/2)]^2/x^2*lim 1/cos4x = 2*(1/2)*(1/2)*lim
1/cos4x

We'll substitute x by
0:

2*(1/2)*(1/2)*lim 1/cos4x = (1/2)*(1/cos 0) = 1/2*1 =
1/2

The limit of the function, if the
accumulation point is x = 0, is: lim f(x) = 1/2.