 Hello and welcome to the session. In this session we discuss the following question which says evaluate the limit, limit x times to pi by 2, 1 minus sine cube x upon cos square x. Before we move on to the solution, let's discuss some important identities. First we have 1 minus cos cube x can be written as 1 minus into cos square x plus cos x plus 1. That's sine square x plus cos square x is equal to 1. This means that sine square x is equal to 1 minus cos square x that we use for this question. Let's now move on to the solution. We are supposed to find the limit. Limit x times 2 pi by 2 equal to cos square x as we have limit x times 2 pi is equal to pi by 2 plus h. So, x equal to pi by 2 h times 2 equal to 0. Time place of x we will put pi by 2 plus h. So, we have 1 minus h would be minus sine h. So, using this we can have x to 0 1 by cos cube h. This way we can use this formula stated in the Q-idea. So, using this we can write the numerator where x to 0 minus cos h this way into cos h plus 1 be whole. And this whole upon sine square x is equal to 1 minus cos. The denominator of this function could be written as x to 0 cos h the whole into cos h plus 1 be whole. And this whole upon the h could be written as 1 minus cos h the whole into 1 plus cos h. 1 minus cos h 1 minus cos h cancels we are left with limit h times 2 0 plus 1 and this whole upon h as 0. So, this is equal to cos 0 plus 1 upon 1 plus cos 0 raise 1. So, this is equal to 1 square which is 1 plus 1 plus 1 upon 1 plus is equal to 3 upon 2. That we have the required limit is equal to 3 upon this should be solution of this question.