 Hello and welcome to the session. In this session we are going to discuss the following question which says that Evaluate limit extends to pi by 2 square root of 2 minus sin of x minus 1 whole upon pi by 2 minus x whole raised to part 2 N-hôpital's rule states that f of x and g of x are the functions such that f of a is equal to 0 and g of a is equal to 0 then limit extends to a f of x by g of x is equal to limit extends to a at dash of x by g dash of x the idea let us proceed with the solution. We are to find the value of the expression limit extends to pi by 2 square root of 2 minus sin of x minus 1 whole upon pi by 2 minus x whole raised to part 2 we can write it as limit extends to pi by 2 2 minus sin of x raised to power 1 by 2 minus 1 upon pi by 2 minus x raised to power 2. Now if we put the value of excess pi by 2 in this expression then we get 2 minus sin of pi by 2 raised to power 1 by 2 minus 1 upon pi by 2 minus pi by 2 the whole square which is equal to 2 minus sin pi by 2 is 1 raised to power 1 by 2 minus 1 upon 0 which is equal to 1 raised to power 1 by 2 which is 1 minus of 1 by 0 that is 0 by 0 So this is the 0 by 0 form. Now according to L-Hopital's rule if f of x and g of x are the 2 functions such that f of a is equal to 0 and g of a is equal to 0 then limit extends to a f of x upon g of x is equal to limit extends to a f dash x upon g dash x Now applying L-Hopital's rule we have limit extends to pi by 2 differential of the numerator with respect to x that is differentiating 2 minus sin of x raised to power 1 by 2 with respect to x we get 1 by 2 into 2 minus sin of x raised to power minus 1 by 2 into differential of minus of sin of x with respect to x that is minus of cos of x minus differential of 1 with respect to x that is 0 whole upon differential of pi by 2 minus x the whole square with respect to x that is 2 into pi by 2 minus x into differential of minus x with respect to x that is minus 1 which can be written as limit extends to pi by 2 minus of 2 minus sin of x raised to power minus 1 by 2 into cos of x whole upon minus 4 into pi by 2 minus of x which is equal to limit extends to pi by 2 2 minus sin of x raised to power minus 1 by 2 into cos of x whole upon 4 into pi by 2 minus x Now if we put the value of x as pi by 2 in this expression we get 2 minus sin pi by 2 whole raised to power minus 1 by 2 into cos of pi by 2 whole upon 4 into pi by 2 minus pi by 2 has 2 minus sin pi by 2 and sin pi by 2 is 1 therefore 2 minus 1 raised to power minus 1 by 2 into cos pi by 2 which is 0 upon 4 into pi by 2 minus pi by 2 that is 4 into 0 therefore in the numerator we have 2 minus 1 raised to power minus 1 by 2 into 0 which is 0 upon 4 into 0 which is also 0 So this is of 0 by 0 form so again applying the L-Hopital's rule limit tends to pi by 2 will take differential of the numerator with respect to x by product's rule We take differential of the numerator with respect to x by product's rule So first we take the differential of 2 minus sin of x raised to power minus 1 by 2 with respect to x that is minus 1 by 2 into 2 minus sin of x raised to power minus 3 by 2 into differential of minus of sin of x that is minus of cos of x into cos x plus sin of x raised to power minus 1 by 2 into differential of cos of x that is minus sin of x whole upon differential of 4 into pi by 2 minus x with respect to x that is 4 into minus 1 so we can write limit x tends to pi by 2 1 by 2 into cos square of x into 2 minus sin of x raised to power minus 3 by 2 minus of sin of x into 2 minus sin of x raised to power minus 1 by 2 whole upon minus of 4 Now if we put the value of x as pi by 2 in this expression we get 1 by 2 cos square of pi by 2 2 minus sin of pi by 2 raised to power minus 3 by 2 minus of sin pi by 2 into 2 minus sin of pi by 2 raised to power minus 1 by 2 whole upon minus 4 which is equal to 1 by 2 into cos square of pi by 2 is 0 into 2 minus sin of pi by 2 is 1 raised to power minus 3 by 2 minus of sin of pi by 2 is 1 into 2 minus sin of pi by 2 is 1 raised to power minus 1 by 2 whole upon minus of 4 which is equal to 1 by 2 into 0 into 2 minus 1 raised to power minus 3 by 2 is 0 minus of 1 into 2 minus 1 that is 1 raised to power minus 1 by 2 by minus 4 which is equal to minus 1 into 1 raised to power minus 1 by 2 is 1 upon minus 4 which is equal to minus 1 by minus 4 that is 1 upon 4 therefore the value of the expression limit x tends to pi by 2 square root of 2 minus sin of x minus 1 whole upon pi by 2 minus x whole square is equal to 1 upon 4 which is the required answer this completes our session hope you enjoyed this session