 Hello and welcome to the session. In this session, we are going to discuss the following question. It says that evaluate limit 1 upon x minus 3 minus of 1 upon log of x minus 2 as x tends to 3. We know that n-hopitors rule states that f of x and g of x are functions such that f of a by g of a is of the sum 0 by 0 of infinity by infinity. 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. With this key idea, let us proceed with the solution. We need to find the value of the expression limit 1 upon x minus 3 minus of 1 upon log of x minus 2 x tends to 3. Now if we put the value of x as 3 in this expression, we get 1 upon 3 minus 3 that is 1 upon 0 which is equal to infinity minus of 1 upon log of 3 minus 2 that is 1 upon log of 1 which is equal to 1 by 0 as log of 1 is equal to 0 which is equal to infinity so this expression is of infinity minus infinity form on taking the ACM we get limit log of x minus 2 minus of x minus 3 upon x minus 3 into log of x minus 2 as x tends to 3 Now if we put the value of x as 3 in this expression, we get log of 3 minus 2 that is log of 1 which is equal to 0 minus 3 minus 3 that is 0 so the numerator becomes 0 and in the denominator we have 3 minus 3 that is 0 into log of 3 minus 2 that is log of 1 which is 0 so the denominator also becomes 0 so this expression is of 0 by 0 form and using n-hospital's rule we know that if f of x and g of x are functions such that f of a by g of a is of the sum 0 by 0 or infinity by infinity 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 here we have to differentiate the numerator and denominator separately so applying n-hospital's rule we get limit by differentiating log of x minus 2 with respect to x we get 1 upon x minus 2 minus of differentiating x minus 3 with respect to x we get 1 upon x minus 3 differentiating x minus 3 into log of x minus 2 with respect to x using product rule we get x minus 3 into differentiation of log of x minus 2 with respect to x that is 1 upon x minus 2 plus log of x minus 2 into differentiating x minus 3 with respect to x that is 1 as x tends to 3 on solving this further we get limit 1 minus x plus 2 by x minus 2 upon x minus 3 plus x minus 2 into log of x minus 2 by x minus 2 as x tends to 3 that is we have taken a theorem in the numerator and denominator this is limit 3 minus x upon x minus 3 plus x minus 2 into log of x minus 2 as x tends to 3 now again if we put the value of x as 3 in this expression we get 3 minus 3 which is equal to 0 in the numerator and the denominator becomes 3 minus 3 that is 0 plus 3 minus 2 that is 1 into log of 3 minus 2 that is log of 1 which is equal to 0 so this expression is of 0 by 0 form so again applying n-hospital rule we get limit now differentiating numerator with respect to x that is differentiating 3 minus x with respect to x we get minus 1 upon differentiating denominator with respect to x and we get differentiating x minus 3 with respect to x we get 1 plus now differentiating x minus 2 into log of x minus 2 using further theorem we have x minus 2 into the derivative of log of x minus 2 with respect to x that is 1 upon x minus 2 plus log of x minus 2 into the configuration of x minus 2 with respect to x that is 1 as x tends to 3 which implies that limit minus 1 upon 1 plus 1 plus log of x minus 2 into 1 that is log of x minus 2 as x tends to 3 now if we put the value of x as 3 in this expression we get minus 1 upon 1 plus 1 that is 2 plus log of 3 minus 2 that is log of 1 which is equal to 0 so we get minus 1 by 2 therefore the value of the expression limits 1 upon x minus 3 minus of 1 upon log of x minus 2 as x tends to 3 if we get minus 1 upon 2 which is the required answer this completes our session hope you enjoyed this session