# What is the value of lim x->pi/2 [(1-sinx)/cos x]

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### 1 Answer

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

Substituting x = pi/2, gives us 0/0, which is indeterminate. This allows us to use the L'Hopital's rule and substitute the numerator and denominator with their derivatives

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

substituting x = pi/2

=> 0

**The required value of the limit is 0.**