Correct Answer - Option 4 : 1/2
Concept:
L'Hospital's Rule:
If the limit becomes \({0\over 0}\) or \({\pm\infty\over \pm\infty}\), it is solved by differentiating numerator and denominator.
Calculation:
Let L = \(\rm\lim_{x\rightarrow 1}{\log x\over x^2-1}\)
Putting the value of the x, we get \({0\over 0}\).
Differentiating numerator and denominator wrt x
L = \(\rm\lim_{x\rightarrow 1}{{1\over x}\over 2x -0} = \rm\lim_{x\rightarrow 1}{1\over 2x^2}\)
Putting the value of x = 1
L = 1/2
