 Hello and welcome to the session. In this session we are going to discuss the following question which says that Determine limit x tends to 3 x minus 3 raise to power x minus 3 An inhibitor's rule states that if f of x and g of x 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 f of x by g of x is equal to limit x tends to a f dash of x upon g dash of x and 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. We are to find the value of the expression limit x tends to 3 x minus 3 raise to power x minus 3 that is of 0 raise to power 0 form. Now let y be equal to x minus 3 raise to power x minus 3 Now taking log on both the sides we have log of y is equal to x minus 3 into log of x minus 3 which can also be written as log of y is equal to log of x minus 3 upon 1 upon x minus 3. Now taking limits on both the sides we have limit x tends to 3 log of y is equal to limit x tends to 3 log of x minus 3 upon 1 by x minus 3 Now if we put the value of x as 3 in this expression we get log of 3 minus 3 upon 1 upon 3 minus 3 that is log of 0 upon 1 by 0 which takes infinity by infinity form since log of 0 is equal to minus infinity and 1 by 0 is infinity. Therefore it takes infinity by infinity form and according to L-Hopital's rule we have 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 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 and this rule is also applicable for f of a is equal to infinity and g of a is equal to infinity. Now since limit x tends to 3 log of x minus 3 upon 1 upon x minus 3 takes infinity by infinity form therefore applying L-Hopital's rule we have limit x tends to 3 log of y is equal to limit x tends to 3 differential of log of x minus 3 with respect to x that is 1 upon x minus 3 upon differential of 1 upon x minus 3 with respect to x differential of 1 upon x minus 3 with respect to x that is minus 1 upon x minus 3 d whole square which is equal to limit x tends to 3 1 upon x minus 3 into x minus 3 d whole square by minus 1 which is equal to limit x tends to 3 minus of x minus 3 now if we put the value of x as 3 we get minus of 3 minus 3 that is 0 so limit x to 3 log of y is equal to 0 we can write log of limit x tends to 3 y is equal to 0 which implies that limit x tends to 3 y is equal to e raise to power 0 which is equal to 1 we have assumed the value of y as x minus 3 raise to power x minus 3 therefore we have limit x tends to 3 y that is x minus 3 raise to power x minus 3 is equal to 1 which is the required answer this completes our session hope you enjoyed this session