 Hello and welcome to the session. In this session we are going to discuss the following question which says that Determine A. Limit of log of sin of x upon log of tan of x as x tends to 0. B. Limit of log of cos of x upon tan of x as x tends to pi by 2. Inhabitance rule states that if f of x and g of x are two functions such that f of a is equal to 0 and g of a is equal to 0, then limit f of x by g of x as x tends to a is equal to the limit f by x by g by x as x tends to a. This rule is also applicable for f of a is equal to infinity and g of a is equal to infinity. With this key idea let us proceed with the solution. Now we have to find the value of the expression limit log of sin of x by log of tan of x extends to 0. Now if we put the value of x as 0 then the expression becomes log of sin of 0 upon log of tan of 0 which is equal to log of 0 upon log of 0. Therefore it takes infinity by infinity form since log of 0 is equal to minus infinity. According to L'Hopital's rule which states that if f of x and g of x are functions such that f of a is equal to infinity and g of a is equal to infinity, then limit f of x upon g of x as x tends to a can be written as limit f dash of x upon g dash of x as x tends to a. Therefore applying the L'Hopital's rule we get limit differential of log of sin of x with respect to x that is 1 upon sin of x into differential of sin of x with respect to x that is cos of x upon differential of log of tan of x with respect to x that is 1 upon tan of x into differential of tan of x with respect to x that is 6 square of x x tends to 0. Now if we put the value of x as 0 in this expression we get cos of 0 upon sin of 0 whole upon 6 square of 0 upon tan of 0 which is equal to cos 0 is 1 and sin 0 is 0. So we have 1 upon 0 that is infinity and in the denominator we have 6 square of 0 that is 1 and tan of 0 which is 0 so this is also infinity. So the expression is of infinity by infinity form and this can be written as limit cos of x upon sin of x is quarter of x upon 1 upon tan of x is quarter of x into 6 square of x as x tends to 0 and this is equal to limit 1 upon 6 square of x to 0. Now putting the value of x as 0 we get 1 upon 6 square of 0 which is equal to 1 upon 1 square that is 1. Therefore the value of the expression limit log of sin of x upon log of tan of x as x tends to 0 is equal to 1 which is the required answer. Now we shall find the value of the expression limit log of cos of x by tan of x as x tends to pi by 2. Now if we put the value of x as pi by 2 in this expression we get log of cos of pi by 2 by tan of pi by 2. Since cos of pi by 2 is 0 therefore in the numerator we have log of 0 that is minus infinity and tan of pi by 2 is not defined so this expression is of infinity by infinity form and according to L-Hopitans' rules we have if f of x and g of x are two functions such that f of a is equal to infinity and g of a is equal to infinity then limit f of x upon g of x as x tends to a is equal to limit f dash of x by g dash of x as x tends to a. Therefore applying the L-Hopitans' rules we have limit differential of log of cos of x with respect to x that is 1 upon cos of x into differential of cos of x with respect to x that is minus of sin of x upon differential of tan of x with respect to x that is 6 square of x as x tends to pi by 2 this can be written as limit 1 upon cos of x can be written as sin of x into minus of sin of x by 6 square of x to pi by 2 which is equal to limit minus of sin of x by sec of x to pi by 2 that is limit minus of sin of x into 1 upon sec of x can be written as cos of x as x tends to pi by 2. If we put the value of x as pi by 2 in this expression we get minus of sin of pi by 2 into cos of pi by 2 which is equal to minus of sin of pi by 2 is 1 into cos of pi by 2 which is 0 that is equal to 0. Therefore the value of the expression limit log of cos of x by tan of x to pi by 2 is equal to 0 which is the required answer this completes our session hope you enjoyed this session.