 Hello and welcome to the session. In this session we are going to discuss the following question which says that Determine a. Limit sin of x into log of x as x tends to 0. b. Limit log of 1 minus x into cos of pi by 4 x as x tends to 0. So, n-heptance rule states that if the surface 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 f of x by 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 this rule is also applicable for f of a is equal to infinity g of a is equal to infinity With this clear idea we shall proceed with the solution. We are to find the value of the expression limit sin of x into log of x as x tends to 0. Now if we put the value of x as 0 in this expression we get sin of 0 into log of 0 which is of 0 into infinity form since sin of 0 is 0 and log of 0 is minus infinity. Now this can also be written as limit log of x upon 1 upon sin of x to 0. Now again if we put the value of x as 0 in this expression we get log of 0 that is minus infinity upon 1 upon sin of 0 which is 0. So we have 1 upon 0 which is also infinity. So this is of infinity by infinity form. Now according to n-hôpital's rule we have 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 limit f of x upon g of x as x tends to a is equal to the limit f dash of x upon g dash of x as x tends to a. Now the expression limit x tends to 0 log of x upon 1 upon sin of x is of infinity by infinity form. Therefore applying n-hôpital's rule we have to log log of x with respect to x that is 1 upon x whole upon differential of 1 upon sin of x with respect to x that is minus of cos of x upon sin square of x tends to 0. Now putting the value of x as 0 in the expression we get 1 upon 0 that is infinity whole upon minus of cos of 0 upon sin square of 0 that is 1 upon 0 which is also infinity. So this is of infinity by infinity form. Now this expression can also be written as limit sin square of x upon minus of x into cos of x as x tends to 0. Now if you put the value of x as 0 in the expression we get sin square of 0 which is equal to 0 upon minus of 0 into cos of 0 that is also 0. So this is of 0 by 0 form and according to n-hôpital's rule we know that if f of x and g of x are functions such that f of a is equal to 0 and g of a is equal to 0 then limit f of x upon g of x as x tends to a is equal to the limit f dash of x upon g dash of x as x tends to a. Therefore applying n-hôpital's rule we have limit differential of sin square of x with respect to x that is 2 sin of x into differential of sin of x with respect to x that is cos of x upon differentiating minus of x into cos of x with respect to x using product rule that is we have minus of x into differential of cos of x with respect to x that is minus of sin of x plus cos of x into differential of minus x with respect to x that is minus 1 as x tends to 0 which can be written as limit 2 sin of x into cos of x upon x into sin of x minus of cos of x as x tends to 0 Now putting the value of x as 0 in the expression we get 2 into sin of 0 into cos of 0 whole upon 0 into sin of 0 minus cos of 0 which is equal to 2 into sin of 0 is 0 into cos of 0 which is 1 upon 0 into sin of 0 is 0 minus of cos of 0 which is 1 which is equal to 0 upon minus 1 that is 0 therefore the value of the expression limit sin of x into log of x as x tends to 0 is equal to 0 which is the required answer Now we shall find the value of the expression limit log of 1 minus x into cos of pi by 4x x tends to 0 if we put the value of x as 0 in the expression we get log of 1 minus 0 that is log of 1 which is equal to 0 into cos of 0 which is not defined so this is a 0 into infinity form and this can be written as limit log of 1 minus x into cos of pi by 4x can also be written as 1 upon sin of pi by 4x as x tends to 0 now if we put the value of x as 0 in the expression we get log of 1 minus 0 that is log of 1 which is equal to 0 and sin of 0 is also 0 so this is of 0 by 0 sum now according to L-Hopital's rule if we have f of x and g of x are the two functions such that f of a is 0 and g of a is 0 then limit f of x upon g of x as x tends to a is equal to the limit f dash of x upon g dash of x as x tends to a so applying L-Hopital's rule we have limit differential of the numerator with respect to x that is differential of log of 1 minus x with respect to x we have 1 upon 1 minus x into differential of 1 minus x with respect to x that is minus 1 upon differential of the denominator with respect to x that is differential of sin of pi by 4x with respect to x is cos of pi by 4 of x into differential of pi by 4 of x with respect to x that is pi by 4 as x tends to 0 this can be written as limit minus 1 upon 1 minus x whole upon pi by 4 into cos of pi by 4x as x tends to 0 now putting the value of x of 0 in the expression we get minus 1 upon 1 minus 0 whole upon pi by 4 into cos of 0 which is equal to minus 1 upon pi by 4 into cos of 0 that is 1 which is equal to minus 1 upon pi by 4 that is minus 1 into 4 upon pi which is equal to minus 4 upon pi therefore the value of the expression limit log of 1 minus x into cos of pi by 4x as x tends to 0 is equal to minus 4 upon pi which is the required answer this completes our session hope you enjoyed this session