limx→1log xx−1
We have, limx→1log xx−1
At x=0,the value of the expression is the form 00
applying L'Hospital Rule
limx→1ddxlog xddx(x−1)
limx→11x1 {ddxlog x=1x}
Putting x=1
=11=1
limx→1logxx−1 is equal to
limx→1 logxx−1 = [Rpet 1996; MP PET 1996; P.CET 2002]