what is the limit of the function (1-sin x)/cos^2 x, x -> 90 degrees?

First, we'll verify if we'll get an indetermination by
substituting x by the value of the accumulation
point.

lim (1-sin x)/(cos x)^2 = (1 - sin 90)/(cos 90)^2 =
(1-1)/0 = 0/0

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

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

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

We'll simplify by -cos x and we'll
get:

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

lim 1/2sin x = 1/2sin 90 =
1/2

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