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

*print*Print*list*Cite

### 2 Answers

We have to find the value of lim x--> 90[ (1- sin x)/(cos x)^2]

substituting x = 90 degrees, we get the indeterminate form 0/0, so we can use l'Hopital's rule and substitute the numerator and denominator by their derivatives.

=> lim x->90 [ (- cos x)/ (-2 cos x * sin x)]

=> lim x->90 [ 1/ (2 sin x)]

substitute x = 90

=> 1/ 2

**The required value of lim x--> 90[ (1- sin x)/(cos x)^2] is (1/2).**

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

limĀ (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**