 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that evaluate 1. Limit theta tends to 0 log of theta upon cot of theta 2. Limit x tends to 2 log of x minus 2 upon log of e raise to power x minus e square. A Hopkins 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 x tends to a plus x by g of x is equal to limit x tends to a f dash of x upon g dash of x. This rule is also applicable for 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 theta tends to 0 log of theta upon cot of theta. Now if we put the value of theta as 0 in this expression we get log of 0 upon cot of 0 which is of infinity by infinity form since log of 0 is equal to minus infinity and cot of 0 is not defined. According to L Hopkins rule 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 x tends to a f of x upon g of x is equal to limit x tends to a f dash of x upon g dash of x. So applying L Hopkins rule we have limit theta tends to 0 differential of log of theta with respect to theta that is 1 by theta upon differential of cot of theta with respect to theta that is minus 1 upon sine square of theta. Now again if we put the value of theta as 0 in this expression we get 1 by 0 upon minus 1 upon sine square of 0 that is equal to 1 by 0 is infinity upon minus 1 upon sine of 0 is 0 so it's minus 1 upon 0 that is infinity. So this is of infinity by infinity form and this can be written as limit theta tends to 0 minus of sine square theta by theta. Now putting the value of theta as 0 in this expression we get minus of sine square of 0 by 0 which is of 0 by 0 form since sine of 0 is equal to 0 and we know that L Hopkins rule states that if f of x and g of x are functions such that f of a is 0 and g of a is 0 then limit x tends to a f of x upon g of x is equal to limit x tends to a f dash of x upon g dash of x. So applying L Hopkins rule we have limit theta tends to 0 differential of minus sine square of theta with respect to theta that is minus 2 sine of theta into differential of sine of theta that is cos of theta by differential of theta with respect to theta that is 1. Now putting the value of theta as 0 in this expression we get minus 2 sine of 0 into quarter of 0 which is equal to minus 2 into sine of 0 is 0 and quarter of 0 is 1. So this is equal to 0 therefore the value of the expression limit of theta tends to 0 log of theta by quarter of theta is equal to 0 which is the required answer. Now we shall find the value of the expression limit x tends to 2 log of x minus 2 upon log of e raise to power x minus e square. Now if we put the value of x as 2 in this expression we get log of 2 minus 2 upon log of e square minus e square which is equal to log of 2 minus 2 is 0 upon log of e square minus e square which is log 0 and this is of infinity by infinity form since log of 0 is equal to minus of infinity. So this is of infinity by infinity form according to L Hopkins rule if x of x and g of x are functions such that x of a is equal to infinity and g of a is equal to infinity then limit x tends to a f of x upon g of x is equal to limit x tends to a f dash of x upon g dash of x. So applying L Hopkins rule we have limit x tends to 2 differential of log of x minus 2 with respect to x that is 1 upon x minus 2 upon differential of log of e raise to power x minus e square with respect to x that is 1 upon e raise to power x minus e square into differential of e raise to power x minus e square that is e raise to power x minus 0 which is equal to limit x tends to 2 1 upon x minus 2 upon e raise to power x upon e raise to power x minus e square. Now if we put the value of x as 2 in this expression we get 1 upon 2 minus 2 upon e square upon e square minus e square that is 1 upon 0 whole upon e square by 0 which is of infinity by infinity form. So this is of infinity by infinity form and this can also be written as limit x tends to 2 e raise to power x minus e square upon e raise to power x into x minus 2. Now if we put the value of x as 2 in this expression we get e square minus e square that is 0 upon e square into 2 minus 2 that is e square into 0 which is 0. So this is of 0 by 0 form so applying l-hopity rule again we have limit x tends to 2 differential of e raise to power x minus e square with respect to x that is e raise to power x upon differential of e raise to power x into x minus 2 with respect to x by product rule we have e raise to power x into differential of x minus 2 with respect to x that is 1 plus x minus 2 into differential of e raise to power x with respect to x that is e raise to power x. So this is equal to limit x tends to 2 e raise to power x upon e raise to power x plus e raise to power x into x minus 2. Now putting the value of x as 2 in this expression we get e square upon e square plus e square into 2 minus 2 that is equal to e square upon e square plus e square into 0 which is equal to e square by e square that is 1. Therefore the value of this expression limit x tends to 2 log of x minus 2 upon log of e raise to power x minus e square is equal to 1 which is the required answer. This completes our session hope you enjoyed this session.