 Hello and welcome to the session. In this session we are going to discuss the following question which says that evaluate 1. Limit x tends to pi by 2 1 minus sin x by cortex 2. Limit x tends to pi by 2 1 minus sin x by cortex n-hôpital's rule states that if fx and gx are the two functions such that fA is equal to 0 and g of A is equal to 0 then limit x tends to A fx upon gx is equal to limit x tends to A f dash x by g dash x. With this key idea we shall proceed with the solution. We need to find the value of the expression limit x tends to pi by 2 1 minus sin x by cortex. Now if we put the value of x as pi by 2 in the given expression we get 1 minus sin of pi by 2 by cortex of pi by 2 which is equal to 1 minus 1 as sin pi by 2 is 1 upon cortex pi by 2 which is 0 which is equal to 0 by 0 so this is as 0 by 0 sin. Now we use n-hôpital's rule which states that if fx and gx are the two functions such that f of A is equal to 0 and g of A is equal to 0 then limit x tends to A fx upon gx can be written as limit x tends to A f dash x upon g dash x. Now using n-hôpital's rule we have limit x tends to pi by 2 Now differentiating the numerator with respect to x we get differentiation of 1 with respect to x is 0 minus of differentiating sin x with respect to x we get cos x upon differentiating cos x with respect to x we get minus of sin x which is equal to limit x tends to pi by 2 minus of cos x by minus of sin x which can be written as limit x tends to pi by 2 cos x upon sin x. Now on putting x is equal to pi by 2 in the given expression we get cos of pi by 2 by sin of pi by 2 which is equal to cos pi by 2 that is 0 upon sin pi by 2 that is 1 0 by 1 is equal to 0. Therefore the value of the expression limit x tends to pi by 2 1 minus sin x by cos x is equal to 0 which is the required answer Next we have limit x tends to pi by 2 1 minus sin x by cos x. Now if we put the value of x is pi by 2 in the given expression we get 1 minus sin of pi by 2 by cos of pi by 2 which is equal to 1 minus sin pi by 2 is 1 and cos pi by 2 is 0 which is equal to 0 by 0. So this is also of 0 by 0 form so using 2-Hopitans rule we have limit tends to pi by 2 Differentiating the numerator with respect to x we have differentiation of 1 with respect to x is 0 minus differentiation of sin x with respect to x is cos x upon Differentiating cot x with respect to x we get minus cot x square x which can be written as limit tends to pi by 2 minus of cot x upon minus of cot x square x that is limit x tends to pi by 2 cos of x into 1 upon cot x square of x can be written as sin square of x. Now putting the value of x as pi by 2 we get cos of pi by 2 into sin square pi by 2 which is equal to cos pi by 2 is 0 into sin square pi by 2 that is 1 square which is equal to 0. Therefore the value of the expression limit x tends to pi by 2 1 minus sin x by cot x is equal to 0 which is the required answer. This completes our session. Hope you enjoyed this session.